mat 1512 (calculus a)

iii MAT1512/1/2010-2012

Contents

PageIntroduction vii

How to use the study guide viii

Keys to success in studying Mathematics ix

Preparing for the examination ix

CHAPTER 0: Preliminaries 1

1. Background 1

2. Learning outcomes 1

3. Principles of problem solving 2

4. Summary 3

5. The way forward 3

CHAPTER 1: Limits and Continuity 4

1. Background 4


3. Prescribed reading 5

4. The limit 5

4.1 Introduction to the limit concept 5

4.2 Definition of a limit: left- and right-hand limits 6

5. Worked examples 6

I. Limits as x→ c (c ∈ R) 7

II. Limits as x→ ±∞ 10

III. Limits involving absolute values 14

IV. Left-hand and right-hand limit 16

V. Limits involving trigonometric functions 19

VI. The Squeeze Theorem 23

VII. The ε− δ definition of a limit 26

VIII. Continuity 29

Key points 38

iv

CHAPTER 2: Differentiation of different types of functions 39

1. Background 39



4. The deriviative 40

4.1 Introducing the derivative 40

4.2 Definition of the derivative 42


I. Differentiation from first principles (definition of the derivative) 44

II. Basic differentiation formulas 46

(a) The power rule 46

(b) The product rule 47

(c) The quotient rule 48

(d) The chain rule 49

(e) Combinations of rules 50

III. Derivatives of trigonometric functions and inverse trigonometric functions 52

(a) Derivatives of trigonometric functions 52

(b) Derivatives of inverse trigonometric functions 56

IV. Derivatives of exponential and logarithmic functions 57

(a) The exponential function 57

(b) The logarithmic function 58

(c) Examples of the exponential function 59

(d) Examples of the logarithmic function 60

V. Logarithmic differentiation 62

(a) The simplification of functions 62

(b) Functions of the form f(x) = g(x)h(x) 64

VI. Implicit differentiation 66

VII. Tangents and normal lines 71

VIII. The Mean Value Theorem 78

Key points 81

v MAT1512/1/2010-2012

CHAPTER 3: Integration 82

1. Background 82




I. Antiderivatives 84

II. The definite integral and the Fundamental Theorem of Calculus — Part I 86

III. The definite integral and the area between the curve and the x-axis 87

IV. The definite integral and the area under the curve 90

V. The Mean Value Theorem for definite integrals 94

VI. The Fundamental Theorem of Calculus — Part II 96

VII. Integration in general 99

VIII. Indefinite integrals 100

IX. Integration by substitution 101

(a) Indefinite integrals 101

(b) Definite integrals 106

(c) Steps for integration by substitution 108

X. Integration of exponential and logarithmic functions 109

XI. Review of formulas and techniques of integration <Author: This heading does not appear in

the text. Please insert in the text. CP>

Key points 115

CHAPTER 4: First—order Differential Equations, Growth and Decayand Partial derivatives

116

1. Background 116




I. First-order differential equations 117

(a) The solution of differential equations 117

(b) Separable differential equations 117

(c) Initial-value problems 122

vi

II. Growth and decay 126

III. Partial derivatives 135

(a) The partial derivatives technique 135

(b) Notations for partial derivatives 136

(c) Rules for finding the partial derivatives of z = f(x, y) 136

(d) Partial differentiation 138

(e) Functions of more than two variables 139

(f) Higher-order derivatives 140

Key points 149

APPENDIX 150

I. Sequences and Summation (Sigma) Notation 150

II. Mathematical Induction 155

III. The Binomial Theorem 159

vii MAT1512/1

Introduction

Calculus is a set of formal rules and procedures. It gives you the tools you need to measure change both qualita-tively and quantitatively. Wikipedia (www.wikepedia.org) defines calculus as a branch of mathematics that includesthe study of limits, derivatives, integrals, and infinite series, which constitutes a major part of modern university edu-cation. Calculus has widespread applications in science and engineering and is used to solve complex and expansiveproblems for which algebra alone is insufficient. It builds on analytical geometry and mathematical analysis and in-cludes two major branches – differential calculus and integral calculus – which are related through the FundamentalTheorem of Calculus. Differential calculus explores and analyses rates of change quantitatively and qualitatively.Integral calculus deals with the analysis (quantitatively and/or qualitatively) of how quantities or measures of valuesaccumulate or diminish over time. The two processes – differentiation and integration – are reciprocal.

The purpose of this module is to equip you, the learner, with those basic skills in differential and integral cal-culus that are essential for the physical, life, and economic sciences. Most of the time you will be dealing withfunctions. Basically, a function is a generalised input-output process that defines the mapping of a set of inputvalues to a set of output values. It is often defined as a rule for obtaining a numerical value from another givennumerical value. You are also going to have to develop a very large repertoire of methods for depicting functionsgraphically/geometrically.

This course is built around your prescribed book. The purpose of this study guide is to guide you through thoseparts of the prescribed book that you must study for this module, and to provide you with many additional workedexamples. The prescribed book is:

Robert T Smith & Roland B MintonCalculus, Early Transcendental FunctionsMcGraw-Hill Higher Education – International EditionThird edition, c2007ISSBN 13:978007110715-8

viii

How to use the study guide

From now on we will refer to the prescribed book as “Smith and Minton”, or simply “Smith”. The study guidemust always be used in conjunction with the prescribed textbook, because it is not a complete set of notes on thebook. Chapters 1 to 4 of this study guide contain many additional worked problems, taken from past examinationpapers and assignments. Before going through them, study the relevant parts in Smith, and do the examples andsome of the exercises in Smith. Also, before going through our solution to a problem in the study guide, try solvingit yourself. Remember that reading maths often means reading the same thing over and over again.

We have included numerous worked examples for you. These are designed to stimulate your thinking in such away that you will come to appreciate and master the delicate beauty and intricacies of the subject. All you have todo is keep on going! Follow all the instructions given. Try to write down all the answers to the activities in full.This is extremely important, as a major part of learning mathematics is thinking and writing down what you think.By writing everything down, you will develop the essential skill of communicating mathematics effectively. Theother reason for writing your answers down, is to prevent you from losing your train of thought about a conceptor mathematical idea. If this happens, it takes a while to get your reasoning back to the same point. If you haveeverything written down, you can also go back the next day and check your reasoning. Learning mathematics is anactivity, and you will only learn by doing. Because you will be thinking about the problems, you will, in most cases,be able to determine by yourself whether you are right or not.

Your assignments are included in Tutorial Letter 101. Attempt these once you have completed all the relatedactivities in your study guide. Spend a part of your time each day doing some of the questions, and a part studyingnew material. Being able to do an assignment is proof that you have mastered the work of that particular section. InTutorial Letter 101 we have indicated which assignments you should submit for evaluation.

Also remember that statements, theorems and definitions are the building blocks of your mathematical language –you cannot learn anything without knowing the basic facts. Begin by reading the preface of your prescribed book.This should give you a good idea of the importance of the subject you are about to study.

ix MAT1512/1

Keys to success in studying Mathematics

Plan Plan each week according to what happens in your life that week. Decide how manystudy hours you need, and decide where you are going to fit those hours in. Writedown what you plan to do in each session. This way you can look forward to thework you have set for yourself and measure your success.

Evaluate Evaluate each study session. Did you enjoy your studies? If not, why not? Whatcan you change to make it better? Did you achieve what you set out to achieve?Did you use your time constructively? What can you do to improve your timemanagement?

Understand Understand the way you learn. One can only learn maths through repetition andpractice. DO IT, even if you think you know the answer. Understanding comesthrough repetition, and this understanding eventually brings the JOY of masteringmathematics.

Structure Structure each study session for best results. A suggestion is to divide your time intofour parts: (1) Do 5 to 10 minutes’ revision and read through the important points.(2) Go through all the questions of the previous activities. (3) Study new materialby following the instructions in the study guide. (4) Attempt the questions in yourassignments about the sections you have completed.

Get organised (1) Make notes of the facts you have to remember. (2) Write down the answers toall your activities. (3) Complete all the ASSIGNMENTS!

Pace yourself Work out how fast you will have to go. Some sections will take longer than others,but assign about one month for each chapter. This will give you 2 to 4 days for eachsection in each chapter. Bear in mind that this is not the only module you haveregistered for.

Preparing for the examinationYou are studying to improve yourself andNOT only to pass the examination. The examination is there toREWARDyou for what you have done and to CONFIRM that you have done your work properly. If you have followed allthe instructions above, you will already be prepared for the examination and have nothing to worry about. Manystudents spend their time thinking that they cannot master mathematics. This is not true. If you put your mind to itand work hard, you will eventually master this subject.

GOOD LUCK!

1 MAT1512/1

CHAPTER 0

PRELIMINARIES

1. Background

This chapter of the study guide deals with Chapter 0 in Smith. It forms a link between school mathematics and cal-culus. Topics in algebra, trigonometry, analytic geometry and functions which are needed for calculus are reviewed.Graphs of functions are studied, since they enable us to see and understand the behaviour of functions. In studyingthis chapter, you should brush up and revise your school mathematics – especially those parts that are needed foran effective study of calculus. The material in this chapter is background material. At the same time, some topicswhich are important to understand calculus are introduced. Therefore, the chapter will contain a few topics whichyou may have touched on only briefly, or not at all, at school.

2. Learning outcomes

At the end of this chapter, you should be able to

• algebraically manipulate real numbers, solve equations and work comfortably with mathematical conceptssuch as variables, inequalities and absolute values

• recognise, demonstrate your knowledge of and work with the different types of functions (polynomial, ratio-nal, trigonometric, exponential and logarithmic functions), their properties and their representations

In calculus, the basic mathematical tool at the root of all we do is the function. That is why, in this preliminarychapter, you have to familiarise yourself with mathematical ideas related to functions, such as the different types offunctions, their graphs and representations of how they behave. First read the preliminaries of this chapter given onpage 1 in Smith. You should study the whole chapter. Remember that problem solving plays a great and crucialrole in the process of studying calculus.

Note: The way to master calculus is to solve lots of calculus problems!

2

3. Principles of problem solving

There are no exact rules for solving problems. However, it is possible to outline some general steps, or to give somehints or principles that may be useful in the solution of certain problems.

The following steps and principles, which have been adapted from George Polya’s book How to solve it, are in factjust common sense made explicit.

Step 1: Understand the problem. Read the problem and make sure that you understand what it is about. Askyourself the following questions: What is the unknown? What are the given quantities? What conditions are given?In many cases it is useful to draw a “given” diagram and identify the given quantities in the diagram. It is usuallynecessary to introduce suitable notation. In choosing symbols for the unknown quantities, we use letters such asa, b, c . . . x and y, and it also helps to use abbreviations or symbols such as V (volume) and t (time).

Step 2: Think of a plan. Try to find the connection between the data and the unknown. You may have to relatea problem to others you have seen before. Have you seen the same problem in a slightly different form? Do youknow a similar/related problem? Do you know a theorem, rule, method or result that could be useful? Could youreformulate the problem? Could you re–state it still differently? Go back to your definitions.

If you cannot solve the proposed problem, try to solve some related problem first. Sometimes you will be able tofind a similar problem, while in other instances you may have to use components, sections, and methods of otherproblems to solve a new one. Every problem you encounter becomes a potential method for solving other problems.

Step 3: Carry out the plan. In order to carry out your plan, you have to check each step. Can you see clearlythat the step is correct? Can you prove that it is correct?

Step 4: Look back. Examine the solution you have obtained. Can you check the result? Can you check theargument? Can you derive the solution differently? Can you see it at a glance? Can you use the result, or themethod, for some other problem?

3 MAT1512/1

4. Summary

Table 1 below is a summary of the mathematical ideas you need to master, with references to the relevant examplesin your textbook. Try to do as many review exercises as you can.

Table 1

Topic(s) Section in Smith andMinton

I. Polynomials and rational Examples 1.1 – 1.23functions

II. Sketching graphs of different Examples 2.2 – 2.7functions

III. Inverse functions Examples 3.1 – 3.8

IV. Trigonometric and inverse Examples 4.1 – 4.10trigonometric functions

V. Exponential and logarithmic Examples 5.1 – 5.11functions

VI. Transformation of functions Examples 6.1 – 6.9

5. The way forward

The topics in the rest of the study guide are:

Chapter 1: Limits and continuityChapter 2: Differentiation of different types of functionsChapter 3: IntegrationChapter 4: First-order differential equations, growth and decay, and partial derivatives.

4

CHAPTER 1

LIMITS AND CONTINUITY

1. Background

The idea of a limit underlies the various branches of calculus. In fact, without limits, calculus simply would notexist. Every single notion of calculus is a limit in one sense or another and the idea of a limit plays a fundamentalrole in concepts such as instantaneous velocity, the slope of a curve, the length of a curve, the sum of infinite series,etc. The role played by limits in both differential and integral calculus is crucial. In this chapter, three main ideaswill be introduced. They are: (a) the basic rules (definitions) of limits and their applications in evaluating limitsof algebraic and trigonometric functions; (b) the application of limits to continuity of functions; and (c) the use ofthe Squeeze Theorem in determining certain limits. Begin by reading through the preview of this chapter on page74 in Smith. You will notice that a mathematical model is used to predict how a function will behave. Graphs arealso used to represent function behaviour and will later be used to solve problems. This approach is used to solvevarious calculus problems.



• demonstrate an understanding of the concepts relating to limits and how these are applied in evaluating limitsof the form lim

x→c f (x), limx→c+f (x), lim

x→c−f (x), lim

x→+∞ f (x) and limx→−∞ f (x) , where c is a real number and f

is an algebraic or trigonometric function

• use the limit definition of continuity• solve problems based on the Squeeze Theorem

5 MAT1512/1

3. Prescribed reading

In Smith & Minton, Chapter 1, Sections 1.1, 1.2, 1.3, 1.4, 1.5 and appendix A, theorem A.1, theorem A.2, theoremA.3, theorem A.4, theorem A.5, and theorem A.6.

Note: The way to master calculus is to solve a lot of calculus problems!

4 The limit

4.1 Introduction to the limit concept

Read carefully through section 1.1 in Smith. You will notice that we have two problems – one dealing with findingthe slope of a curve at a specific point and the other one dealing with finding the length of a curve. In the firstproblem, the closer the second point gets to a certain point, the closer the computed value gets to the actual valuewe are looking for. This calculus problem involves a process called the limit. You can estimate the slope of thecurve using a sequence of approximations. The limit allows you to compute the slope exactly. Similarly, in thesecond problem – in trying to compute the distance along a curved path – you would have to go through betterapproximations successively until you reach the actual value. The actual arc length is obtained by using the limit.The limit value is the height or the y-value of a function.

A left-hand limit is the y-value you obtain by approaching x from the left side.A right-hand limit is the y-value you obtain by approaching x from the right side.

Let us use two functions to explore the limit concept. (These are the same functions as those used in section 1.2

of your prescribed textbook.) The first function, f (x) = x2 − 4x − 2 , is undefined at x = 2, but its behaviour can be

examined in the area close to x = 2. From the value tables and the accompanying graph on page 80 in Smith, youcan see that the limit of f (x) as x approaches 2 from the left or right is 4.

x f (x) = x2 − 4x − 2

1.9 3.91.99 3.991.999 3.9991.9999 3.9999

2

4

22_x

x f (x) = x2 − 4x − 2

2.1 4.12.01 4.012.001 4.0012.0001 4.0001

limx→2−

f (x) = 4 limx→2+

f (x) = 4

This can be written mathematically as follows:

limx→2

f (x) = 4.

6

Consider another function g (x) = x2 − 5x − 2 or y =

x2 − 5x − 2 .

x g (x) = x2 − 5x − 2

1.9 13.91.99 103.991.999 1003.9991.9999 10,003.9999

5_

10_

y

x

10

5

5 10

x y = x2 − 5x − 2

2.1 -5.12.01 -95.992.001 -995.9992.0001 -9,995.9999

limx→2−

g (x) = 10, 004 limx→2+

g (x) = −9, 996The limit of g(x) as x approaches 2 from the left, and the limit of g(x) as x approaches 2 from the right are notequal. Mathematically, this can be written as follows: lim

x→2−g (x) = lim

x→2+g (x) and this statement means that

limx→2

g (x) does not exist.

4.2. Definition of a limit: left- and right-hand limitsWe say a limit exists if and only if the corresponding one-sided limits are equal. That is, iflimx→a−

f (x) = limx→a+

f (x) . In general, limx→a f (x) = L for some number L if and only if

limx→a−

f (x) = limx→a+

f (x) = L .

Let us now consider some worked examples.

5. Worked examplesOur collection of worked examples of the work covered in this chapter of Smith can be divided into eight sets (withsome overlap), namely:

I. Limits as x → c, c ∈ R (cancellations)II. Limits as x → ±∞III. Limits involving absolute values

IV. Left-hand and right-hand limits

V. Limits involving trigonometric functions

VI. The Squeeze Theorem

VII. The − δ definition of a limitVIII. Continuity

7 MAT1512/1

The table below (Table 2) shows which sections of Smith, and which worked examples in the study guide you mustconsult.

Table 2

Topic(s) Sections in Smith & Minton Study guideexamples

I. Limits as x → c, c ∈ R Sections 1.1, 1.2 and 1.3 1–12

II. Limits as x → ±∞ Section 1.5 (examples 5.6 – 5.12) 13–27

III. Limits involving absolute values Section 1.2 (example 2.5) 22–27

IV. Left-hand and right-hand limits Sections 1.2, 1.3 (example 3.9), 1.4 28–40and 1.5 (examples 5.3 – 5.5 & 5.9)

V. Limits involving trigonometric Section 1.2 (example 2.4) and 1.3 41–52functions (example 3.6)

VI. The Squeeze Theorem Section 1.3 (example 3.8) 53–57

VII. The − δ definition of a limit Sections 1.6 and 1.7 (read only) 58–63

VIII. Continuity Section 1.4 (examples 4.1 – 4.5) 64–70

In the following sections, we present a number of worked examples based on the mathematical concepts and ideasin each section of this chapter, together with appropriate teaching texts and references to Smith. The solutions tothe problems appear immediately after each set of examples. Unlike the exercises in Smith, the examples in thestudy guide are not arranged in order of difficulty. Only attempt them once you have studied the relevant parts anddone some of the exercises in Smith.

I. Limits as x → c (c ∈ R)

Determine the following limits:

1. limx→−1

x3 + 1−x2 + x + 2 2. lim

x→−3x + 3

x2 + 4x + 3

3. limx→1

x2 − 2x + 1x3 − 2x2 + x 4. lim

x→05x3 + 8x23x4 − 16x2

8

5. limx→−1

x2 − 2x − 3x2 − 1 6. lim

x→3x2 − 9x3 − 27

7. limx→−1

x4 − 1x + 1 8. lim

x→3x3 − 27x2 − 3x

9. limx→1

x − 1√x2 + 24− 5 10. lim

h→0(x + h)2 + 1−√x2 + 1

h

11. limx→−2

x2 − 2x 13 − 2

x2 − 2x − 8 12. limx→10

(x − 2) 13 − 2x − 10

Solutions:

1. limx→−1

x3 + 1−x2 + x + 2 2. lim

x→−3x + 3

x2 + 4x + 3= lim

x→−1(x + 1) x2 − x + 1(x + 1) (2− x) = lim

x→−3x + 3

(x + 3) (x + 1)= lim

x→−1x2 − x + 12− x = lim

x→−31

x + 1= 3

3 = 1. = 1−3+ 1 = −

12 .

3. = limx→1

x2 − 2x + 1x3 − 2x2 + x 4. lim

x→05x3 + 8x23x4 − 16x2

= limx→1

(x − 1)2x (x − 1)2 = lim

x→0x2 (5x + 8)x2 3x2 − 16

= limx→1

1x= 1. = lim

x→05x + 83x2 − 16

= = − 816 = − 1

2 .

5. limx→−1

x2 − 2x − 3x2 − 1 6. lim

x→3x2 − 9x3 − 27

= limx→−1

(x + 1) (x − 3)(x + 1) (x − 1) = lim

x→3(x − 3) (x + 3)

(x − 3) x2 + 3x + 9= lim

x→−1x − 3x − 1 = lim

x→3x + 3

x2 + 3x + 9= −4

−2 = 2. = 627 = 2

9 .

9 MAT1512/1

7. limx→−1

x4 − 1x + 1 8. = lim

x→3x3 − 27x2 − 3x

= limx→−1

x2 − 1 x2 + 1x + 1 = lim

x→3(x − 3) x2 + 3x + 9

x (x − 3)= lim

x→−1(x − 1) (x + 1) x2 + 1

x + 1 = limx→3

x2 + 3x + 9x

= limx→−1 (

x − 1) x2 + 1 = 9.

= (−1− 1) (−1)2 + 1= −4.

In the following two solutions we make use of the identity (i) and rationalising (ii):

(i) A2 − B2 = (A − B) (A + B)(ii) Rationalising the denominator, eg

1√x= 1√

x·√x√x=√xx

(A − B)(A + B) = A2 − B2.9.

limx→1

x − 1√x2 + 24− 5 = lim

x→1x − 1√

x2 + 24− 5 ·√x2 + 24+ 5√x2 + 24+ 5

= limx→1

(x − 1) √x2 + 24+ 5x2 + 24− 25

= limx→1

(x − 1) √x2 + 24+ 5x2 − 1

= limx→1

(x − 1) √x2 + 24+ 5(x − 1) (x + 1)

= limx→1

√x2 + 24+ 5x + 1 = 5.

10.

limh→0

(x + h)2 + 1−√x2 + 1h

= limh→0

(x + h)2 + 1−√x2 + 1 (x + h)2 + 1+√x2 + 1h (x + h)2 + 1+√x2 + 1

= limh→0

(x + h)2 + 1− x2 + 1h (x + h)2 + 1+√x2 + 1

= limh→0

x2 + 2xh + h2 + 1− x2 − 1h (x + h)2 + 1+√x2 + 1

= limh→0

2x + h(x + h)2 + 1+√x2 + 1

= x√x2 + 1 .

10

In the following two solutions we make use of the identity

(A − B)(A2 + AB + B2) = A3 − B3.

11.

limx→−2

x2 − 2x 13 − 2

x2 − 2x − 8 = limx→−2

x2 − 2x 13 − 2

x2 − 2x − 8 · x2 − 2x 2

3 + 2 x2 − 2x 13 + 4

x2 − 2x 23 + 2 x2 − 2x 1

3 + 4

= lim

x→−2x2 − 2x − 8

x2 − 2x − 8 x2 − 2x 23 + 2 x2 − 2x 1

3 + 4

= limx→−2

1

x2 − 2x 23 + 2 x2 − 2x 1

3 + 4= 1

4+ 4+ 4 =112.

(Here we put A = (x2−2x) 13 and B = 2, and multiplied both numerator and denominator by A2+ AB+B2.)

12.

limx→10

(x − 2) 13 − 2x − 10

= limx→10

(x − 2) 13 − 2x − 10 · (x − 2)

23 + 2 (x − 2) 13 + 4

(x − 2) 23 + 2 (x − 2) 13 + 4= lim

x→10(x − 2)− 8

(x − 10) (x − 2) 23 + 2 (x − 2) 13 + 4

= limx→10

1

(x − 2) 23 + 2 (x − 2) 13 + 4= 1

4+ 4+ 4 =112.

II. Limits as x→ ±∞In this case, if you are required to find the limit of a function which is a polynomial divided by another polynomial,find the highest power of the variable in the denominator and divide each term in both numerator and denominatorby that power.

Note:

limx→±∞

1x = 0

Example:

Find the following limit: limx→∞

6x2 − 12x2 + 7x

11 MAT1512/1

Solution:

limx→∞

6x2 − 12x2 + 7x = lim

x→∞

6x2x2 − 1

x22x2x2 + 7x

x2

Notice that the highest power in thedenominator is x2.

= limx→∞

6− 1x2

2+ 7x2

Divide each term in both numeratorand denominator by x2.

= 6− 02+ 0

= 62= 3.

Now determine the following limits:

13. limx→∞

x4 + x312x3 + 128 14. lim

x→∞7x3 + 1x2 + x

15. limx→−∞

7x3 + 1x2 + x 16. lim

x→∞9x4 + x

2x4 + 5x2 − x + 6

17. limx→−∞

√x2 + 4x − 2x

2x18. lim

x→−∞4x + 8√

x2 + 7x − 2x

19. limx→−∞

√x2 + 2x + x 20. lim

x→−∞1√

x2 + 5x + x

21. limx→∞ x3 + x2 1

3 − x

Solutions:

13.

limx→∞

x4 + x312x3 + 128

= limx→∞

x + 112+ 128/x3

= ∞.

14.

limx→∞

7x3 + 1x2 + x

= limx→∞

7x + 1/x21+ 1/x

= ∞.

12

15.

limx→−∞

7x3 + 1x2 + x

= limx→−∞

7x + 1x2

1+ 1x

= −∞.

16.

limx→∞

9x4 + x2x4 + 5x2 − x + 6

= limx→∞

9+ 1/x3

2+ 5/x2 − 1/x3 + 6/x4

= limx→∞

9+ 1/x3

2+ 5/x2 − 1/x3 + 6/x4

= 9+ 02+ 0− 0+ 0

= 92.

17.

limx→−∞

√x2 + 4x − 2x

2x

= limx→−∞

x2 (1+ 4/x)− 2x2x

= limx→−∞

|x |√1+ 4/x − 2x2x

= limx→−∞

−x√1+ 4/x − 2x2x

(|x | = −x since x < 0)

= limx→−∞

−x √1+ 4/x + 22x

= limx→−∞

− √1+ 4/x + 22

= −32.

18.

limx→−∞

4x + 8√x2 + 7x − 2x

= limx→−∞

4x + 8x2 (1+ (7/x))− 2x

= limx→−∞

4x + 8|x |√1+ (7/x)− 2x (

√x2 = |x |)

= limx→−∞

4x + 8−x√1+ (7/x)− 2x (|x | = −x since x < 0)

= limx→−∞

4+ (8/x)−√1+ (7/x)− 2

= 4+ 0−√1+ 0− 2 = −

43.

13 MAT1512/1

19.

limx→−∞ x2 + 2x + x

= limx→−∞

√x2 + 2x + x √

x2 + 2x − x√x2 + 2x − x

= limx→−∞

x2 + 2x − x2√x2 + 2x − x

= limx→−∞

2x|x |√1+ 2/x − x

= limx→−∞

2x−x√1+ 2/x − x (|x| = −x since x < 0)

= limx→−∞

2−√1+ 2/x − 1 = −1.

20.

limx→−∞

1√x2 + 5x + x

= limx→−∞

1√x2 + 5x + x ·

√x2 + 5x − x√x2 + 5x − x

= limx→−∞

√x2 + 5x − x5x

= limx→−∞

x2 (1+ 5/x)− x5x

= limx→−∞

|x |√1+ 5/x − x5x

√x2 = |x |

= limx→−∞

−x√1+ 5/x − x5x

(|x| = −x since x < 0)

= limx→−∞

−√1+ 5/x − 15

= −√1+ 0− 15

= −25.

21.

limx→∞ x3 + x2 1

3 − x

= limx→∞

x3 + x2 13 − x x3 + x2 2

3 + x x3 + x2 13 + x2

x3 + x2 23 + x x3 + x2 1

3 + x2

= limx→∞

x3 + x2 − x3x3 + x2 2

3 + x x3 + x2 13 + x2

= limx→∞

x2

x2 (1+ 1/x) 23 + x2 (1+ 1/x) 13 + x2= lim

x→∞1

(1+ 1/x) 23 + (1+ 1/x) 13 + 1= 13.

14

III. Limits involving absolute values

Remember from high school that

|x | = x if x ≥ 0−x if x < 0

The definition of absolute value gives a function in two sections (parts) with corresponding intervals,ie if y = |x | then

y = x if x ≥ 0−x if x < 0

Another example:

If

f (x) = |3− 2x | = 3− 2x, if 3− 2x ≥ 0⇒ 3 ≥ 2x ⇒ 32 ≥ x ⇒ x ≤ 3

2− (3− 2x) if 3− 2x < 0⇒ 3 < 2x ⇒ 3

2 < x ⇒ x > 32 .

Thus,

f (x) = 3− 2x, if x ≤ 32

−3+ 2x, if x > 32

Let us look at one example:

Find the following limit:

limx→∞

x − 5|3− 2x |

Solution:

limx→∞

x − 5|3− 2x | = lim

x→∞x − 5−3+ 2x = lim

x→∞

xx − 5

x−3x + 2x

x

|3− 2x | = − (3− 2x) since x > 32

= limx→∞

1− 5x

−3x + 2

= 1− 0−0+ 2

= 12

Note: Since we have x →∞, x > 32 and this is why we have |3− 2x | = −3+ 2x in our example above.

15 MAT1512/1

Now determine the rest of the limits below.

22. limx→∞

3x + 1|2− x | 23. lim

x→−∞|x |

|x | + 1

24. limx→−∞

x|x | 25. lim

x→−∞|x + 1|x + 1

26. limx→−∞

|x + 1|2x + 3 27. lim

x→−∞2+ |2x + 1|2x + 1

Solutions:

22.

|2− x | = 2− x if 2− x ≥ 0− (2− x) if 2− x < 0

= 2− x if x ≤ 2x − 2 if x > 2

Hence

limx→∞

3x + 1|2− x |

= limx→∞

3x + 1x − 2 ( |2− x | since x > 2)

= limx→∞

3+ 1/x1− 2/x

= 3+ 01− 0 = 3.

23.

limx→−∞

|x ||x | + 1

= limx→−∞

−x−x + 1 (|x | = −x since x < 0)

= limx→−∞

−1−1+ 1

x= 1.

24.

limx→−∞

x|x |

= limx→−∞

x−x (|x| = −x since x < 0)

= limx→−∞ (−1) = −1.

25.

limx→−∞

|x + 1|x + 1

= limx→−∞

− (x + 1)x + 1 (|x + 1| = − (x + 1) since x < −1)

= limx→−∞ (−1) = −1.

16

26.

limx→−∞

|x + 1|2x + 3

= limx→−∞

− (x + 1)2x + 3 (since x < −1)

= limx→−∞

−1− 1/x2+ 3/x = −1

2.

27.

limx→−∞

2+ |2x + 1|2x + 1

= limx→−∞

2− (2x + 1)2x + 1 (|2x + 1| = − (2x + 1) since x < −1

2)

= limx→−∞

1− 2x2x + 1

= limx→−∞

1x − 22+ 1

x= −1.

IV. Left-hand and right-hand limits

A left–hand limit is the y-value you obtain by approaching x from the left side.A right–hand limit is the y-value you obtain by approaching x from the right side.

Let us try out one example:Suppose

h (x) =

x if x < 0x2 if 0 < x ≤ 2

8− x if x > 2.

Evaluate the following limits:

28. limx→0−

h (x)

29. limx→0+

h (x)

30. limx→2−

h (x)

31. limx→2+

h (x)

32. limx→2

h (x)

Solutions:

28. limx→0−

h (x) = limx→0−

x = 0.

17 MAT1512/1

29. limx→0+

h (x) = limx→0+

x2 = 02 = 0.

30. limx→2−

h (x) = limx→2−

x2 = 22 = 4.

31. limx→2+

h (x) = limx→2+

(8− x) = 8− 2 = 6.

32. limx→2

h (x) does not exist, since limx→2−

h (x) = limx→2+

h (x)

Now determine the following limits:

33. limx→2−

1x − 2 .

34. limx→2−

f (x), where

f (x) =

x + 2 if x > 2x2 if x < 20 if x = 2.

35. limx→0−

1|x | .

36. limx→0−

x|x | .

37. Let

f (x) = 2 if x > 11− x2 if x < 1.

(a) Draw the graph of f.

(b) Determine limx→1−

f (x) and limx→1+

f (x).

(c) Does limx→1

f (x) exist? Give a reason for your answer.

38. Let

f (x) =

x2 + 1 if x > 20 if x = 2 or x = 01x if x < 2 and x = 0.

Determine limx→2+

f (x) , limx→2−

f (x) , limx→2

f (x), limx→∞ f (x) and lim

x→−∞ f (x), if these limits exist.

39. Determine limx→0−

xtan x

.

40. Let

f (x) =

3− x if x < 22 if x = 2x2 if x > 2.

Find limx→2+

f (x) and limx→2−

f (x). Does limx→2 f (x) exist? If so, what is it?

18

Solutions:

33. limx→2−

1x − 2

= −∞.

34. limx→2−

f (x) = limx→2−

x2 (since x < 2)

= 4.

35. limx→0−

1|x |

= limx→0−

1−x (|x | = −x since x < 0)

=∞ .

36. If x tends to 0−, then x is negative. According to the definition of the absolute value, we then have that|x| = −x .

Consequently,

limx→0−

x|x |

= limx→0−

x−x = lim

x→0−(−1) = −1.

37. (a)

( )x

0

1_

1

1

2

y

x

y = f

(b) limx→1−

f (x) = limx→1−

(1− x2) = 1− 1 = 0.limx→1+

f (x) = limx→1+

(2) = 2.

19 MAT1512/1

(c) limx→1

f (x) does not exist, becauselimx→1+

f (x) = limx→1−

f (x) .

38. limx→2+

f (x) = limx→2+

(x2 + 1) = 5.

limx→2−

f (x) = limx→2−

1x

= 12.

limx→2

f (x) does not exist, since limx→2+

f (x) = limx→2−

f (x).

limx→∞ f (x) = lim

x→∞(x2 + 1) =∞.

limx→−∞ f (x) = lim

x→−∞1x

= 0.

39. limx→0−

xtan x = lim

x→0−xsin x · cos x

= limx→0−

xsin x · limx→0− cos x

= 1limx→0−

sin xx· limx→0−

cos x

= 11 · 1 = 1.

40. limx→2+

f (x) = limx→2+

x2= 22= 1

while

limx→2−

f (x) = limx→2−

(3− x) = 3− 2 = 1.limx→2

f (x) exists, since limx→2+

f (x) = limx→2−

f (x).

We have limx→2

f (x) = 1.

V. Limits involving trigonometric functionsIn order to compute the derivatives of the trigonometric functions (for example, by using the first principle ofdifferentiation), we first have to know how to evaluate the limits of trigonometric functions.

Below are some of the theorems you need to know:

(i) limx→0

sin x = 0,

(ii) limx→0

cos x = 1,

(iii) limx→0

sin xx

= limx→0

xsin x

= 1,

(iv) limx→0

cos x − 1x

= limx→0

1− cos xx

= 0.

20

Examples:

Find the limit of the following:

limθ→0

sin (5θ)3θ

Solution:

In order to apply theorem (iii) above, we first have to re-write the function as follows:

sin 5θ3θ

= sin 5θ3θ

.55= 5 sin 5θ

3.5θ= 53sin 5θ5θ

Notice that as θ → 0 we have 5θ → 0. So by theorem (iii) above, with x = 5θ,we have lim

θ→0sin 5θ5θ

= lim5θ→0

sin (5θ)5θ

= 1.

Therefore,

limθ→0

sin 5θ3θ

= limθ→0

53sin (5θ)5θ

= 53limθ→0

sin 5θ5θ

= 53.1

= 53.

Also find the limit oflimx→0

x cot x .

Solution:

limx→0 x cot x = lim

x→0 xcos xsin x

= limx→0

cos xsin xx

=limx→0

cos x

limx→0

sin xx

= 11= 1.

21 MAT1512/1

Also, determine the following limits:

41. limθ→0

θ2

tan θ42. lim

θ→0sin (5θ)2θ

43. limx→0

x + sin xx

44. limt→0

tan (2t)t

45. limθ→0

sin (5θ)sin (2θ)

46. limx→0 (tan (2x) · csc (4x))

47. limθ→5

sin (2θ − 10)sin (θ − 5) 48. lim

θ→01− cos θ

θ

49. limx→−∞ x sin 1x 50. lim

x→∞ x sin 1x

51. limx→−∞ x2 sin 1x 52. lim

θ→01− cos θ1− cos (3θ)

Solutions:

41. limθ→0

θ2

tan θ= limθ→0

θ2 cos θsin θ 42. lim

θ→0sin (5θ)2θ

= limθ→0

θ

sin θθ cos θ = lim

θ→05 sin (5θ)2.5θ

= limθ→0

θ

sin θ· limθ→0 (

θ cos θ) = 52× 1 = 5/2.

= limθ→0

θsin θ · limθ→0 θ · limθ→0 cos θ

= 1× 0× 1 = 0.

43. limx→0

x + sin xx

44. limt→0

tan (2t)t

= limx→0

1+ sin xx

= limt→0

sin (2t)t · cos (2t)

= limx→0

1+ limx→0

sin xx

= limt→0

sin (2t)2t

· 2cos (2t)

= 1+ 1 = 2. = 1× 21= 2.

45. limθ→0

sin (5θ)sin (2θ)

46. limx→0 (

tan (2x) · csc (4x))= lim

θ→0sin (5θ)5θ

· 5θ2θ· 2θsin (2θ)

= limx→0

sin (2x)cos (2x) · sin (4x)

= limθ→0

sin (5θ)5θ

× 52× limθ→0

2θsin (2θ)

= limx→0

sin (2x)cos (2x)

· 12 sin (2x) cos (2x)

= 1× 52× 1 = 5

2. = lim

x→012· 1cos2 (2x)

= 12.

22

47. Put x = θ − 5. Then 2θ − 10 = 2x and x → 0⇔ θ → 5. Now,

limθ→5

sin (2θ − 10)sin (θ − 5) = lim

x→0sin (2x)sin x

= limx→0

sin (2x)2x

· 2xx· xsin x

= 1× 2× 1 = 2.

48.

limθ→0

1− cos θθ

= limθ→0

(1− cos θ) (1+ cos θ)θ (1+ cos θ)

= limθ→0

1− cos2 θθ (1+ cos θ)

= limθ→0

sin2 θθ (1+ cos θ)

= limθ→0

sin θθ

· limθ→0

sin θ1+ cos θ

= 1.0 = 0.

49. Put θ = 1x. Then x = 1

θand x →−∞⇔ θ → 0−. Hence,

limx→−∞ x sin

1x

= limθ→0−

sin θθ

= 1.

50.

limx→∞ x sin

1x

= limx→0+

sin θθ

(where θ = 1x)

= 1.

51. Put θ = 1x . Then x →−∞⇔ θ → 0−. Thus

limx→−∞ x2 sin

1x

= limθ→0−

sin θθ

· 1θ

= −∞.

23 MAT1512/1

52.

limθ→0

1− cos θ1− cos (3θ)

= limθ→0

1− cos θ1− cos (3θ) ·

1+ cos θ1+ cos θ ·

1+ cos (3θ)1+ cos (3θ)

= limθ→0

1− cos2 θ (1+ cos (3θ))1− cos2 (3θ) (1+ cos θ)

= limθ→0

sin2 θ · (1+ cos (3θ))sin2 (3θ) · (1+ cos θ)

= limθ→0

sin θθ

2

· θ2

9θ2· 3θsin (3θ)

2

· 1+ cos (3θ)1+ cos θ

= 12 · 19· 12 · 2

2= 19.

VI. The Squeeze Theorem

Note: The Squeeze Theorem is also known as the Sandwich Theorem or Pinching Theorem.The Squeeze Theorem is used to evaluate the limit values of a complicated function, or the limit values of anunknown function. The Squeeze Theorem is stated as follows:-

If f (x) ≤ g (x) ≤ h (x) for all x in an open interval that contains a (except possibly at a)and lim

x→a f (x) = limx→a h (x) = L , then limx→a g (x) = L .

The Sandwich Theorem is illustrated by the sketches below.

Imagine you have a sandwich as shown in the following figure:

y g( )x=

y f= ( )x y h= x( )

y h= x( )

y f= ( )x

The second slice istaken as a biggerfunction,

The filling is taken as a’’complicated’’ function,

The first slice istaken as asmaller function,

y g= x( )

The theorem says that if g(x) is squeezed or pinched between f (x) and h(x) near a, and if f and h have the samelimit value L at a, then g is forced to have the same limit value L at a.

Suppose you have to find the following limit:

limx→0

x2 sin1x

24

This is a complicated function, in other words, it is difficult to determine the limit of the function. In such a case wecan use the Squeeze (Sandwich) Theorem.

Squeeze Theorem

If we have f (x) ≤ g (x) ≤ h (x) and for some value L ,

limx→c f (x) = L and lim

x→c h (x) = L ,

then, by the Sandwich (Squeeze) Theorem

limx→c g (x) = L

Now let us find the limit of

limx→0

x2 sin 1x . (This is an example of a complicated function.)

Solution:

To use the Squeeze Theorem, we first have to find functions f and h such that

f (x) ≤ x2 sin 1x≤ h (x)

wherelimx→0

f (x) = limx→0

h (x) .

Recall that−1 ≤ sin 1

x≤ 1 for all x = 0.

Now we multiply by x2 (note that for x = 0, x2 > 0).This means that the inequality signs will not change in direction, and we getx2 (−1) ≤ x2. sin 1

x ≤ x2.1, which means that −x2 ≤ x2. sin 1x ≤ x2 for all x = 0.

Then limx→0

−x2 = 0 and limx→0

x2 = 0.

So from the Sandwhich Theorem, it follows that limx→0

x2. sin 1x = 0.

Work through examples 3.7 and 3.8 on pages 92 to 94 in your prescribed textbook (Smith & Minton).After that, go through the given worked examples below.

53. Find limx→∞ f (x) if

2x2

x2 + 1 < f (x) <2x2 + 5x2

.

25 MAT1512/1

54. Suppose that

1− x2

6<

x sin x2− 2 cos x < 1

for x close to zero. Determinelimx→0

x sin x2− 2 cos x .

55. Use the Squeeze Theorem to determine

limx→−∞

sin xx.


limx→∞ ( x /x) and lim

x→−∞ ( x /x) .


limx→∞

1− cos x21+ x3 .

Solutions:

53. limx→∞

2x2

x2 + 1 = limx→∞

21+ 1/x2 = 2 and

limx→∞

2x2 + 5x2

= limx→∞

2+ 5/x21

= 2.

By the Squeeze Theorem, it follows that limx→∞ f (x) = 2.

54. limx→0

1− x2

6= 1− 0

6= 1,

whilelimx→0

1 = 1;

so, by the Squeeze Theorem

limx→0

x sin x2− 2 cos x = 1.

55. For all x it holds that−1 ≤ sin x ≤ 1.

If x < 0, we can divide by x to obtain1x≤ sin x

x≤ −1

x.

Now limx→−∞

1x= 0, and lim

x→−∞ −1x

= − limx→−∞

1x= 0. It follows by the Squeeze Theorem that

limx→−∞

sin xx

= 0.

26

56. For all x it holds thatx − 1 < x ≤ x . (*)

If x > 0 thenx − 1x

< x /x ≤ 1.

Now

limx→∞

x − 1x

= limx→∞

1− 1x

1= 1;

while limx→∞ 1 = 1. From the Squeeze Theorem it follows that

limx→∞ ( x /x) = 1.

If x < 0, then we can divide (∗) by x to obtainx − 1x

> x /x ≥ 1.

Similarly, as above, it follows thatlimx→−∞ ( x /x) = 1.

57. For all x it holds that−1 ≤ − cos x2 ≤ 1

Add 1:0 ≤ 1− cos x2 ≤ 2

For x > 0 (because x →∞), we divide by 1+ x3 > 0 to obtain0

1+ x3 ≤1− cos x21+ x3 ≤ 2

1+ x3 .

Now limx→∞ 0 = 0 and limx→∞

21+ x3 = 0. It follows by the Squeeze Theorem that

limx→∞

1− cos x21+ x3 = 0.

VII. The ε − δ definition of a limit (read only)This section is not for examination purposes; therefore you should not spend too much time on it. However, thoseof you who intend to continue with pure mathematics should be familiar with the contents of this section.

58. Determine δ > 0 so that0 < x − 1

4< δ⇒ 1

4x− 1 < 1

20.

59. Prove from the ε − δ definition that limx→2(7x − 1) = 13.

27 MAT1512/1

60. Prove from the ε − δ definition that limx→−1

f (x) = 10, where f (x) = −3x + 7.

61. Prove from the ε − δ definition that limx→2

x2 − 4x − 2 = 4.

62. Prove from the ε − δ definition that limx→1

x2 − 1x − 1 = 2.

63. Prove from the ε − δ definition that limx→−3

x2 − 9x + 3 = −6.

Solutions:

58.14x− 1 < 1

20

⇒− 120<14x− 1 < 1

20

⇒ 1− 120<14x− 1+ 1 < 1

20+ 1

⇒ 1920<14x<2120

⇒ 2019> 4x >

2021

⇒ 521< x <

519

⇒ 521− 14< x − 1

4<519− 14

⇒− 184< x − 1

4<176.

If we choose δ = 184, we should have that

0 < x − 14< δ⇒ 1

4x− 1 < 1

20.

59. According to the definition of a limit, let f (x) = 7x − 1, xo = 2 and L = 13. Given > 0, we need to findδ > 0 so that for all x

0 < |x − 2| < δ ⇒ | f (x)− 13| < .

Now

| f (x)− 13| = |7x − 1− 13|= |7x − 14|= 7 |x − 2|

So let δ =7. Then

0 < |x − 2| < δ ⇒ | f (x)− 13| = 7 |x − 2| (from above)< 7δ

= 7 ·7= .

28

60. We have that

| f (x)− 10| = |−3x + 7− 10| = |−3x − 3| = |−3(x + 1)|= −3|x + 1|= 3 |x − (−1)| .

If > 0 is given, put δ =3. Then

0 < |x − (−1)| < δ ⇒ | f (x)− 10| = 3 |x − (−1)|< 3δ = 3 ·

3= .

This shows thatlimx→−1

f (x) = 10.

61. Let > 0 be given. We have that

x2 − 4x − 2 − 4

= (x + 2) (x − 2)x − 2 − 4

= |x + 2− 4| (if x = 2)= |x − 2|.

Put δ = . Then, if 0 < |x − 2| < δ, it follows that x = 2, and sox2 − 4x − 2 − 4 = |x − 2| < δ = .

62. Let > 0 be given. Then

x2 − 1x − 1 − 2 = (x − 1) (x + 1)

x − 1 − 2= |x + 1− 2| (if x = 1)= |x − 1|.

Put δ = . Then0 < |x − 1| < δ ⇒ x2 − 1

x − 1 − 2 < .

Hence,

limx→1

x2 − 1x − 1 = 2.

63. In the definition of a limit, we put f (x) = (x2− 9)/(x + 3), xo = −3 and L = −6. Let > 0 be given. Then

x2 − 9x + 3 − (−6) = (x + 3) (x − 3)

x + 3 + 6= |x − 3+ 6| (if x = −3)= |x + 3|= |x − (−3)|.

29 MAT1512/1

Let δ = . If0 < |x − (−3)| < δ

then x = −3, and then it follows from the above thatx2 − 9x + 3 − (−6) = |x − (−3)| < .

Hence,

limx→−3

x2 − 9x + 3 = −6.

VIII. Continuity

In the previous topics, notice that the limit of a function as x approaches a can sometimes be found by just calculatingthe value of the function (y) at x = a. The functions with this property are said to be continuous at a. A continuousprocess is one that takes place gradually, without interruption or abrupt changes.

Definition 1

A function f is continuous at a number a if limx→a f (x) = f (a) .

Definition 2

A function f is continuous from the right at a number a if limx→a+

f (x) = f (a)

and f is continuous from the left at a number a if limx→a−

f (x) = f (a) .

If f is not continuous at a, we say that f is discontinuous at a, or that f has a discontinuity at a. Whenwe consider the two definitions above, we notice that three things (or conditions) are required for f to becontinuous at a.

A function f is continuous at x = a when the following conditions are satisfied:

(i) if f (a) is defined (that is, a is in the domain of f )

(ii) if limx→a f (x) exists [that is, f must be defined in an open interval that contains a] and limx→a−

f (x) = limx→a+

f (x)

(iii) if limx→a f (x) = f (a).

If one or more of the three conditions above are not satisfied, then f is said to be discontinuous at x = a. [Try tothink of the process of finding a limit as making a prediction as to what the y-value will be when x finally gets toa.]

30

In figures (a), (b), (c) and (d) below, three main types of discontinuities – removable, jump and non-removablediscontinuities – are shown. Are you able to explain why the function f is discontinuous at x = a in all fourdiagrams?

Figure (a) Figure (b)

f ( )af ( )ax alim f ( )x

x af ( )x

f ( )x = f ( )ax alimx = a

hole at x = a

[REMOVABLE]

ax

y

ax

y

[REMOVABLE]

is undefined f ( )abut exists,

_

the graph has ahole at ;

lim exists

defined but

the graph has a

f ( )aand is

Figure (c) Figure (d)

f ( )

xa

x

y

[JUMP]

a

y

a

[NON−REMOVABLE]

The limit does notexist, but is defined, the

graph has a jump atx = a

The limit does notexist, the function valueor y−valueextends to infinity atx = a.

expands and

Let us now work through some examples.

Example 1:

(a) Determine if the function f (x) = 1− x, if x ≤ 2x2 − 2x, if x > 2

is continuous or discontinuous at a = 2,and explain why.

(b) Sketch the graph of the function.

31 MAT1512/1

Solution:

(a)

limx→2−

f (x) = limx→2−

(1− x) = 1− 2 = −1limx→2+

f (x) = limx→2+

x2 − 2x = (2)2 − 2 (2) = 4− 4 = 0

Since

limx→2−

f (x) = limx→2+

f (x)

limx→2

f (x) does not exist [that is, condition (ii) has not been met].

Therefore the function is discontinuous at x = 2.

(b) The key points on the graph are:

(i) the vertical intercept(s) (where x = 0)(ii) the horizontal intercept(s) (where f (x) = 0)(iii) the turning points

2_

1_

2_

f ( )x = 1 _ x

y

x0 1 2 3

1

2

3

1_

f ( )x = x2 _ 2 x

f (x) = 1− x is a straight line with intercepts at (0, 1) and (1, 0) for the interval where x ≤ 2.f (x) = x2−2x is a curve with intercepts where x(x−2) = 0 that is (0, 0) and (2, 0) are the x-intercepts,but they do not fall in the interval x > 2.

32

Example 2:

If

f (x) =x2−x−2x−2 if x = 21, if x = 2,

determine if the function f (x) is continuous at x = 2.

Solution:

For the limx→2

f (x) to exist, we must first find limx→2−

f (x) and limx→2+

f (x) . Since

limx→2−

f (x) = limx→2+

f (x) = limx→2

x2 − x − 2x − 2

= limx→2

(x − 2) (x + 1)x − 2 = lim

x→2 (x + 1) = 3,

the limit exists.

But f (2) = 1 is also defined and limx→2

f (x) = f (2) .So f (x) is not continuous at x = 2 (this is the “removable” type of discontinuity).

Example 3:

Find the value of the constant c that would make the function f continuous at x = 3, where:

f (x) = cx + 1, if x ≤ 3cx2 − 1, if x > 3.

Solution:

We determine the left- and right-hand limits as follows:

limx→3−

f (x) = limx→3−

(cx + 1) = 3c + 1

andlimx→3+

f (x) = limx→3+

cx2 − 1 = 9c − 1.

The condition for f to be continuous at x = 3 is that limx→3−

f (x) = limx→3+

f (x) ,that is

3c + 1 = 9c − 1⇔ 9c − 3c = 1+ 1⇔ 6c = 2⇔ c = 1

3 .

33 MAT1512/1

Thus f (x) is continuous at x = 3, forc = 1

3.

You can now work through the last examples of this chapter.

64. Let

f (x) = x if x ≤ 1(x − 1)2 if x > 1.

(a) Draw the graph of f (x) .

(b) Determine limx→1+


f (x). Does limx→1

f (x) exist? Give a reason for your answer.

(c) Is f (x) continuous at x = 1? Give a reason for your answer.

65. Let

f (x) = 4− x2 if x > 13x2 if x < 1.

(a) Draw the graph of f (x) .

(b) Calculate limx→1+


f (x).

(c) What value, if any, must be assigned to f (1) to make f (x) continuous at x = 1? Give reasons for youranswer.

66. (a) At which points (if any) is the function

f (x) = x + ax2 + (a + b) x + ab

discontinuous? (a and b are constants.)

(b) At which points (if any) canf (x) = x + a

x2 + (a + b) x + abbe made continuous by assigning a certain value to f (x), and what is the value?

67. Let

g(x) =x2+2x−15x−3 if x = 3k if x = 3.

What value, if any, must k be for g to be continuous at x = 3? Give reasons for your answer.

68. Let

g(x) = bx2 if x ≥ 12

x3 if x < 12 .

What value must b be for g to be continuous at x = 12? Give reasons for your answer.

34

69. Let

g(x) = x3 + 4 if x ≤ 0b + 1x + 2 if x > 0.

What value must b be for g to be continuous at x = 0? Give reasons for your answer.

70. Let

g(x) = αx2 if x ≥ 2x + 1 if x < 2.

What value must α be for g to be continuous at x = 2? Give reasons for your answer.

Solutions:

64. (a)

1

1

y

x0

y = f ( )x

(b) (i) limx→1+

f (x) = limx→1+

(x − 1)2 = 0.(ii) lim

x→1−f (x) = lim

x→1−(x) = 1.

(iii) limx→1 f (x) does not exist, because limx→1+

f (x) = limx→1−

f (x).

(c) The function f (x) is thus not continuous at x = 1, because limx→1

f (x) does not exist.

35 MAT1512/1

65. (a)

y = f ( )x

y

0x

3

1 2

(b) limx→1+

f (x) = limx→1+

(4− x2) = 4− 1 = 3;limx→1−

f (x) = limx→1−

(3x2) = 3.

(c) Let f (1) = 3, since limx→1

f (x) = 3.

66. (a) f (x) = x + ax2 + (a + b) x + ab

= x + a(x + a) (x + b) .

f (x) is discontinuous where

(x + a)(x + b) = 0

that is at x = −a or x = −b.

(b) The case a = b

We calculate the limits of f (x) when x tends to −a and to −b.

limx→−b+

x + a(x + a) (x + b)

= limx→−b+

1x + b

= ∞.

36

The discontinuity at x = −b cannot be removed. Now,

limx→−a

x + a(x + a) (x + b)

= limx→−a

1x + b

= 1b − a .

The discontinuity at x = −a can be removed. We can make f (x) continuous at x = −a by putting

f (−a) = 1b − a .

The case a = b

In this case there is a non-removable discontinuity at x = −b.Note: Removable discontinuity is when a discontinuity can be removed by redefining the function at thatpoint. In the case of a non–removable discontinuity, the limits do not exist, so there is no way to redefine thefunction at that point of discontinuity. See pages 99-100 of Smith.

67. limx→3

x2 + 2x − 15x − 3

= limx→3

(x − 3) (x + 5)x − 3

= limx→3 (

x + 5)= 8.From the definition of continuity it follows that k must equal to 8 to make g continuous at x = 3.

68. For g (x) to be continuous at x = 12, limx→ 1

2

g(x) must exist (and be equal to g12). For this limit to exist

limx→ 1

2+ g(x) and limx→ 1

2− g(x) must exist and be equal. Now

limx→ 1

2+ g(x) = lim

x→ 12+(bx

2) = b4

andlimx→ 1

2− g(x) = lim

x→ 12−(x

3) = 18.

Thus we must have thatb4= 18

orb = 1

2.

37 MAT1512/1

69. The function g (x) will be continuous at x = 0 if

g (0) = limx→0

g (x) .

Also, limx→0

g(x)will exist provided that limx→0−

g(x) and limx→0+

g(x) exist and are equal. Hence wemust determinethe latter two limits, set them equal, and solve for b. Now

limx→0−

g(x) = limx→0−

(x3 + 4) = 0+ 4 = 4,

whilelimx→0+

g(x) = limx→0+

b + 1x + 2 = b + 1

0+ 2 =b + 12.

Letb+ 12

= 4.

Thenb = 7.

Hence if b = 7, then g (x) is continuous at x = 0.

70. If g (x) is to be continuous at x = 2, then it must hold that

g(2) = limx→2

g(x),

that is

4α = limx→2+

g(x). ...(∗)Now lim

x→2g(x) will exist provided lim

x→2+g(x) and lim

x→2−g(x) exist, and

limx→2+

g(x) = limx→2−

g(x). ...(∗∗)We have that

limx→2+

g(x) = limx→2+

(αx2) = 4α,

whilelimx→2−

g(x) = limx→2−

(x + 1) = 3.

Hence both (∗) and (∗∗) will be satisfied if 4α = 3. Thus if α = 34 , then g (x) is continuous at x = 2.

38

Key points

You should now be comfortable with the notion of the limit of a function and its use. This includes one-sided limits,infinite limits (limits resulting in infinity) and limits at infinity (limits of functions as x approaches infinity).

At this stage you should be able to

• evaluate limits of the form limx→c f (x), limx→c−

f (x) , limx→c+

f (x) and limx→±∞ f (x) , where c is a real number and

f (x) is an algebraic or trigonometric function

• provide a formal definition of the limit, using the correct mathematical notation• apply the laws governing limits• determine and evaluate the limits of sums, products, quotients and compositions of functions• evaluate the limits of functions analytically, graphically and numerically• use the limit definition of continuity to determine whether a function is continuous or discontinuous at a point• use the Squeeze Theorem to determine certain limits

At the end of each exercise set in Smith & Minton, there is a section entitled “Exploratory exercises”. Try thequestions in these sections – they may help you order your thoughts. Also try the questions in the section “For yourreview” at the end of the chapter.

39 MAT1512/1

CHAPTER 2

DIFFERENTIATION OF DIFFERENTTYPES OF FUNCTIONS

1. Background

Differentiation is widely applied in engineering, chemistry, physics, biology and many other disciplines in scienceand technology. It is basically a concept that deals with rate of change. Among other things, it is used to define theslopes of curves, to calculate the velocities and accelerations of moving objects, to find the firing angle that gives acannon its greatest range, and to predict the times when planets will be closest to each other or farthest apart. In thischapter we begin the study of differential calculus. The central concept of differential calculus is the derivative.After learning how to calculate derivatives, you will use them to solve problems involving rates of change. Thischapter is devoted to (a) the definition of the derivative, or differentiation from first principles; (b) the rules usedto find derivatives of algebraic and trigonometric functions; (c) problems involving tangent and normal lines; (d)techniques such as logarithmic differentiation and implicit differentiation; and (e) the application of the Mean ValueTheorem and Rolle’s Theorem.



• demonstrate an understanding of the first principles of differentiation

• apply basic differential formulas such as the power rule, product rule, quotient rule, chain rule, and combina-tions of these rules to differentiate a variety of algebraic and trigonometric functions

• compute the derivatives of trigonometric functions and inverse trigonometric functions, and of exponentialand logarithmic functions

• use the methods of logarithmic differentiation and implicit differentiation to find derivatives

40

• solve problems involving tangent and normal lines and the Mean Value Theorem



The overview of chapter 2 of Smith on page 145 and sections 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8 and 2.9.

4. The derivative

4.1 Introducing the derivative

The definition of the derivative can be approached in two different ways. One is geometrical (as a slope of a curve)and the other is physical (as a rate of change). Both of these definitions can be used, but the emphasis should be onusing the derivative as a tool for solving calculus problems.

Remember, in high school, you used the first principle of differentiation to find the slope of a tangent line at aparticular point.

Definition: The slope mtan of the tangent line y = f (x) at x = a is:mtan = lim

h→0f (a + h)− f (a)

hExample:

Find the slope of the tangent line to the curve y = x2 + 1 at x = 1.

Solution:y

x

8

4

_ 2 2 4

mtan = limh→0

f (1+ h)− f (1)h

= limh→0

(1+ h)2 + 1 − (1+ 1)h

= limh→0

1+ 2h + h2 + 1− 2h

= limh→0

2h + h2h

= limh→0

h (2+ h)h

= limh→0 (

2+ h) = 2

Thus at x = 1, mtan = 2.

41 MAT1512/1

You should also be familiar with the approach used in mechanics when dealing with problems involving averageand instantaneous velocity (rate of change) .

Average velocity :

vave = total distancetotal time

Instantaneous velocity :

v = limh→0

f (t + h)− f (t)h

Example:

Consider an object that is dropped from a height of 64 m. Its position is given as f (t) = 64−16t2. Find its averagevelocity between

(a) t = 1 and t = 2

(b) t = 1.5 and t = 2

(c) t = 1.9 and t = 2

Also find its instantaneous velocity at t = 2.

Solution:

(a) vave = f (2)− f (1)2− 1 = 64− 16 (2)

2 − 64− 16(1)21

= −48 m/s.

(b) vave = f (2)− f (1.5)2− 1.5 = 64− 16 (2)2 − 64− 16(1.5)2

0.5= −56 m/s.

(c) vave = f (2)− f (1.9)2− 1.9 = 64− 16 (2)2 − 64− 16(1.9)2

0.1= −48 m/s.

(d) The instantaneous velocity is given by:

vinstantaneous = limh→0

f (2+ h)− f (2)(2+ h)− 2

= limh→0

64− 16 (2+ h)2 − 64−−16 (2)2h

= limh→0

64− 16 4+ 4h + h2 − 64−−16 (2)2h

= limh→0

−64h − 16h2h

= limh→0

−16h (h + 4)h

= limh→0

−16 (h + 4) = −64 m/s.

42

4.2 Definition of the derivative

The derivative of the function f (x) (at a particular point where x = a) is defined asf (a) = lim

h→0f (a + h)− f (a)

h,

provided that the limit exists. We say that f is differentiable at x = a when the above limit exists.In general, the derivative of f (x) is the function f (x) given by

f (x) = limh→0

f (x + h)− f (x)h

,

provided that this limit exists. Finding the first derivative by using the above method is called the first principle ofdifferentiation.

Note: If we write x = a+h, then h = x −a and h will approach 0 if and only if x approaches a. Therefore anotherway of defining the derivative is as follows:

f (a) = limx→a

f (x)− f (a)x − a .

Let us try an example:

Use the definition of the derivative to find g (x) if g (x) = x4.

g (x) = limh→0

g (x + h)− g (x)h

= limh→0

(x + h)4 − x4h

= limh→0

(x + h)2 − x2 (x + h)2 + x2h

= limh→0

h [2x + h] (x + h)2 + x2h

= limh→0

[2x + h] (x + h)2 + x2

= (2x) x2 + x2= 4x3.

5. Worked examples

In some instances, we do compute derivatives directly from the definition. However, such computations are tediousand very time-consuming. Fortunately, several differential rules or basic differentiation formulas have been devel-oped to find derivatives without using the definition directly. In order to introduce you to the basic differentiationtechniques, these rules are presented together with worked examples. You should go and read about these rules inSmith and work through the examples, following each rule. You will find the rules on the following pages in Smith:

Derivative of a constant p170The power rule for positive integer powers of x p 171

The constant multiple rule p 173The sum and difference rule p 173The power rule for negative integer powers of x p 184

The product rule p181The quotient rule p182The chain rule p189

43 MAT1512/1

Our collection of worked examples relating to chapter 2 of Smith divides naturally into eight sets, with some overlap,namely:

I. Differentiation from first principles

II. Basic differentiation formulas (use of the power rule, product rule, quotient rule, chain rule, and combinationsof these rules to differentiate a variety of functions)

III. Derivatives of trigonometric functions and inverse trigonometric functions

IV. Derivatives of exponential and logarithmic functions

V. Logarithmic differentiation

VI. Implicit differentiation

VII. Tangent and normal lines

VIII. The Mean Value theorem

Attempt the problems appearing immediately after each set, once you have studied the relevant parts of Smith anddone some of the exercises there.

44

The table below (Table 3) shows how these worked examples and their solutions are organised.

Table 3

Topic(s) Sections in Study guideSmith & Minton examples

I. Differentiation from first Section 2.2 (examples 1–5principles (or by using the 2.1 – 2.7)definition of the derivative)

II. Basic differentiation Sections 2.2 – 2.5, 2.7, 2.8 6–31(use of the power rule, Sections 3.1 – 3.5higher-order Sections 3.6 – 3.7product rule, Sections 4.1 – 4.2quotient rule, Sections 4.3chain rule, Sections 5.1 – 5.4and combinations of these rulesto differentiate a variety offunctions)

III. Derivatives of trigonometric Section 2.6 (examples 32–49functions and 6.1 – 6.4 & 6.7)inverse trigonometric functions Section 2.8 (examples 50–52

8.5 – 8.6)IV. Derivatives of exponential and Section 2.7 (examples 53–63

logarithmic functions 7.2 – 7.4V. Logarithmic differentiation Section 2.7 (example 7.6) 64–74

VI. Implicit differentiation Section 2.8: (examples 75–82(second order) 8.1 – 8.3 & 8.4)

VII. Tangent and normal lines Sections 1.2 – 1.5 83–92VIII. The Mean Value Theorem Section 2.8 (example 9.3) 93–95

I. Differentiation from first principles (definition of the derivative)

1. Find the derivative of 3x2 + 2x − 1 at x = 1.2. If f (x) = 1

x (x = 0), find f (x) .3. If f (x) = √x for (x ≥ 0), find f (x) .

45 MAT1512/1

Solutions:

1.

f (x) = limh→0

f (x + h)− f (x)h

= limh→0

3 (x + h)2 + 2 (x + h)− 1− 3x2 + 2x − 1h

= limh→0

3x2 + 6xh + h2 + 2x + 2h − 1− 3x2 − 2x + 1h

= limh→0

6xh + h2 + 2hh

= limh→0

h (6x + h + 2)h

= limh→0

6x + h + 2 = 6x + 2.At x = 1, f (1) = 6 (1)+ 2 = 8.

2.

f (x) = limh→0

f (x + h)− f (x)h

= limh→0

1x+h − 1

x

h= limh→0

x − (x + h)x (x + h) · 1

h

= limh→0

−hx2 + xh ·

1h= limh→0

−1x2 + xh = −

1x2.

3.

f (x) = limh→0

f (x + h)− f (x)h

= limh→0

√x + h −√x

h= limh→0

√x + h −√x

h·√x + h +√x√x + h +√x

= limh→0

x + h − xh√x + h +√x = lim

h→0h

h√x + h +√x = lim

h→01√

x + h +√x= 1

2√x.

4. Use the definition of the derivative to prove that g (0) = 0 if g (x) = x2 |x | .

Solution:

By definition,

g (0) = limh→0

g (0+ h)− g (0)h

,

(a) provided this limit exists. Now

limh→0

g (0+ h)− h (0)h

= limh→0

h2 |h|− 0h

= limh→0 (h |h|)

= 0.

5. Use the definition of the derivative to show that f (2) does not exist when f (x) = |x − 2|.

46

(a)

limh→0+

f (2+ h)− f (2)h

= limh→0+

|(2+ h)− 2|− |2− 2|h

= limh→0+

|h|h

= limh→0+

hh

(|h| = h since h > 0)= 1,

(b) while similarly

limh→0−

f (2+ h)− f (2)h

= limh→0−

|h|h

= limh→0

−hh

(|h| = −h since h < 0)= −1.

Since the left-hand and right-hand limits are not equal, it follows that

limh→0

f (2+ h)− f (2)h

does not exist, that is, f is not differentiable at 2.

II. Basic differentiation formulas

(a) The Power RuleIf y = xn, then

ddx(xn) = nxn−1

i.e.dydx

= nxn−1

Examples:

Find the derivatives of the following:

6. f (x) = x8 7.ddx

3√x2 8. m (x) = x 12 + 3x− 12

Solutions:

7. f (x) = 8x8−1 = 8x7 7.ddx

3√x2 = ddxx 23 = 2

3x 23−1 = 2

3x− 13 = 2

3x 13

8. m (x) = x 12 + 3x− 12 ⇒ m (x) = 12x 12

− 32x 32

Examples involving absolute values:


9. k (a) = 1|a| 10. k (s) = |s|

47 MAT1512/1

Solutions:

9. k(a) = 1|a|

=

1a

if a > 0

−1a

if a < 0.

Hence,

k (a) =

− 1a2

if a > 0

1a2

if a < 0.

In fact, k (a) can be written as a single expression, namely k (a) = − 1a |a| .

10.

k (s) = |s|⇒ k (s) = 1 if s > 0

−1 if s < 0.

Note: k (s) = |s| is not differentiable at s = 0.(b) The Product Rule

If y = f.g, then y = f .g + f.g orddx

f (x) g (x) = f (x) g (x)+ f (x) g (x)

Examples:

11. Find f (x) if f (x) = 2x4 − 3x + 5 x2 −√x + 2x

Solution:

f (x) = ddx

2x4 − 3x + 5 x2 −√x + 2x

+ 2x4 − 3x + 5 ddx

x2 −√x + 2x

= 8x3 − 3 x2 −√x + 2x

+ 2x4 − 3x + 5 2x − 12√x− 2x2

12. Differentiate the following function:

y = (x + 1)10 (2x + 3)11 (4− x)12

48

Solution:

y = (x + 1)10(2x + 3)11(4− x)12

⇒ dydx

= (x + 1)10(2x + 3)11 ddx(4− x)12 + (x + 1)10(4− x)12 d

dx(2x + 3)11

+ (2x + 3)11(4− x)12 ddx(x + 1)10

= (x + 1)10(2x + 3)11 · 12(4− x)11 ddx(4− x)+ (x + 1)10(4− x)12 · 11(2x + 3)10 · d

dx(2x + 3)

+ (2x + 3)11(4− x)12.10(x + 1)9= −12(x + 1)10(2x + 3)11(4− x)11 + 22(x + 1)10(2x + 3)10(4− x)12

+ 10(x + 1)9(2x + 3)11(4− x)12.

(c) The Quotient Rule:

If y = fg, then

f .g − f.gg2

= y

orddx

f (x)g (x)

= f (x) g (x)− f (x) g (x)

g (x) 2

13. Compute the derivative ofx2 − 2x2 + 1 .

Solution:

f (x) =ddx x

2 − 2 x2 + 1 − x2 − 2 ddx x

2 + 1x2 + 1 2

= 2x x2 + 1 − x2 − 2 (2x)x2 + 1 2

= 6x

x2 + 1 2 .

More examples:

Differentiate the following functions:

14. g (t) =√2t + 1√

2t + 1+ 1 15. g (x) = x + 1(x + 2)10 16. y = 2x + 5

3x − 2

Solutions:

49 MAT1512/1

14.

g(t) =√2t + 1√

2t + 1+ 1 =(2t + 1) 12

(2t + 1) 12 + 1⇒ g (t) = 1

2(2t + 1)− 12 · 2 · (2t + 1) 12 + 1 −

(2t + 1) 12 · 12(2t + 1)− 12 · 2 / (2t + 1) 12 + 1 2 = 1√

2t + 1 √2t + 1+ 1 2 .

15.

g (x) = x + 1(x + 2)10

⇒ g (x) = 1. (x + 2)10 − (x + 1) (x + 2)9 .10.1(x + 2)20 = 1− 9x

(x + 2)11 .

16.

y = 2x + 53x − 2

⇒ dydx

= (3x − 2) 2− (2x + 5) 3(3x − 2)2 = −19

(3x − 2)2 .

(d) The Chain Rule

If y = f ◦ g, then y = f (g) .g

orddx

f (g (x)) = f (g (x)) g (x)

Examples:

17. Differentiate y = x3 + x + 1 5. 18. Findddt

√100+ 8t .

19. Finddydx

if y = (4− 3x)9 .

Solutions:

17. Let u = x3 + x + 1, then y = u5, anddydx

= dydududx

= ddu

u5dudx

= 5u4ddx

x3 + x + 1 = 5 x3 + x + 1 4 3x2 + 1

18. Let u = 100+ 8t, then √100+ 8t = (100+ 8t) 12 = u 12 , and

ddu(u)

12 = 1

2u−

12dudt= 12√100+ 8t

ddt(100+ 8t) = 4√

100+ 8t

19. y = (4− 3x)9 ⇒ dydx

= 9 (4− 3x)8 (−3) .

50

(e) Combinations of rules

In the following worked examples, combinations of rules are used to determine the required derivatives.Find the first derivatives of the following. Do not simplify.

20. f (x) = (5x + 11)20(4− x)30 21. f (w) = (3w2 + 2) 23 (9w − 1)− 13

22. f (x) = 1− x1+ x2 23. f (x) = √

x + 1√x

100

24. θ(y) = 13(2y2 + 5y) 32 25. f (x) = (2x3 + 3)10(3x2 + 2)20

26. h(θ) = (θ 13 + 7)100(2θ− 13 + 7)200 27. y = x5+ 15x

5

28. y = (1− 6x) 23

Solutions:

20. f (x) = (5x + 11)20(4− x)30⇒ f (x) = 20(5x + 11)195(4− x)30 + (5x + 11)2030(4− x)29(−1).

21. f (w) = (3w2 + 2) 23 (9w − 1)− 13⇒ f (w) = 2

3(3w2 + 2)− 13 (6w)(9w − 1)− 13 + (3w2 + 2) 23 − 1

3 (9w − 1)−43 9.

22. f (x) = 1− x1+ x2 =

1− x1+ x2

12

⇒ f (x) = 12

1− x1+ x2

− 12· (−1) 1+ x

2 − (1− x) · 2x1+ x2 2

.

23. f (x) = √x + 1√

x

100

⇒ f (x) = 100 √x + 1√

x

99

· 12√x− 12x 32

.

24. θ(y) = 13(2y2 + 5y) 32

⇒ θ (y) = 13· 32(2y2 + 5y) 12 (4y + 5).

51 MAT1512/1

25. f (x) = (2x3 + 3)10(3x2 + 2)20⇒ f (x) = 10(2x3 + 3)9 · 6x2 · (3x2 + 2)20 + 20(3x2 + 2)19 · 6x · (2x3 + 3)10.

26. h(θ) = (θ 13 + 7)100(2θ− 13 + 7)200

⇒ h (θ) = 100(θ 13 + 7)99 · 13θ−

23 · (2θ− 13 + 7)200 + (θ 13 + 7)100 · 200(2θ− 13 + 7)199 · (−2

3θ−

43 ).

27. y = x5+ 15x

5

⇒ dydx

= 5 x5+ 15x

4 15− 15x−2

28. y = (1− 6x) 23⇒ dydx

= 23

1

(1− 6x) 13(−6).

Also do the following exercises:

29. Show that y = 2 |x | if y = x |x | .30. Let g (x) , h (x) and k (x) be differentiable functions such that

f (x) = g (x) h (x) k (x)

for all x . Use the product rule for the derivative of the product of two functions to write down the derivativeof f in terms of g (x) , h (x) and k (x) and their derivatives.

31. Use the chain rule and product rule to derive the quotient rule for derivatives.

Solutions:

29. Case I Suppose x > 0. Then |x | = x , soy = x2

andy = 2x = 2 |x | .

Case II Suppose x < 0. Then |x | = −x , so

y = −x2

andy = −2x = 2 (−x) = 2 |x | .

52

Case III Suppose x = 0. If f (x) = x |x|, then

limh→0

f (0+ h)− f (0)h

= limh→0

h |h|− 0h

= limh→0

|h|= 0 = 2 |0| .

Hence, for all values of x we have thaty = 2 |x | .

30. Letf (x) = (x)k(x),

where(x) = g(x)h(x).

Then

f (x) = (x)k(x)+ (x)k (x)

= (g (x)h(x)+ g(x)h (x))k(x)+ (x)k (x)

= g (x)h(x)k(x)+ g(x)h (x)k(x)+ g(x)h(x)k (x).

31. Let f (x) = g (x)h (x)

, where g and h are differentiable functions. Now

f (x) = g(x) · (h(x))−1

so

f (x) = g (x) · (h(x))−1 + g(x) · − 1(h (x))2

· h (x)(by the product and chain rules)

= g (x)h (x)

− g (x) · h (x)(h (x))2

= h (x) g (x)− g (x) h (x)(h (x))2

.

III. Derivatives of trigonometric functions and inverse trigonometric functions

(a) Derivatives of trigonometric functions

Below are limits which will help you find some of the trigonometric derivatives.

limθ→0

cos θ = 1, limθ→0

1− cos θθ

= limθ→0

cos θ − 1θ

= 0

limθ→0

sin θ = 0, limθ→0

sin θθ

= limθ→0

θ

sin θ= 1

53 MAT1512/1

A very important table

You should know these derivative formulae by heart!

Note: These derivative formulae can also be derived from first principles.

ddx(sin x) = cos x d

dx(cot x) = − csc2 x

ddx(cos x) = − sin x d

dx(sec x) = sec x tan x

ddx(tan x) = sec2 x d

dx(csc x) = − csc x cot x

Here is an example worked out from first principles:

Prove thatddxsin x = cos x .

Solution:

If f (x) = sin x, then

f (x) = limh→0

f (x + h)− f (x)h

= limh→0

sin (x + h)− sin (x)h

= limh→0

sin x cos h + cos x sin h − sin xh

= limh→0

sin xcos h − 1

h+ cos x sin h

h

= limh→0

sin x . limh→0

cos h − 1h

+ limh→0

cos x . limh→0

sin hh

= sin x .0+ cos x .1= cos x

Now determine the derivatives of the following functions:

32. f (x) = x5 cos x 33. f (x) = sin 2xx + 1

Solutions:

32.

f (x) = ddx

x5 cos x = ddx

x5 cos x + x5 ddx(cos x) {product rule}

= 5x4 cos x − x5 sin x .

54

33. The derivative of f (x) = sin 2xx + 1 is

f (x) = cos2xx + 1

ddx

2xx + 1 {chain rule}

= cos2xx + 1

2 (x + 1)− 2x (1)(x + 1)2 {quotient rule}

= cos2xx + 1

2(x + 1)2 .

Now go through the rest of the worked examples. Find the first derivative. Do not simplify.

34. h(t) = sec t + tan t 35. y = x csc x

36. g (θ) = sin θθ

37. h(t) = sin ttan t + 1

38. f (t) =√3t + cos tt + 3 39. y = cos x

1+ sin x

40. h (w) = sin3 (cotw) 41. k (x) = cot2 (3x)42. g(x) = √cos cx + cos cα,where α is a constant43. g(θ) = sec 32 (1− θ) 44. g(x) = tan4(tan x)

45. h(t) = (t− 13 + 1). tan t 46. y = 2x + 53x − 2

47. y = cos(3πx/2) 48. y = cos3 x

49. y = [sin(x + 5)] 54

Solutions:

34. h(t) = sec t + tan t⇒ h (t) = (sec t)(tan t)+ sec2 t.

35. y = x cosc x⇒ y = x .(− cosc x cot x)+ (cosc x).1.

36. g(θ) = sin θθ

⇒ g (θ) = θ cos θ − (sin θ) (1)θ2

.

55 MAT1512/1

37. h(t) = sin ttan t + 1

⇒ h (t) = cos t (tan t + 1)− sin t sec2 t(tan t + 1)2 .

38. f (t) =√3t + cos tt + 3

⇒ f (t) =1

2√3t+cos t (3− sin t) (t + 3)−

√3t + cos t · 1

(t + 3)2 .

39. y = cos x1+ sin x

⇒ y = (1+ sin x) (− sin x)− cos x (0+ cos x)(1+ sin x)2 .

40. h(w) = sin3(cot w)⇒ h (w) = 3 sin2(cot w). cos(cot w).(−cosec2w).

41. k(x) = cot2(3x)⇒ k (x) = 2 cot(3x).(−cosec2(3x)).3.

42. g(x) = √cos cx + cos cα,where α is a constant⇒ g (x) = 1

2√cos cx + cos cα · (− cos cx . cot cx).

43. g(θ) = sec32 (1− θ)⇒ g (θ) = 3

2sec 12 (1− θ) · sec(1− θ) · tan(1− θ) · (−1).

44. g(x) = tan4(tan x)⇒ g (x) = 4 tan3(tan x). sec2(tan x). sec2 x .

45. h(t) = (t− 13 + 1) · tan t⇒ h (t) = −1

3t− 43 · tan t + (t− 13 + 1) · sec2 t.

46. g(θ) = cos2(θ2)⇒ g (θ) = 2 cos(θ2) · (− sin(θ2)) · 2θ .

47. y = cos(3πx/2)⇒ dydx

= − sin(3πx/2) ddx(3πx/2)

= −(3π/2) sin(3πx/2).

56

48. y = cos3 x⇒ dydx

= 3(cos2 x)(− sin x).

49. y = [sin(x + 5)] 54⇒ dydx

= 54[sin(x + 5)] 14 [cos(x + 5)].

(b) Derivatives of inverse trigonometric functions

These are useful in applications and are essential for solving problems. The rules or definitions are as follows:

ddxsin−1 x = 1√

1− x2 , for −1 < x < 1

ddxcos−1 x = −1√

1− x2 , for −1 < x < 1

ddxtan−1 x = 1

1+ x2

ddxcot−1 x = −1

1+ x2

ddxsec−1 x = 1

|x |√x2 − 1 , for |x | > 1

ddxcsc−1 x = −1

|x |√x2 − 1 , for |x | > 1

Examples:

Compute the derivatives of the following functions:

50. cos−1 3x2 51. sec−1 x 2 52. tan−1 x3

Solutions:

50.ddxcos−1 3x2 = −1

1− 3x2 2ddx

3x2 = −6x√1− 9x4 .

51.ddx

sec−1 x 2 = 2 sec−1 x ddxsec−1 x = 2 sec−1 x 1

|x |√x2 − 1 .

52.ddxtan−1 x3 = 1

1+ x3 2ddxx3 = 3x2

1+ x6 .

57 MAT1512/1

IV. Derivatives of exponential and logarithmic functions

(a) The exponential function

Definition:

The inverse function of the function ln x is known as the expo–nential function and is indicated by exp x (or ex).

Properties of the exponential function

(a) The graph of ex is the mirror image of the graph of ln x in the line y = x . Therefore the graph looks like this:y

x

y = ex y = x

y = ln x

10

1

(b) exp x is defined for all x ∈ R, in other words

Dom (exp x) = R.

(c) Im (exp x) = {y|y > 0}

(d)ddx(ex) = ex

(e) ex+y = ex · ey

(f) ex−y = ex/ey

(g) erx = (ex)r

(h) If x →∞, then ex →∞.

58

(i) If x →−∞, then ex → 0.

(b) The logarithmic function

Definition:

The natural logarithmic function y = ln x is the inversefunction of the exponential function.

Properties of the ln function

(a) The graph of ln x looks like this:

y

x

y = x

y = ln x

10

(b) ln x is defined for all x > 0, in other words

Dom (ln x) = {x | x > 0} .

(c) Im (ln x) = R(d) ln x < 0 if 0 < x < 1,

ln 1 = 0,ln x > 0 if x > 1.

(e)ddx(ln x) = 1

x(f) ln(xy) = ln x + ln y, if x, y > 0(g) ln

xy= ln x − ln y, if x, y > 0

59 MAT1512/1

(h) ln xr = r ln x , if x > 0, r rational

(i) logb a =logc alogc b

(j) If x →∞, then ln x →∞.(k) If x → 0+, then ln x →−∞.

Remark:

The following mistake is often made when applying rule (h):

ln xr = r ln x (this is correct)

but(ln x)r = r ln x (incorrect).

The index r must be an index of x, but not of ln x .

The inverse of the exponential function ex is called the natural logarithmic function y = ln x . In this section, westudy the natural logarithm both as a differentiable function and as a device for simplifying calculations.

(c) Examples of the exponential function


53. f (x) = e√x

54. y = ex cos x

55. y = tan e3x−2

56. y = e3x

1+ ex57. y = ex+ex

Solutions:

53. f (x) = e√x

f (x) = e√x · 12 x−12 = e

√x

2√x.

54. y = ex cos x

y = ex cos x (1. cos x + x (− sin x)) = ex cos x (cos x − x sin x) .

60

55. y = tan e3x−2

y = sec2 e3x−2 e3x−2.3 = 3e3x−2 sec2 e3x−2 .

56. y = e3x

1+ ex

y = e3x

1+ ex =(1+ ex) e3x .3 − e3x (0+ ex (1))

(1+ ex)2 = 3e3x + 3e4x − e4x(1+ ex)2 = 3e

3x + 2e4x(1+ ex)2 .

57. y = ex+ex

y = ex+ex (1+ ex (1)) = ex+ex (1+ ex) .

(d) Examples of the logarithmic function


58. f (x) = x2 ln 1− x2

59. f (x) = log3 x2 − 4

60. f (x) = log10x

x − 161. y = ln x

√1− x2 sin x

62. y = x sin x

63. y = (sin x)x

Solutions:

58. f (x) = x2 ln 1− x2

f (x) = 2x ln 1− x2 + x2 · 11− x2 (0− 2x) = 2x ln 1− x

2 − 2x3

1− x2 .

59. f (x) = log3 x2 − 4Note that (or recall that) loga x =

ln xln a

,

so f (x) = log3 x2 − 4 = ln x2 − 4ln 3

∴ f (x) = 1ln 3

1x2 − 4 (2x − 0) = 1

ln 32xx2 − 4 = 2x

x2 − 4 ln 3 .

61 MAT1512/1

60. f (x) = log10x

x − 1Recall: loga x =

ln xln a

,

so f (x) = log10x

x − 1 =ln

xx − 1ln 10

= 1ln 10

lnx

x − 1 = 1ln 10

[ln x − ln (x − 1)]

∴ f (x) = 1ln 10

1 (x − 1)− 1.xx (x − 1) = 1

ln 10x − 1− xx (x − 1) = −1

x (x − 1) ln 10

61. y = ln x√1− x2 sin x

y = ln x + 12ln 1− x2 + ln sin x

so y = 1x+ 12

11− x2 (0− 2x) + 1

sin x(cos x (1))

that is y = 1x+ 12

−2x1− x2 + cos x

sin x

= 1x+ −x1− x2 + cot x .

62. y = x sin xYou should take natural logs on both sides first and get:

ln y = ln x sin xln y = sin x ln x

Now take the derivatives of both sides:

1yy = cos x . ln x + sin x .1

x

y = y cos x ln x + sin xx

= x sin x cos x . ln x + sin xx

.

63. y = (sin x)xTake logs on both sides and get:

ln y = x ln (sin x)and

1yy = 1. ln (sin x)+ x 1

sin x. cos x ,

so y = y ln (sin x)+ x .cos xsin x

= (sin x)x [ln (sin x)+ x . cot x] .

62

V. Logarithmic differentiation

(a) The simplification of functions

As a result of its properties, a function involving the logarithmic function can be simplified before it is differentiated,as becomes clear in the following example:

Example:

Determine f (x) if f (x) = ln 3 x−1x2 .

Solution:

Now

f (x) = lnx − 1x2

1/3

(using properties (f), (g) and (h) of logarithmic functions described earlier on)

= 13ln (x − 1)− ln x2

= 13ln(x − 1)− 2

3ln x .

Therefore

f (x) = 13

1(x − 1) − 2

31x

= −x + 23x (x − 1) .

Even if the logarithmic function itself does not occur in the expression, it can be used for the simplification offunctions. We apply the logarithmic function to both sides of the equation and then differentiate (see below).

Remark:

Always remember that when applying the chain rule to ddx (ln f (x)) , it becomes

ddx(ln f (x)) = 1

f (x)f (x) .

(the derivative of the first function multiplied by the derivative of the second function).

For example:

ddx(ln sin x)

= 1sin x

· cos x .

Examples:

64. Determine y if y = x2e2x cos x3.

63 MAT1512/1

65. Determine f (x) if f (x) = x(1−x2)2√1+x2

.

66. Determine f (x) if f (x) = log3√x2+4

(x−1) 32.

Solutions:

64. We have that

y = x2e2x cos x3

By applying the logarithmic function on both sides, we find that

ln y = ln x2e2x cos x3

= 2 ln x + 2x + ln cos x3.

By differentiating we obtain

1y· y = 2

x+ 2+ 1

cos x3· − sin x3 · 3x2.

Therefore

y = y2x+ 2− 3x2 tan x3

= x2e2x cos x32x+ 2− 3x2 tan x3 .

65.

f (x) = x 1− x2 2√1+ x2

Once again we apply the logarithmic function on both sides and thus obtain

ln f (x) = lnx 1− x2 2√1+ x2

= ln x + 2 ln(1− x2)− 12ln(1+ x2).

By differentiation we therefore obtain that

1f (x)

· f (x) = 1x+ −4x1− x2 −

x1+ x2 .

64

Therefore

f (x) = x 1− x2 2√1+ x2

1x− 4x1− x2 −

x1+ x2

= 1− x2 2√1+ x2 −

4x2 1− x2√1+ x2 − x

2 1− x2 2

1+ x2 32

= 1− x2 2 1+ x2 − 4x2 1− x2 1+ x2 − x2 1− x2 2

1+ x2 32

= 1− x2 1− x2 1+ x2 − 4x2 1+ x2 − x2 1− x21+ x2 3

2

= 1− 5x2 − 4x4 1− x21+ x2 3

2.

66.

f (x) = log3

√x2 + 4

(x − 1) 32= 1

2log3(x2 + 4)−

32log3(x − 1)

Now using the fact that

loga x =logb xlogb a

[Formula 5.4 on p 55 in Smith]

we find that

f (x) = 12loge x2 + 4loge 3

− 32loge (x − 1)loge 3

= 12ln x2 + 4ln 3

− 32ln (x − 1)ln 3

.

If we keep in mind that ln 3 is a constant, it follows that

f (x) = 12 ln 3

· 1x2 + 4 · 2x −

32 ln 3

· 1x − 1

= xx2 + 4 ln 3 −

32 (x − 1) ln 3 .

(b) Functions of the form f (x) = g(x)h(x)

We employ logarithmic differentiation to differentiate functions of the form

f (x) = g(x)h(x)

Just as in (a), we do this by applying the logarithmic function on both sides and then making use of the propertiesof the logarithmic function.

Thenln f (x) = ln g(x)h(x).

65 MAT1512/1

Thereforeln f (x) = h(x). ln g(x).

Example:

Determine f (x) if f (x) = √x√x.

Solution:

By applying the logarithmic function on both sides, we obtain

ln f (x) = ln√x√x

= √x · ln √x (See the remark earlier on under (a) The simplification of functions.)

By differentiating, we obtain from the chain rule and the product rule that

1f (x)

f (x) = √x · 1√

x· 12x−

12 + ln √x · 1

2x−

12

= 12√x+ 12√xln√x .

Therefore

f (x) = f (x)12√x+ 12√xln√x

= √x√x 12√x1+ ln √x

=√x√x

2√x1+ ln √x .

Now attempt the following yourselves. (The answers are given at the end, so that you can make sure that youunderstand this important section well.)

Determine f (x) below:

67. f (x) = ln x +√x2 + 1 (be careful with this exercise!)

68. f (x) = (ln x)√x

69. f (x) = ln 3 x2 − 1x2 + 1

70. f (x) = x ln x + 7x23

66

71. f (x) = x (x − 1) 13(x + 1) 23

72. f (x) = a3x2

73. f (x) = xe−x2

74. f (x) = log10(log10 x)

Solutions:

67.1√x2 + 1 (Use the chain rule on the right-hand side and simplify.)

68.(ln x)

√x

√x

1ln x

+ ln ln x2

69.4x

3 x4 − 1

70.2 ln xx

× x ln x + 2 ln 73x 13

× 7x23

71.2x2 + 3x − 3

3 (x − 1) 23 (x + 1) 53

72. 6x a3x2 ln a

73. xe−x2 · e−x2 1

x − 2x ln x

74.1

x ln x ln 10

VI. Implicit differentiation

This technique is used when the function contains variables that cannot be separated from each other. This processof differentiating both sides of the equation with respect to x and then solving for y (x) is called implicit differen-tiation.

Example:

Find y (x) for x2 + y3 − 2y = 3. Then find the slope of the tangent line at the point (2, 1).

67 MAT1512/1

Solution:

Differentiate both sidesddx

x2 + y3 − 2y = ddx(3)

with respect to x : 2x + 3y2y (x)− 2y (x) = 0

3y2y (x)− 2y (x) = −2xSolve for y (x) :

y (x) = −2x3y2 − 2

Find the slope of the

tangent line at (2, 1) : y (2) = −43− 2 = −4

Therefore the equation of the tangent line is

y − 1 = −4 (x − 2)

Now try the following exercises:

75. Find y (x) for x2y2 − 2x = 4− 4y. Then find the slope of the tangent line at the point (2,−2) .76. Determine the derivative of tan−1 x .

77. Van der Waal’s equation for a specific gas is P + 5V 2

(V − 0.03) = 9.7.

Using the volume V as a function of pressure, use implicit differentiation to find the derivativedVdP

at thepoint (2,−2).

78. Find the equations of the tangent and the normal lines to the curve of

2xy + π sin y = 2π at the point 1,π

2.

79. Finddydx

if tan xy2 + 3y = 2xy.

80. Finddydx

if y2x2 + y + cos (xy) = 0.

81. Finddydx

if sin (x + y) = y2 cos x .82. Find the equations for the tangent and the normal lines to the curve of

x sin (2y) = y cos (2x)

at the point π4 ,

π2 .

68

Solutions:

75.

ddx

x2y2 − 2x = ddx(4− 4y)

⇒ 2xy2 + x2 2ydydx

− 2 = 0− 4dydx

⇒ 2x2y + 4 dydx

= 2− 2xy2

⇒ dydx

= 2− 2xy22x2y + 4

∴ dydx (2,−2)

= mT = 2− 2 (2) (−2)22 (−2) (2)2 + 4 =

−14−12 =

76

The equation of the tangent line is

y − (−2) = 76(x − 2)

that is y + 2 = 76x − 14

6

that is y = 76x − 14

6− 2

= 76x − 14

6− 126

= 76x − 26

6

= 76x − 13

3.

76. The derivative of tan−1 x :

Note: Since tan x is differentiable, tan−1 x is also differentiable. To find its derivative, let y = tan−1 x . Thentan y = tan tan−1 x ⇒ tan y = x.

Differentiate this equation implicitly with respect to x to get:

sec2 ydydx

= 1.

Thendydx

= 1sec2 y

= 11+ tan2 x =

11+ x2 .

Thereforeddx

tan−1 x = 11+ x2 .

69 MAT1512/1

77.

P + 5V 2 (V − 0.03) = 9.7

⇒ PV − 0.03P + 5V −

0.15V 2

= 9.7

⇒ ddP

PV − 0.03P + 5V−1 − 0.15V−2 = 0

⇒ 1V + P dVdP

− 0.03+ 5 (−1) V−2 dVdP

+ 2 · (0.15) V−3 dVdP

= 0

⇒ P − 5V 2+ 0.3V 3

dVdP

= 0.03− V

⇒ dVdP

= 0.03− VP − 5

V 2+ 0.3V 3

∴ dVdP (2,−2)

= 0.03− (−2)2− 5

(−2)2 +0.3(−2)3

= 0.03+ 22− 5

4− 0.38

= 203/10057/80

= 812150

.

78. 2xy + π sin y = 2πDifferentiate both sides:

2y + 2x dydx + π cos ydydx

= 0

⇒ (2x + π cos y)dydx

= −2y

⇒ dydx

= −2y2x + π cos y .

At (1, π2 ):dydx

= −2 · π22+ π cos π2

= −π2.

Equation of the tangent line:

y − π2 = −π

2 (x − 1)⇒ y = −π2 x + π .

Equation of the normal line:

y − π2 = 2

π(x − 1)⇒ y = 2

πx − 2

π+ π

2 .

70

79. tan xy2 + 3y = 2xyDifferentiate both sides:

sec2(xy2)[y2 + 2xy dydx]+ 3dy

dx= 2y + 2x dydx

⇒ 2xy sec2(xy2)− 2x + 3 dydx

= 2y − y2 sec2(xy2)

⇒ dydx

= 2y − y2 sec2(xy2)2xy sec2(xy2)− 2x + 3 .

80. y2x2 + y + cos (xy) = 0Differentiate both sides:

2ydydxx2 + 2xy2 + dy

dx− sin(xy) y + x dy

dx= 0

⇒ 2yx2 + 1− x sin(xy) dydx

= −2xy2 + y sin(xy)

⇒ dydx

= y sin(xy)− 2xy22x2y + 1− x sin(xy) .

81. sin (x + y) = y2 cos xDifferentiate on both sides:

cos(x + y) · 1+ dydx

= 2y dydx cos x + y2(− sin x)

⇒ cos(x + y)+ cos(x + y)dydx

= 2y dydx cos x − y2 sin x

⇒ dydx

cos(x + y)− 2y cos x = −y2 sin x − cos(x + y)

⇒ dydx

= −y2 sin x − cos(x + y)cos(x + y)− 2y cos x .

82. x sin (2y) = y cos (2x)Differentiate both sides:

sin 2y + x cos 2y dydx

= dydxcos 2x − 2y sin 2x

⇒ dydx

2x cos 2y − cos 2x = −2y sin 2x − sin 2y

⇒ dydx

= 2y sin 2x + sin 2ycos 2x − 2x cos 2y .

At the point (π4 ,π2 ) the slope is:

dydx

= π sin(π2 )+ sinπcos(π2 )− 2π

4 cosπ= π

2π4= 2.

Therefore the equation of the tangent at the point (π4 ,π2 ) is:

y − π2

= 2(x − π4)

⇒ y = 2x,

71 MAT1512/1

and the equation of the normal line is:

y = −12x + 5π

8.

VII. Tangents and normal lines

From your high-school mathematics, you should know that differentiating the equation of a curve gives you aformula for the gradient (m) of the curve. The gradient of a curve at a point is equal to the gradient of the tangent atthat point. For example, in order to find the equation of the tangent to the curve y = x3 at the point (2, 8), you firsthave to determine the derivative: m = y = 3x2.

The gradient of the tangent at the point (2, 8) when x = 2 is 3 (2)2 = 12.The equation for the tangent line is: y = 12x − 16 since y

x= 8− y2− x = 12 .

The normal to the curve is the line which is perpendicular (at right angles) to the tangent to the curve at that point.If two lines are perpendicular, then the product of their gradients is−1. So if the gradient of the tangent at the point(2, 8) of the curve y = x3 is 12, then the gradient of the normal is − 1

12, since − 1

12× 12 = −1. Now try to work

through the following problems involving tangents and normal lines.

Examples:

83. Two curves are said to be orthogonal at a point where the curves intersect if their tangents are perpendicular toeach other at the point of intersection. Show that the curves y = sin(2x) and y = − sin(x/2) are orthogonalat the origin.

84. Find the x- and y-intercepts of the line that is tangent to the curve y = x3 at the point (−2,−8).85. Does the graph of the function

y = 2x + sin x

have any horizontal tangents in the interval 0 ≤ x ≤ 2π? If so, where? If not, why not?86. Find equations for the lines that are tangent and normal to the curve f (x) = 1+ cos x at the point (π/2, 1).87. (a) Find the equation for the tangent to the graph of

h(x) = 4(1+ x)2

which makes an angle of 45◦ with the x-axis.

[NBAssume that the scales along the x- and y-axes are the same. Angles are measured counterclockwisefrom the positive x-axis.]

(b) What is the equation of the normal to the curve of h at the point of tangency for the tangent consideredin (a) above?

88. Find the points on the curve y = 2x3 − 3x2 − 12x + 20 where the tangent is parallel to the x-axis. What arethe equations for these tangents?

72

89. Determine the equation for the normal to the curve of

f (x) = 1x2 + 1

at the point x = 1.90. Determine the equations for the tangents to the curve of y = x3 − 6x + 2 which are parallel to the line

y = 6x − 2.91. The line normal to the curve y = x2 + 2x − 3 at (1, 0) intersects the curve at what other point?92. What are the equations for the tangents to the curve f (x) = x2 + 5x + 9 which pass through the origin?

Solutions:

83.

y = sin(2x)

⇒ dydx

= (cos(2x))ddx(2x)

= 2 cos(2x),

sodydx x=0

= 2× 1 = 2.

Also,

y = − sin(x2)

⇒ dydx

= − cos(x2)ddx

xx

= −12cos(

x2), 2

sodydx x=0

= −12× 1 = −1

2.

Since −12(2) = −1, the given curves are orthogonal at the origin.

84. Letf (x) = x3.

Thenf (x) = 3x2,

so the gradient at the point (−2,−8) is f (−2) , that is 12. The equation of the tangent at the point (−2,−8)is therefore given by

y − (−8) = 12(x − (−2))

73 MAT1512/1

or

y = 12x + 16= 4(3x + 4).

The x-intercept of this tangent is −43and its y-intercept is 16.

85. Letf (x) = 2x + sin x .

Thenf (x) = 2+ cos x .

Now the gradient of a horizontal line is 0, so if the graph of f is to have a horizontal tangent at the point(x, f (x)), then the value of f (x) must be 0. We must therefore solve the equation

f (x) = 0

for all values of x occurring in the interval 0 ≤ x ≤ 2π . But if

2+ cos x = 0

thencos x = −2,

which is impossible. Hence the graph of f has no horizontal tangents.

86.

f (x) = 1+ cos x⇒ f (x) = − sin x.

Hence the gradient at the pointπ

2, 1 is f

π

2, that is−1. The equation of the tangent at the point π

2, 1

is thereforey − 1 = −1 x − π

2

ory = −x + π

2+ 1.

The gradient of the normal at the pointπ

2, 1 is 1. The equation for the normal at the point

π

2, 1 is

thereforey − 1 = 1 x − π

2

ory = x − π

2+ 1.

74

87. (a) The tangent which makes an angle of 45◦ with the x-axis has a gradient of 1. To find the x-coordinateof the corresponding point of tangency we let

h (x) = 1

and solve for x . Now

h(x) = 4(1+ x)−2⇒ h (x) = −8(1+ x)−3,

and

− 8(1+ x)3 = 1

⇔ x = −3.

We have thath(−3) = 4

(1− 3)2 = 1.

The tangent thus passes through the point (−3, 1) and has a gradient of 1. Its equation isy = 1.(x + 3)+ 1

that isy = x + 4.

(b) We obtain the equation for the required normal:

y = −1.(x + 3)+ 1

that isy = −x − 2.

88.

y = 2x3 − 3x2 − 12x + 20⇒ y = 6x2 − 6x − 12.

Now

6x2 − 6x − 12 = 0

⇒ 6(x2 − x − 2) = 0

⇒ 6(x − 2)(x + 1) = 0

⇒ x = 2 or x = −1.

If x = 2 then y = 2× 8− 3× 4− 24+ 20 = 0.If x = −1 then y = −2− 3+ 12+ 20 = 27.The required points are (2, 0) and (−1, 27). The equations for the tangents are y = 0 and y = 27 respectively.

75 MAT1512/1

89.

f (x) = 1x2 + 1 = x2 + 1 −1

⇒ f (x) = −(x2 + 1)−2 · 2x.∴ f (1) = −1

2,

so the gradient of the normal at x = 1 is 2. Now

f (1) = 12,

so the equation for the normal at x = 1 is

y − 12= 2(x − 1).

90. The slope of the tangent to the curve of y = x3 − 6x + 2 is given by y = dydx. Now

y = 3x2 − 6.

Since the tangent must be parallel to y = 6x−2,we must have that y is equal to the slope of y = 6x−2, thatis

y = 6⇔ 3x2 − 6 = 6⇔ 3x2 − 12 = 0⇔ x2 − 4 = 0⇔ x = ±2.

The corresponding y-values are

y = (−2)3 − 6(−2)+ 2= −8+ 12+ 2 = 6

for x = −2, and

y = (2)3 − 6(2)+ 2= 8− 12+ 2 = −2

for x = 2.

Therefore the tangent points are (−2, 6) and (2,−2).

The equation for a line with slope m which passes through (xo, yo) is given by

y − yo = m(x − xo).

76

The tangent through (−2, 6) therefore has the equation

y − 6 = 6(x − (−2))= 6x + 12

that isy = 6x + 18.

The tangent through (2,−2) has the equation

y − (−2) = 6(x − 2)

that isy + 2 = 6x − 12

that isy = 6x − 14.

91. The normal to the curve y = x2 + 2x − 3 at the point (1, 0) is perpendicular to the tangent at that point.Hence, if we know the gradient of the tangent we can calculate the gradient of the normal. Now if

f (x) = x2 + 2x − 3

thenf (x) = 2x + 2.

Thusf (1) = 4,

which is the gradient of the tangent. Hence the gradient of the normal is−14. Since the normal passes through

the point (1, 0), its equation is

y − 0 = −14(x − 1)

that isy = −1

4x + 1

4.

To find the points of intersection of the curve y = x2 + 2x − 3 and the line y = −14x + 1

4, let

x2 + 2x − 3 = −14x + 1

4.

Then4x2 + 8x − 12 = −x + 1

so4x2 + 9x − 13 = 0,

77 MAT1512/1

that is(x − 1)(4x + 13) = 0.

Hence x = 1 or x = −134. If x = −13

4then

−14x + 1

4= 1316+ 14= 1716.

The other point of intersection of the normal and the curve is therefore −134,1716

.

92. The equation of the tangent at a point (α, f (α)) is

y − f (α) = f (α)(x − α),

that isy = f (α) · x − α · f (α)+ f (α).

The tangent will pass through the origin only if the y–intercept is 0, that is

f (α)− α · f (α) = 0,

and then the equation of the tangent becomes

y = f (α) · x . ...(∗)Hence the solutions of the equation

f (x)− x · f (x) = 0

will give the x-coordinates α of the points of contact between the tangents and the curve, and then the equa-tions of the corresponding tangents can be determined from (∗).

Now

f (x) = x2 + 5x + 9⇒ f (x) = 2x + 5;

and

f (x)− x · f (x) = 0

⇒ x2 + 5x + 9− x(2x + 5) = 9− x2 = 0⇒ x = 3 or x = −3.

We have that f (3) = 11 and f (−3) = −1. Hence the required tangents have the formulae y = 11x andy = −x .

78

VIII. The Mean Value Theorem

In this section we discuss the Mean Value Theorem. Please read carefully through sections 2.9 and 4.4 of chapter2 in Smith. We would advise you to work through the examples in order to become familiar with how the MeanValue Theorem is applied. In section 2.9 you have to understand theorem 9.1 and theorem 9.4. You have to be ableto apply these theorems. Please work through examples 9.1 to 9.3 in this section.

The Mean Value Theorem is one of the most important tools in calculus. It states that if f (x) is defined andcontinuous on the interval [a, b] and differentiable on (a, b), then there is at least one number c in the interval (a, b)(that is a < c < b) such that

f (c) = f (b)− f (a)b − a .

The special case when f (a) = f (b) is known as Rolle’s Theorem. In that case, we have f (c) = 0. In other words,there exists a point in the interval (a, b) which has a horizontal tangent. In fact, the Mean Value Theorem can bestated in terms of slopes.

The numberf (b)− f (a)b − a

is the slope of the line passing through (a, f (a)) and (b, f (b)), (see Figure 4).

Figure 4

cb

y

xa

Example:

Let f (x) = 1x, a = −1 and b = 1. We have f (b)− f (a)

b − a = 11− −1

1= 22= 1.

On the other hand, for any c ∈ [−1, 1] , not equal to 0, we have f (c) = − 1c2= 1.

So the equation f (c) = f (b)− f (a)b − a does not have a solution for c. This does not contradict the Mean Value

Theorem since f (x) is not continuous on [−1, 1].

79 MAT1512/1

Note: The derivative of a constant function is 0. You may wonder whether a function with a derivative of zero isconstant – the answer is yes. To show this, we let f (x) be a differentiable function on an interval I, with f (x) = 0,for every x ∈ I . Then for a and b in I , the Mean Value Theorem implies that f (b)− f (a)

b − a = f (c) for somepoint c between a and b. So our assumption implies that f (b)− f (a) = 0. (b − a). Thus f (b) = f (a) for any aand b in I, which means that f (x) is a constant function.

Worked Examples:

For the following functions, find a value of c which complies with the Mean Value Theorem.

93. f (x) = x2 + 1, on the interval [−2, 2] .94. f (x) = x3 + x2, on the interval [0, 1] .95. f (x) = sin x , on the interval 0, π2 .

Solutions:

93. Since f (x) = x2+1 is a polynomial, f (x) is continuous on [−2, 2] and differentiable on (−2, 2). Accordingto the Mean Value Theorem, this means that there is a number c in (−2, 2) for which

f (c) = f (2)− f (−2)2− (−2) = 22 + 1 − (−2)2 + 1

4= 5− 5

4= 04= 0.

To find this number c, we obtain f (c) and then set f (c) = 0.f (c) = 2c and 2c = 0, so c = 0Notice that c = 0 ∈ [−2, 2] ∴ c = 0.

94. f (x) = x3 + x2 is a polynomial, continuous on [0, 1] and differentiable on (0, 1). According to the MeanValue Theorem, there is a number c in (0, 1) for which

f (c) = f (1)− f (0)1− (0) = 2− 0

1= 2.

But f (c) = 3c2 + 2c.So, f (c) = 3c2 + 2c = 2⇒ 3c2 + 2c − 2 = 0.Using the quadratic formula we get

c = −2± 22 − 4 (3) (−2)2 (3)

= −2±√28

6= −2± 2

√7

6

c = −1±√73

⇒ c = −1−√7

3or c = −1+

√7

3⇒ c ≈ −1.22 or c ≈ 0.55.

But since c ≈ −1.22 /∈ [0, 1] , we only accept the alternative c ≈ 0.55 that is

c = −1+√7

3.

80

95. f (x) = sin x, on the interval 0, π2 .

As f (x) = sin x is a trigonometric function, it means that f (x) is continuous on 0, π2 and differentiable on0, π2 . The conditions for the Mean Value Theorem hold, and so there exists c ∈ 0, π2 such that

f (c) = f π2 − f (0)π2 − 0

= sinπ2 − 2 sin (0)

π2

= 1− 0π2

= 1π2= 2π.

So c ≈ 0.88 ∈ 0,π

2.

But f (x) = cos x , f (c) = cos c and c has to be in the first quadrant. Thusc = cos−1 2

π≈ 0.88,

so c = 0.88 ∈ 0,π

2,

that is c = cos−1 2π

≈ 0.88.

81 MAT1512/1

Key points

You should by now have an understanding of the notion of differentiability and should also be able to compute thederivatives of different functions using basic differential formulas such as the power, product, quotient and chainrules. Think of differentiation as a process during which a function is “processed” to produce another function –the derivative. This process (differentiation) is carried out using different formulas and rules. We have introducedthe basic ones in this chapter. We have also demonstrated some techniques you can use to compute the derivativesof different algebraic and trigonometric functions. In addition to this, we have demonstrated how some theorems,such as the Mean Value Theorem, can be applied.


• compute the derivative of a function from first principles (using the definition of the derivative)• distinguish between the continuity and differentiability of a function• use the basic differentiation formulas to compute derivatives of algebraic and trigonometric functions• compute the derivatives of trigonometric functions and inverse trigonometric functions, as well as exponentialand logarithmic functions

• use the methods of logarithmic differentiation and implicit differentiation to find derivatives• solve problems involving tangents and normal lines and apply the Mean Value Theorem

Continue practising solving these problems until you have mastered the basic techniques! Go through the section“For your review” at the end of each chapter to consolidate what you have learnt. Also use other textbooks and theinternet.

82

CHAPTER 3

INTEGRATION

1. Background

In principle, calculus deals with two geometric problems: the one is to find the tangent line to a curve and the otheris one to find the area of a region under a curve. Both these problems are limiting processes. The first one is calleddifferentiation and the second process is integration. In this chapter we turn to the latter process.

To find the area of the region under the graph of a positive continuous function f defined on an interval [a, b] wesubdivide the interval [a, b] into a finite number of subintervals, say n, the kth subinterval having length xk , andthen we consider the sum of the form n

k=1 f (tk) xk , where tk is some point in the kth subinterval. This sum is anapproximation of the area that is arrived at by adding up n rectangles. Making the subdivisions finer and finer, thissum will tend to a limit as we let n → ∞ and, roughly speaking, this limit is the definition of the definite integralba f (x)dx (Riemann’s definition).

The two concepts “derivative” and “integral”, arise in entirely different ways and it is a remarkable fact indeedthat the two are intimately connected. We will show that differentiation and integration are, in a sense, inverseoperations.

There is a connection between differential calculus and integral calculus. This connection is called the FundamentalTheorem of Calculus and we shall see in this chapter that it greatly simplifies the solution of many problems.



• show that you understand the notion of the antiderivative by finding the antiderivative of basic algebraic,trigonometric, exponential and logarithmic functions,

• use the Fundamental Theorem of Calculus to find the derivatives of functions of the form F (x) = g(x)a f (t) dt

83 MAT1512/1

• evaluate definite integrals and use them to determine the area between a curve and the x-axis, and the areabetween curves

• use substitution or term-by-term integration techniques to integrate basic algebraic, trigonometric, exponentialand logarithmic functions

• solve problems involving the Mean Value Theorem for definite integrals



The overview of chapter 4 of Smith on pages 344 to 428 – especially sections 4.1, 4.4, 4.5 and 4.6, as well as chapter5, section 5.1 and chapter 6, section 6.1.

4. Worked examples

Our collection of worked examples of the work covered in chapter 4 and sections 5.1 and 6.1 of chapters 5 and 6 ofSmith can be divided into eleven sets (with some overlap), namely:

I. Antiderivatives

II. The definite integral and the Fundamental Theorem of Calculus – Part I

III. The definite integral and the area between the curve and the x-axis

IV. The definite integral and the area under the curve

V. The Mean Value Theorem for definite integrals

VI. The Fundamental Theorem of Calculus – Part II

VII. Integration in general

VIII. Indefinite integrals

IX. Integration by substitution

X. Integration of exponential and logarithmic functions

XI. Review of formulas and techniques of integration

Attempt the problems appearing immediately after each set once you have studied the relevant parts of Smith anddone some of the exercises there.

The table below (Table 4) shows how these worked examples and their solutions are organised.

84

Table 4

Topic(s) Section in Study guideSmith & Minton examples

I. Antiderivatives Section 4.1 1–10II. The definite integral and the Section 4.5 11–14

Fundamental Theorem of pages 383–386Calculus – Part I

III. The definite integral and the area Section 5.1 15–17between the curve and the x-axis

IV. The definite integral and the Section 5.1 18–21area under the curve Section 4.3 (read only)

V. The Mean Value Theorem for Section 4.4 22–23definite integrals page 379

VI. The Fundamental Theorem of Section 4.5 24–26Calculus – Part II pages 387–388 Exercises 27–30

VII. Integration in general Chapter 4 & 6, Section 6.1 11–52VIII. Indefinite integrals Section 4.1, pages 346–351 31–34IX. Integration by substitution Section 4.6 35–37X. Integration of exponential and Section 4.6, Example 6.11 38–39; 43,45,51

logarithmic functions Section 4.8XI. Review of formulas and Section 6.1 40–52

techniques of integration Examples 1.1–1.5pages 510–513

I. Antiderivatives

If we have been given a function, we can find its derivative. But many mathematical problems and their applicationsrequire us to solve the inverse (the reverse) of the derivative problem. This means that, given a function f , youmay be required to find a function F of which the derivative is f . If such a function F exists, it is called theantiderivative.The process of undoing differentiation (in other words, the reverse of differentiation) is known as integration. Theantiderivative is known as an integral.

Find the general antiderivatives of the following functions:

1. x−3 + x2

2. 12 6√x

3. x 23 + 2x− 13

4. θ + sec θ tan θ

85 MAT1512/1

5. −32cosec2

3θ2

6. cos πx2 + π cos xFind the function f if f is given as follows:

7. f (x) = sin x − 5√x2

8. f (x) = 3 cos x + 5 sin x

9. f (t) = sin 4t − 2√t

10. f (t) = 7√t4 + t−6

Solutions:

1. x−3 + x2General antiderivative:

−12x−2 + x

3

3+ c

by Rule 4 on page 255 in Smith (the power rule).

2. 12 6√x = 1

2x− 16

General antiderivative:

12· 65x56 + c (power rule)

= 35x56 + c.

3. x 23 + 2x− 13General antiderivative:

35x53 + 2.3

2x23 + c (power rule)

= 35x53 + 3x 23 + c.

4. θ + sec θ tan θGeneral antiderivative:

12θ2 + sec θ + c.

5. −32cosec2

32θ

General antiderivative:cot32θ + c.

86

6. cos π2 x + π cos xGeneral antiderivative:

2πsinπ

2x + π sin x + c.

7.

f (x) = sin x − 5√x2= sin x − x 25

⇒ f (x) = − cos x − 57x75 + c.

8.

f (x) = 3 cos x + 5 sin x⇒ f (x) = 3 sin x − 5 cos x + c.

9.

f (t) = sin 4t − 2√t = sin 4t − 2t 12⇒ f (t) = −1

4cos 4t − 2 · 2

3t32 + c

= −14cos 4t − 4

3t32 + c.

10.

f (t) = 7√t4 + t−6 = t 47 + t−6

⇒ f (t) = 711t117 − 1

5t−5 + c.

II. The definite integral and the Fundamental Theorem of Calculus – Part I

The Fundamental Theorem of Calculus, Part I is stated as follows:

If f is continuous on [a, b] and F (x) is any antiderivative of f (x) , thenba f (x) dx = F (b)− F (a)

Examples:

Compute the following integrals using the Fundamental Theorem of Calculus:

11.2

0x2 − 2x dx = 1

3x3 − x2

2

0

= 83− 4 = −4

3.

87 MAT1512/1

12.4

1

√x − 1

x2dx =

4

1x12 − x−2dx

= 23x32 + x−1

4

1= 2

3(4)

32 + 4−1 − 2

3+ 1−1

= 163+ 14− 23− 1 = 47

12.

13.4

0e−2xdx

= −12e−2x

4

0= −1

2e8 − −1

2e0

= −12e−8 + 1

2.

Note that in the example above, the integral of an e function is just the e function with its index divided bythe derivative of its index.

14.−1

−32xdx

= 2 ln |x |−1−3 = 2 (ln |−1|− ln |−3|)= 2 (ln 1− ln 3) = −2 ln 3.

III. The definite integral and the area between the curve and the x-axis

It is important to understand that the definite integral of a continuous function f evaluates the area between thisfunction f and the x-axis. See the following examples:

Examples:

15. Find the area under the curve f (x) = sin x on the interval [0,π] if sin x ≥ 0.

Solution:Area =

π

0sin xdx = (cosπ)− (cos 0) = (−1)− (−1) = 2.

16. [The connection between area and calculus]

Calculate the value of 52 (2x − 4) dx geometrically by using a suitable sketch. Then test your answer with

the aid of integration.

88

Solution:

(a) Geometrically:

The graph of y = 2x − 4 looks like this:y

x

B

A

(0, 4)_

(2, 0) (5, 0)

(5, 6)

y = 2 x _ 4

Therefore5

2(2x − 4)dx = area of triangle A

= 1/2× base × height

= 1/2× 3× 6= 9.

(b) Test with the aid of integration:

F(x) = x2 − 4x is an antiderivative of y = 2x − 4 and hence5

2(2x − 4)dx = F(5)− F(2)

= (25− 20)− (4− 8)= 9,

which is the same as in (a) above.

89 MAT1512/1

17. Find the area of the region between the x-axis and the graph of f (x) = 2x + x2 − x3, −1 ≤ x ≤ 2.

Solution:

_ 1

y = f ( x)

y

x210

First find the zeros of f . Since

f (x) = 2x + x2 − x3 = −x(−2− x + x2) = −x(x + 1)(x − 2),

the zeros are x = 0,−1, and 2. It is clear that the zeros partition the interval [−1, 2] into two subintervals:[−1, 0]and [0, 2] with f (x) ≤ 0 for all x ∈ [−1, 0] and f (x) ≥ 0 for all x ∈ [0, 2]. We integrate over each subintervaland add the absolute values of the calculated values.

Integral over [−1, 0]:0

−1(2x + x2 − x3)dx = x2 + x

3

3− x

4

40

−1

= 0− (−1)2 + (−1)3

3− (−1)

4

4

= − 512.

Integral over [0, 2]:

2

0(2x + x2 − x3)dx = x2 + x

3

3− x

4

42

0

= (2)2 + (2)3

3− (2)

4

4− 0

= [4+ 83− 4]

= 83.

Enclosed area:Total enclosed area = −5

12 + 83 = 37

12 .

90

IV. The definite integral and the area under the curve

If f and g are continuous functions with f (x) ≥ g (x) throughout [a, b], then the area of the region between thecurves of f and g from a to b is the integral of [ f − g] from a to b. The area bounded by the two curves y = f (x)and y = g (x) on the interval [a, b] where f and g are continuous, is written as follows:

A =b

af (x)− g (x) dx .

This is shown in the sketch below:

y = f ( )x

y = g( )x

y

x

ab

This is only valid if f (x) ≥ g (x) on [a, b] , that iswhen f (x) is above g (x) or f (x) is tothe right of g (x) .

Examples:

18. Evaluate the area between the curves f (x) = 5− x2 and g (x) = |x + 1|.

Solution:

5_ 1

5_

_ 2

y = f ( x)

y = g ( x)

5

y

xa = b = _

1+

2

17

1

91 MAT1512/1

Points of intersection:

−(x + 1) = 5− x2⇒ x2 − x − 6 = 0

⇒ (x − 3)(x + 2) = 0⇒ x = 3 or x = −2,

and

x + 1 = 5− x2⇒ x2 + x − 4 = 0

⇒ x = −1±√172

.

It is clear that the limits of integration are the x-values of the points of intersection of the two curves: a = −2 andb = −1+√17

2 . Now we evaluate the area of the region by integrating:

A =−1+√17

2

−2[5− x2 − |x + 1|]dx

=−1

−2[5− x2 − (−x − 1)]dx +

−1+√172

−1[5− x2 − (x + 1)]dx

=−1

−2[−x2 + x + 6]dx +

−1+√172

−1[−x2 − x + 4]dx

= − x3

3+ x

2

2+ 6x −1

−2 + − x3

3− x

2

2+ 4x −1+√17

2−1

= 13+ 12− 6− 8

3+ 2− 12 +

− −1+√17 3

24+− −1+√17 2

8+ 4

− −1+√172

− 13− 12− 4

= −83+ 9− (−1+

√17)3

24− (−1+

√17)2

8+ 2(−1+√17)

We would like to encourage you to leave your answers unsimplified, as you are not allowed to use a calculatorin your examinations!

19. Find the area bounded by the graphs of y = 3− x and y = x2 − 9.

Solution:

Find the points of intersection by equating the two functions:

3− x = x2 − 9⇒ 0 = x2 + x − 12 = (x − 3) (x + 4)

that is x = 3 or x = −4.

92

The two curves intersect at x = 3 and x = −4 (see figure below).

y = 3 _ x

y

x4_

3

y = x2 9_

The height of the bounded area is h (x) = (3− x)− x2 − 9 .

Therefore

A =3

−4(3− x)− x2 − 9 dx

=3

−4−x2 − x + 12 dx = −x

3

3− x

2

2+ 12x

3

−4

= −33

3− 3

2

2+ 12 (3) − −(−4)

3

3− (−4)

2

2+ 12 (−4)

= 3436.

20. Find the area bounded by the graphs of y = x2 and y = 2− x2 for 0 ≤ x ≤ 2.

Solution:

y = 2 _ x

y

x1 2

y = x2

2

2

93 MAT1512/1

Since the curves intersect in the middle of the interval, you have to compute two integrals, namely for 2− x2 ≥ x2and for x2 ≥ 2− x2.

To find the point of intersection, solve for x :

x2 = 2− x2 ⇒ 2x2 = 2⇒ x = ±1.

Note that only x = +1 is relevant since x = −1 is outside the given interval.

A =1

02− x2 − x2 dx +

2

1x2 − 2− x2 dx

=1

0(2− 2x) dx +

2

12x2 − 2 dx

= 2x − 2x3

3

1

0+ 2x3

3− 2x

2

1= 2− 2

3− (0− 0)+ 16

3− 4 − 2

3− 2

= 43+ 43+ 43= 4.

21. Find the area bounded by the graphs of x = y2 and x = 2− y2.

Solution:

y2x =

x = 2 _ y2

y

x1 2

2

1

1

2

_

_

It is easier to compute this area by integrating with respect to y. Therefore we find the points of intersection bysolving for y.

y2 = 2− y2 ⇒ y2 = 1⇒ y = ±1.

94

In the interval [−1, 1] ,

A =1

−12− y2 − y2 dy =

1

−12− 2y2 dy

= 2y − 23y3

1

−1= 2− 2

3− −2+ 2

3= 83.

V. The Mean Value Theorem for definite integrals

Now we state the Mean Value Theorem for definite integrals as follows (analogous to theorem 4.4 on p 379 inSmith):

If f is continuous on the interval [a, b] there is atleast one number c between a and b such that

the area under the curvebetween a and b

= ba f (x) dx = (b − a) f (c)

or f (c) = 1b−a

ba f (x) dx .

22. [Finding c using the Mean Value Theorem for integrals]

Find a value c guaranteed by the Mean Value Theorem for integrals for f (x) = x2 + 2x + 3 on [0, 2]

Solution:

2

0x2 + 2x + 3 dx = 1

3x3 + x2 + 3x

2

0

= 83+ 4+ 6

= 8+ 12+ 183

= 383.

95 MAT1512/1

The region bounded by f and the x-axis on [0, 2] is shaded in the next figure.

2_ 1_

2

4

6

8

10

12

14

1 2x

y

0

The Mean Value Theorem for integrals asserts the existence of a number c on [0, 2] such that f (c) (b − a) =383 . We solve this equation to find c:

f (c) (b − a) = 383

⇒ c2 + 2c + 3 (2− 0) = 383

⇒ 2c2 + 4c + 6 = 383

⇒ 6c2 + 12c − 20 = 0

∴ 3c2 + 6c − 10 = 0

∴ c = −6±√36+ 4 (3) (10)6

= −6±√1566

= −1±√393.

Since only one of these points, that is c = −1+√393 ≈ 1, 08, lies within the interval [0, 2] , this value of c satisfies

the Mean Value Theorem for integrals.

The following is the definition of the average value of a function:

If f is integrable on [a, b] , then the average valueor mean value of f on [a, b] denoted by av ( f ) is

av ( f ) = 1b−a

ba f (x) dx

96

23. [Finding the average value of a function on an interval]

Find the average value of f (x) = 2x on the interval [1, 5] .

Solution:The average value is

av ( f ) = 15− 1

5

12xdx = 1

4x2

5

1

= 14(25− 1)

= 6.

Obviously this average value occurs when f (x) = 6 and when x = 3.

VI. The Fundamental Theorem of Calculus – Part II

If f is continuous on [a, b] and F (x) = x0 f (t) dt, then F (x) = f (x) on [a, b].

In other words, if F (x) = xa f (t) dt where a is any real number, then

ddx(F (x)) = d

dx

x

af (t) dt

that is F (x) = f (x) ddx (x) = f (x) (1); therefore F (x) = f (x) .

To apply Part II of the Fundamental Theorem of Calculus, just leave the function f (t) as it is, but replace thevariable t with the variable x times the derivative of x .

Let us write down the two parts of the Fundamental Theorem of Calculus again:

Part I

If G(x) is any antiderivative of f (x) on [a, b] so that G (x) = f (x), thenb

af (x) dx = G (b)− G (a) .

Part II (the Fundamental Theorem of Calculus stated in another way)

Suppose that the function f is continuous on the interval [a, b].

Let the function F be defined on [a, b] by F (x) = xa f (t) dt.

Then F is differentiable on [a, b] and F (x) = f (x), which means that F is an antiderivative of f on [a, b] .

Expressed in symbols:ddx

x

af (t) dt = f (x) = F (x) .

97 MAT1512/1

Remarks:

(1) You should remember that both conclusions of the Fundamental Theorem are useful. Part II concerns thederivative of an integral – it tells you how to differentiate a definite integral with respect to its upper limit.Part I concerns the integral of a derivative – it tells you how to evaluate a definite integral if you can find anantiderivative of the integrand.

(2) Remember that differentiation and integration are inverse operations and cancel each other to a certain extent.

(3) Notice that the lower limit of the integral may be any constant in the interval [a, b] and it does not affect theanswer after differentiation. Thus

ddx

x

0f (t)dt = d

dx

x

−10f (t)dt = d

dx

x

513f (t)dt

etc. This is so because the derivative of a constant is always zero.

(4) Note that the upper limit must be the same as the variable with respect to which we differentiate, in other

words, if we determineddx(the integral) then the upper limit must be x (or a function of x).

(5) The dt in the integral shows you that integration is done with respect to the variable t . Hence you must changeall t’s in the given function f (t) to x’s to obtain f (x), or dF/dx .

(6) If the upper limit of the integral (which you must differentiate) does not consist of the variable of interestonly, but is in fact a function, then you must remember to use the chain rule for differentiation. Thus:

ddx

g(x)

af (t)dt = f (g(x)).g (x).

In this case you replace all t’s by g(x) and then multiply by the derivative of g(x) (look at examples 25 and26 below).

Keeping the above-mentioned in mind, consider the following examples:

Examples:

Determine the derivatives of F(x) in each of the following:

24. F(x) =x

a

11+ sin2 t dt

25. F(x) =x3

a

11+ sin2 t dt

26. F(x) =cos x

sin xt (5− t)dt

98

Solutions:

24. F(x) =x

a

11+ sin2 t dt

It follows from part Ie of the theorem that

F (x) = 11+ sin2 x .

25. F(x) =x3

a

11+ sin2 t dt

In this case we must apply the chain rule, because the upper limit is a function of x . Let f (t) = 11+sin2 t and

g(x) = x3. Then

F (x) = f (g (x)) · g (x)= f x3 · d

dxx3

= 11+ sin2 x3 · 3x2.

(In other words, substitute t in the function f (t) = 11+sin2 t with x

3, but also remember to multiply thederivative of x3 with this answer.)

26. In this example see that both the upper limit and lower limits are functions of x, so that it follows from remark3 above that we may introduce any constant in this case. We call it a.

F (x) =cos x

sin xt (5− t) dt.

=a

sin xt (5− t) dt +

cos x

at (5− t) dt

(from 10, table 5.1 in Smith)[from 5, table 5.3 in Smith]

= −sin x

at (5− t) dt +

cos x

at (5− t) dt

(from 2, table 5.1 in Smith)[from 2, table 5.3 in Smith].

By again applying the chain rule as in example 25, we obtain

F (x) = − sin x (5− sin x) · ddxsin x + cos x (5− cos x) · d

dxcos x

= − sin x cos x (5− sin x)− cos x sin x (5− cos x) .

Now attempt the following yourselves.

99 MAT1512/1

Find the derivatives of F in each of the following cases:

27. F(x) =10

xt + 1

tdt.

28. F(x) =x3

2xcos t dt .

29. F(x) =x3

3sin3 t dt.

30. F(x) =sin x

0

√1− t2dt.

Solution:

27. −x − 1x.

(This is plainly part II of the Fundamental Theorem of Calculus with10

xf (x) dx = −

x

10f (x) dx .)

28. 3x2 cos x3 − 2 cos 2x .

(Again wemust introduce constants for the lower limits, for example0

2x. . .+

x3

0. . . = −

2x

0. . .+

x3

0. . ..)

29. 3x2 sin3 x3.

(Use the chain rule, in other words first substitute t with x3 and then multiply the answer with the derivativeof x3.)

30. |cos x | cos x .

VII. Integration in general

The process of computing an integral (integration) is written as follows:

If F (x) = f (x) , then F (x) dx = f (x) dx,that is F (x) = f (x) dx .

f (x) is the integrand and the term dx identifies x as the variable of integration.

Given a function f , find, if possible, a function F such that:

F (x) = f (x)

A function F defined on an interval which has the property that F (x) = f (x) is called an indefinite integral orantiderivative of f on the same interval for all x, and we write:

100

F (x) = f (x) dx (for indefinite integrals)

or F (x) =b

af (x) dx (for definite integrals)

The term “dx” is part of the notation. It represents the statement “We integrate with respect to x”.

VIII. Indefinite integrals

For indefinite integrals, if F and G are both antiderivatives of f on the interval I, then G (x) = F (x)+ c for someconstant c.

If F is any antiderivative of f , the indefinite integral of f (x)with respect to x is defined by f (x) dx = F (x)+c(where c is the constant of integration).

Example:

27. Evaluate t5dt.

Solution:

We know thatddtt6 = 6t5, so d

dt16t6 = t5. Therefore t5dt = 1

6t6 + c.

xndx = xn+1

n + 1 + c sinceddx

1n + 1 x

n+1 + c = xn.

However, we could also have xndx = xn+1

n + 1 + 5, becauseddx

xn+1

n + 1 + 5 = xn.

So, in general, xndx = xn+1

n + 1 + c, where c is the constant of integration.

Note: This rule does not work for n = −1 since it would produce a division by 0.

Examples of indefinite integrals:

31.

x17dx

= x17+1

17+ 1 + c =x18

18+ c.

32.13√xdx

= x−13 dx = x− 13+1

− 13 + 1

+ c

= x 2323+ c = 3

2x23 + c.

101 MAT1512/1

33.

3 cos x + 4x8 dx

= 3 cos xdx + 4 x8dx

= 3 sin x + 4 x9

9+ c = 3 sin x + 4

9x9 + c.

34.

3ex − 21+ x2 dx

= 3 exdx − 2 11+ x2 dx

= 3ex − 2 tan−1 x + c.

IX. Integration by substitution

(a). Indefinite integralsThe method we are going to discuss now is the principal method by which integrals are evaluated.When the integrand is not in standard form, first of all determine whether the integrand does not consist of twofactors, one of which is the derivative of the other. The method of substitution is the integration version of the chainrule.

Remember that, according to the chain rule, the derivative of 3x2 − 5x + 2 4 is

ddx

3x2 − 5x + 2 4 = 4 3x2 − 5x + 2 3(6x − 5) .

On the right-hand side we now have two expressions, namely

3x2 − 5x + 2 and 6x − 5, which are such thatddx

3x2 − 5x + 2 = 6x − 5.

Since we know that differentiation and integration are inverse operations, we see that

4 3x2 − 5x + 2 3 · (6x − 5) dx = 3x2 − 5x + 2 4 + c.

Example:

Integration of an indefinite integral by substituion

Find7 x2 + 3x + 5 6

(2x + 3) dx .Solution:

102

Look at the problem and make the observation as shown in the boxes:

7 x2 + 3x + 5 6(2x + 3) dx

The derivative of the function in the left-hand boxwhich is (2x + 3) dx is sitting in the right-hand box.

So let u = x2 + 3x + 5.

(NB: Note that the function u appears without the index 6.)

Thendudx

= 2x + 3, that is (2x + 3) dx = du.

(This is the form in the right-hand box.)

Now substitute everything under the integral sign to express the integral in terms of u. We have

7 u6du = 7 17u7 + c = u7 + c.

Then back-substitute to express u in terms of x:

7 u6du = x2 + 3x + 5 7 + c.Remarks:

1. Sometimes the two functions under the integral sign are linked, but a constant might be missing.

If, however, you follow the procedure of substitution correctly, the end product will be correct.

Example:

Findx2 + 2x + 5 4

(x + 1) dx

Solution:

Letu = x2 + 2x + 5.

Thendudx

= 2x + 2 = 2 (x + 1) .

Thus (x + 1) dx = du2.

Substitution then gives

u4du2

= 12

u4du

= 110u5 + c

= 110

x2 + 2x + 5 5 + c.

103 MAT1512/1

If you differentiate this using the chain rule, you will obtain the function under the integral sign.

2. After you have done your substitution and simplified, there should be no left-over x-values in the integrand.

Example:

Substitution with leftover x-values

Findx (3x − 5)3 dx

Solution:

In this case it is so that the derivative of the function with index 3, that is (3x − 5) , is not linked to the otherfunction in the ordinary way. However, we follow the same method.

Let u = 3x − 5, so du = 3dx and dx = du3 .

Now substitute:x (3x − 5)3 dx = xu3

du3= 13

xu3du

It is clear that there is a left-over x-value.

We now follow the next procedure to eliminate the left-over x-term:

Because u = 3x − 5, we can solve for x :x = u + 5

3,

so13

xu3du = 13

u + 53

u3du

= 19

u4 + 5u3 du

= 19u5

5+ 54u4 + c

= 145(3x − 5)5 + 5

36(3x − 5)4 + c.

Examples:

Determine the following integrals:

35. 3x√1− 2x2 dx

36. cos3 x sin5 x dx

37.dx

x2 1+ 1x3 .

104

Solutions:

35. We see that x is the derivative (except for a constant factor) of 1− 2x2. Therefore set

u = 1− 2x2.

Thendu = −4x dx .

In other words−14du = x dx .

Now

3x 1− 2x2dx = 3 1− 2x2 x dx

= 3√u −1

4du

= −34

u12 du

= −34u 323/2

+ c

= −12u32 + c

= −121− 2x2 3

2 + c.

Note: You must not give your answer in terms of u. You must give your answer in terms of x , since that wasthe original variable!

36. In this case we use trigonometric identities. We do this with two different methods.

Method 1cos3 x sin5 x dx

= cos x cos2 x sin5 x dx .

The part in the block, without its power, has the derivative cos x .

Now we must also rewrite the other extra term in terms of sin x, that is cos2 x = 1− sin2 x .

The integral then becomes

cos x 1− sin2 x sin5 x dx

= cos x sin5 x dx − cos x sin7 x dx .

Since cos x is the derivative of sin x we let u = sin x .

105 MAT1512/1

Thendu = cos x dx

and so it follows that

cos3 x sin5 x dx = u5 − u7 du

= u6

6− u

8

8+ c

= sin6 x6

− sin8 x8

+ c.

Method 2cos3 x sin5 x dx

= (cos x)3 sin4 x sin x dx .

The part in the block, without the power, has the derivative (− sin x) (thus the constant (−1) is missing). Nowrewrite sin4 x in terms of cos x :

sin4 x = 1− cos2 x 2 = 1− 2 cos2 x + cos4 x .

Thus

cos3 x sin5 x dx = (cos x)3 1− 2 cos2 x + cos4 x sin x dx

= cos3 x sin x dx − 2 cos5 x sin x dx + cos7 x sin x dx.

Put u = cos x, so that du = (− sin x) dx .

Then

cos3 x sin5 x dx = − u3du + 2 u5du − u7du

= −14u4 + 2

6u6 − 1

8u8 + c

= −14(cos x)4 + 1

3(cos x)6 − 1

8(cos x)8 + c.

37. In this case we see that − 1x2 is the derivative of 1+ 1

x and hence we set

u = 1+ 1x,

and thereforedu = − 1

x2dx

106

and

dx

x2 1+ 1x3 = − du

u3

= −u−2

−2 + c

= 12

1

1+ 1x2 + c

= 12

xx + 1

2

+ c.

(b) Definite integrals

The section on changing the limits of integration on pages 398 to 399 in Smith is very important and students tendto have problems with this. We shall discuss it in detail and show you three different methods you may use. Workthoroughly through all three and make very sure that you understand each step.

Example:

Calculate1

0

x3

x2 + 1 32dx .

Solution:

Method 1

When you have done substitution with the variable u, also change the limits of integration in terms of the variableu, and do not change back to the variable x .

1

0

x3

x2 + 1 32dx :

Letu = (x2 + 1) 12 .

Thenu2 = x2 + 1

and2udu = 2xdx, that is u du = x dx

Now we also change the limits of integration:

Ifx = 0 then u = (02 + 1) 12 = 1

and ifx = 1 then u = (12 + 1) 12 = √2.

107 MAT1512/1

Hence,

1

0

x3

x2 + 1 32dx =

1

0

x2x dx

x2 + 1 32

=√2

1

u2 − 1u3

u du

=√2

11− 1

u2du

= u + 1u

√2

1.

Now substitute the values of the limits directly and do not change the variable back to x :

1

0

x3

x2 + 1 32dx = √

2+ 1√2− (1+ 1)

= √2+

√22− 2

= 32√2− 2.

Method 2

With this method we first determine the indefinite integral, change the answer back to the variable x after substitu-tion, and then put in the values of the limits of integration in terms of the variable x .

We first determinex3

x2 + 1 32dx .

As before, we let u = (x2 + 1) 12 . Then u du = x dx andx3

x2 + 1 32dx = u + 1

u+ c

= x2 + 1 12 + 1

x2 + 1 12+ c.

Now we calculate the definite integral in terms of x :

1

0

x3

x2 + 1 32dx = x2 + 1 1

2 + 1

x2 + 1 12

10

= √2+ 1√

2− (1+ 1)

= 32√2− 2.

108

Method 3

With this method we do not change the limits when we make the substitution, but we show clearly in terms of whichvariable the limits are given:

1

0

x3

x2 + 1 32dx

As before, we let u = (x2 + 1) 12 . Then u du = x dx and1

0

x3

x2 + 1 32dx =

x=1

x=0

u2 − 1u3

u du

= u + 1u

x=1

x=0

= x2 + 1 12 + 1

x2 + 1 12

10

= 32√2− 2.

Remarks:

1. You must decide for yourself which of these three methods you want to use. Please note the incorrectness ofwriting the following:

1

0

x3

x2 + 1 32dx =

1

01− 1

u2du

This is wrong, because the limits are not given in terms of the variable u.

The answer you obtain may be correct, but the method is totally wrong. In the examinations or an assignmentyou will get no marks for an answer like this.

2. There are only two general methods for integration: integration by parts (which is not part of this module)and integration by substitution. All other methods boil down to algebraic manipulation, followed by theapplication of one of the above-mentioned two methods. Keeping this in mind, you will realise the importanceof this section. Now you must try to do all the exercises in Exercise 4.6.

(c) Steps for Integration by substitution

1. Choose a new variable u (usually the innermost expression).

2. Computedudx.

3. Replace all terms in the original integrand with an expression involving u and du.

4. Evaluate the resulting (u) integral. (You may have to try a different u if the first one doesn’t work.)

5. Replace each occurrence of u in the antiderivative with the corresponding expression in x .

109 MAT1512/1

X. Integration of exponential and logarithmic functions

We state the following integration formulas again:

1. exdx = ex + c2. axdx = ax

ln a+ c

3.1xdx = ln |x| + c

which has the following forms when we work with functions f (x) .Following normal substitution, we have

1 e f (x) f (x) dx = e f (x) + c

2 a f (x) f (x) dx = a f (x)

ln a+ c

31f (x)

f (x) dx = ln | f (x)| + c.

Now we shall do one more example of integration by substitution to show you how the logarithmic function is usedin this case:

Example:

Determine the following integral:x2

1− 2x3 dx .

Note that this is of the form 3 above, where f (x) = 1− 2x3 and f (x) = −6x2. (Remember that the constant(−6) will only make a difference to the answer as seen in Remark 1 on page 102 of the student guide.)

Solution:

In this case we letu = 1− 2x3.

Thendu = −6x2dx .

In other words−16du = x2dx .

Then

x2

1− 2x3 dx = −16

duu

= −16ln |u| + c

= −16ln |1− 2x3| + c.

110

Remark:

When working with exponential and logarithmic functions, never approximate your answers, but please giveexact answers. Consider example 5.6 on page 53 of Smith. The answer to the problem is ln 7− 4. This is the exactanswer and if you were to use your pocket calculator to find an approximation of the answer, it would be WRONG.You need not use the pocket calculator at all when doing your assignments!

Example:

Find the following integrals:

38.sec xtan x

dx = ln |tan x | + x .

39.2xx2 + 1dx = ln x

2 + 1 + c. Note that:ddx

x2 + 1 = 2x

XI. Review of formulas and techniques of integration

Go through the rest of the examples carefully. Evaluate:

40. 3 cos 4xdx

41. sec 2x tan 2xdx

42. e3−2xdx

43.4

x 13 1+ x 23dx

44.sin√x√xdx

45.π

0cos xesin xdx

46.0

− π4

sin xcos2 x

dx

47.3

16+ x2 dx

48.x√1− x4 dx

49.1+ x1+ x2 dx

50.2

0f (x) dx, where f (x) =

xx2+1 if x ≤ 1,x

x2+1 if x > 1.

111 MAT1512/1

51.2

−2f (x) dx, where f (x) = xex2 if x < 0,

x2ex3 if x > 0.

52.2+√x3−√x dx

Solutions:

40. 3 cos 4xdx = 3 sin 4x 14+ c = 3

4sin 4x + c.

41. sec 2x tan 2xdx = sec 2x, 12+ c = 1

2sec 2x + c.

42. e3−2xdx = e3−2x , −12+ c = −1

2e3−2x + c.

43.4

x 13 1+ x 23dx :

Let u = 1+ x 23 then du = 23x− 13 dx = 2

3x 13dx ⇒ dx = 3x

13

2du.

So4

x 13 1+ x 23dx = 4 1

x 13 .u.3x 132du = 4 3

21udu

= 6 ln |u| + c = 6 ln 1+ x 23 + c.

44.sin√x√xdx :

Let u = √x then du = 12√xdx ⇒ 2

√xdu = 2udu = dx .

Sosin√x√xdx = sin u

u2udu = 2 sin udu

= 2 (− cos u)+ c = −2 cos√x + c.

45.π

0cos xesin xdx :

Let u = sin x then du = cos xdx ⇒ dx = ducos x

.

We have to change the limits, that is, if x = 0, u = sin 0 = 0 and if x = π , u = sinπ = 0.

Soπ

0

cos xesin xdx =0

0

cos xeu.ducos x

=0

0

eudu = [eu]00

= e0 − e0 = 1− 1 = 0.

46.0

− π4

sin xcos2 x

dx

112

=0

− π4

sec x tan xdx = sec x0

− π4

= sec (0)− sec −π4

= 1cos (0)

− 1cos −π

4= 11− 1

1√2

= 1−√2.

47.3

16+ x2 dx = 31

16+ x2 dx :

Recall:1

a2 + x2 dx =1atan−1

xa+ c

∴ 31

42 + x2 dx = 3.14tan−1

x4+ c

= 34tan−1

x4+ c.

48.x√1− x4 dx =

x

1− x2 2dx :

Let u = x2 then du = 2xdx ⇒ dx = du2x.

Sox√1− u2 .

du2x= 12

x√1− u2

dux= 12

du√1− u2

= 12sin−1 (u)+ c = 1

2sin−1 x2 + c.

49.1+ x1+ x2 dx =

11+ x2 dx +

x1+ x2 dx

= 11+ x2 dx +

x1+ x2 .

22dx Recall that

f (x)f (x)

dx = ln | f (x)| + c

= 11+ x2 dx +

12

2x1+ x2 dx

= tan−1 x + 12ln 1+ x2 + c.

50.2

0

f (x) dx, where f (x) =x

x2+1 if x ≤ 1,x

x2+1 if x > 1.

2

0

f (x) dx =1

0

xx2 + 1dx +

2

1

x2

x2 + 1dx .

113 MAT1512/1

Note: Rationalise the first integral by22and use long division on the second integral to obtain:

12

1

0

2xx2 + 1dx +

2

1

1− 1x2 + 1 dx

= 12

1

0

2xx2 + 1dx +

2

1

1dx −2

1

1x2 + 1dx

= 12ln x2 + 1 1

0 + x]21 − arctan x]21

= 12ln 2+ 1+ π

4− arctan 2.

51.2

−2f (x) dx, where f (x) = xex2 if x < 0,

x2ex3 if y > 0.

2

−2f (x) dx =

0

−2xex2dx +

2

0

x2ex3dx .

For the first integral,0

−2xex2dx :

Let u1 = x2. Thendu1 = 2xdx anddx = du1

2x.

For the second integral,2

0x2ex3dx :

Let u2 = x3.Then du2 = 3x2dx anddx = du2

3x2.

So we have

xex2dx + x2ex3dx

= xeu du12x + x2eu · du23x2 =12

eudu1 + 13 eudu2

= 12eu + c1 + 13e

u + c2.When considering the given definite integral, we ignore the constants c1 and c2 and get:0

−2xex2dx +

2

0x2ex3dx

= ex3

2

0

−2+ ex3

3

2

0

114

= 12ex2

0

−2+ 13ex3

2

0= 12e0 − 1

2e(−2)2 + 1

3e(2)3 − 1

3e(0)3

= 12(1)− 1

2e4 + 1

3e8 − 1

3e0 = 1

2− e

4

2+ e

8

3− 13(1)

= 1− e4

2+ e

8 − 13

.

52.2+√x3−√x dx

Solution:2+√x3−√x dx =

√x + 2

−√x + 3dxThis expression is an improper fraction, so we apply long division first.

−1−√x + 3 √

x + 2+√x − 3

5

So

2+√x3−√x dx = −1+ 5

−√x + 3 dx

= − 1dx + 5 1−√x + 3dx

Now find1

−√x + 3dx :Let u = −√x + 3⇒√

x = 3− u.Then

dudx

= − 12√xand

du = −12√x dx ⇒ dx = −2√xdu.

So dx = −2 (3− u) du,that is dx = (−6+ 2u) du.So

1−√x + 3dx =

1u(−6+ 2u) du

= −6 1udu + 2u

udu

= −6 1udu + 2du

= −6 ln |u| + 2u + c.Therefore,√

x + 2−√x + 3dx = − 1dx + 5 1

−√x + 3dx= −x + 5 (−6 ln |u| + 2u + c)= −x − 30 ln |u| + 10u + 5c= −x − 30 ln 3−√x + 10 3−√x + k, where C = 5c = k= x + 10 3−√x − 30 ln 3−√x + k.

115 MAT1512/1

Key points

You should by now have an understanding of the notion of integration as the reverse of differentiation. You shouldparticularly be able to link it to determining areas between different curves and the x-axis, and the area betweencurves. You should now be comfortable with calculating the integrals of several basic algebraic, trigonometric,exponential and logarithmic functions.


• show that you understand the notion of the antiderivative by finding the antiderivative of basic algebraic,trigonometric, exponential and logarithmic functions

• use the Fundamental Theorem of Calculus to find the derivatives of functions of the form F (x) =g(x)

a

f (t) dt

• evaluate definite integrals and use them to determine the areas between a curve and the x-axis, and the areabetween curves

• use substitution or term – by – term integration techniques to integrate basic algebraic, trigonometric, expo-nential and logarithmic functions

• solve problems involving the Mean Value Theorem for integrals

Continue practising solving these problems until you have mastered the basic techniques! Go through the “For yourReview” section at the end of each chapter to consolidate what you have learnt. Also use other textbooks and theinternet.

116

CHAPTER 4

FIRST–ORDERDIFFERENTIALEQUATIONS,GROWTH AND DECAY, AND PARTIALDERIVATIVES

1. Background

In this chapter we start by looking at first-order differential equations and then we move on to growth and decayproblems, followed by partial derivatives. In many natural processes, quantities grow and decay at a rate proportionalto their size. This is known as growth and decay. It has applications in chemistry, physics, biology, economics andmany other disciplines. In real life, physical quantities usually depend on two or more variables, so we shall alsoturn our attention to functions with several variables and extend the basic ideas of differential calculus to thesefunctions.



• solve first-order differential equations with initial values• solve basic real–life problems involving exponential growth and decay• determine the partial derivatives of functions with several variables


117 MAT1512/1


In chapter 7 we cover sections 7.1 and 7.2 only (see pages 565 to 584 in Smith). Please also study some sections inChapter 12.

4. Worked examples

The greater part of the collection of worked examples is taken from chapter 7 and chapter 12 of the prescribedtextbook by Smith and Minton. They can be divided into three sets (with some overlap), namely,

I. First-order differential equations

II. Growth and decay

III. Partial derivatives

Attempt the problems appearing immediately after each set, once you have studied the relevant parts of Smith anddone some of the exercises there. The table below (table 5) shows how these worked examples and their solutionsare organised.

Table 5

Topic(s) Section in Study guideSmith & Minton examples

I. First-order differential equations Sections 7.1 & 7.2 1-9II. Growth and decay Sections 7.1 to 7.2 10-20III. Partial derivatives Sections 12.3 & 12.5 21-40

I. First-order differential equations

An equation that contains an unknown function and some of its derivatives is called a differential equation (DE).Differential equations are of the type:

dydx

= f (x)

ordydx

= y.

A first-order differential equation is an equation involving the first derivative.The general first-order differential equation is of the type:

dydx

= g (x, y) ,

where g (x, y) is a function with two variables. We refer to “first-order” because of the presence of the first derivativeonly. In general, a differential equation is one that contains an unknown function and one or more of its derivatives.The order of the differential equation is the order of the highest derivative that occurs in the equation.

118

(a) The solution of differential equations

A function f is called a solution for a differential equation if the equation is satisfied when y = f (x) and itsderivatives are substituted into the equation. When solving a differential equation, we are expected to find allpossible solutions for the equation.

Example:

1. Solve the differential equation y = x2 + 7x + 3y2

.

Solution:

(i) Separate the variables: y2y = x2 + 7x + 3.(ii) Integrate both sides with respect to x :

y2y (x) dx = x2 + 7x + 3 dx or

y2dy = x2 + 7x + 3 dx .

Result of integration:y3

3= x3

3+ 7x

2

2+ 3x + c.

(iii) Solve for y: y = 3 x3 + 212 x2 + 9x + 3c.

(b) Separable differential equations

A differential equation is separable if you are able to group the y’s with dy and x’s with dx . For example, y =xy2 − 2xy is a separable DE because you can rewrite it as y = x y2 − 2y or

dydx

= x y2 − 2y .

Divide both sides by y2 − 2y and multiply both sides by dx and obtain:1

y2 − 2y dy = xdx.

On the other hand, an equation like y = xy2 − 2x2y is not separable.

A separable equation is a first-order differential equation in which the expression fordydx

can be factored as afunction of x times a function of y. That is, it can be written in the form

dydx

= g (x) f (y) .

The term separable comes from the fact that the expression on the right side can be “separated” into a function ofx and a function of y. If f (y) = 0, we could re-write the equation as

dydx

= g (x)h (y)

= g (x)h (y (x))

(1)

119 MAT1512/1

where h (y) = 1f (y)

.

To solve the equationdydx

= g (x)h (y)

, we rewrite it in the differential form as

h (y) dy = g (x) dx,

so that all y’s are one side of the equation with dy and all x’s are on the other side of the equation with dx . Thenwe integrate both sides of the equation:

h (y) dy = g (x) dx.

This equation defines y as a function of x . In some cases we may be able to solve for y in terms of x .

Note: The justification for the equation above, h (y) dy = g (x) dx, comes from the substitution rule, as

follows:

L.H.S. = h (y) dy = h (y (x))dydxdx

= h (y (x))g (x)h (y (x))

dx (substituting (1))

= g (x) dx = R.H.S.

In general, the first-order differential equation is

dydx

= h (x, y) .

We consider the special case in which h (x, y) = f (y) g (x) , the product of a function of y alone and a function ofx alone.Examples are:

dydx

= x2y3,

dydx

= ex cos y,

anddydx

= y√1+ x2 .

The examples below, are not included in this special case:

dydx

= x + y; dydx

= sin (xy) , dydx

= 1+ x2 + y2

since in each of these equations the right-handside is not of the form f (y) g (x).

We solvedydx

= f (y) g (x) by first separating the variables, putting all the y’s on one side and all the x’s on theother side.Write

dydx

= f (y) g (x)

asdyf (y)

= g (x) dx .

120

Then integrate both sides (if you can) to get:

F (y) = G (x)+ c,

where F (y) is an antiderivative of1f (y)

and G (x) is an antiderivative of g (x) .

If this equation can be solved for y in terms of x , the resulting function is a solution of the differential equation.

Note that in separating the variables, you must assume that f (y) = 0.

If f (y) = 0 for y = y0, it can be checked that the constant function y (x) = y0 is a solution for the equation.

Examples of general solutions:

2. Solvedydx

= xy.Solution:

It can be noted that y (x) = 0 is a solution. Now assume y (x) = 0 and separate the variables:

dyy

= x dx

⇒ dyy

= x dx

⇒ ln |y| = 12x2 + c

⇒ eln|y| = ex22 +c

⇒ |y| = ex22 · ec

or |y| = kex22 where k = ec is a positive constant.

If y ≥ 0, then |y| = y = ke x22 .If y < 0, then |y| = −y = +ke x22 .Therefore,

y = ke x22 if y ≥ 0,

−ke x22 if y < 0.

This is equivalent to:

y = ae x22

where a is any constant.

Note that the solution y (x) = 0 is included.

121 MAT1512/1

To check if the answer obtained is the solution to the original differential equation, just differentiate thisanswer and you should obtain the original differential equation, namely:

y = aex22

⇒ dydx

= ddx

aex22 = ae x22 · 1

2· 2x

= axex22

= x · ae x22= xy (Proved).

3. Solvedydx

= 2x (y − 1)x2 + 1 .

Solution:

Note that y (x) = 1 is a solution.Now assume that y (x) = 1 and separate the variables:

dyy − 1 = 2x

x2 + 1dx

⇒ 1y − 1dy = 2x

x2 + 1dx⇒ ln |y − 1| = ln x2 + 1 + c⇒ eln|y−1| = eln(x

2+1)+c

⇒ |y − 1| = x2 + 1 ec = k x2 + 1 , where k = ec.

Thereforey − 1 = ±k x2 + 1

where k is a positive constant.

The answer may be written as:

y = 1+ a x2 + 1 = 1+ ax2 + a = ax2 + (a + 1)

where a is an arbitrary constant and a = ±k. This answer includes the special solution y (x) = 1.

Check:

dydx

= 2ax + 0

⇒ dydx

= 2xy − 1x2 + 1 where a = y − 1

x2 + 1 since y = 1+ a x2 + 1 .

122

(c) Initial-value problems (IVPs)

In many physical (real-world) problems, we need to find the particular solution that satisfies a condition of the formy (t0) = y0. This is called the initial condition. The problem of finding a solution for the differential equationthat satisfies the initial condition is called the initial-value problem. In such problems, you are asked to find thesolution y = y (x) or y = f (x) for the differential equation

dydx

= g (x, y) ,

the graph of which passes through a given point (a, b), that is, which satisfies the initial condition

y (a) = b.

Examples:

4. Find a solution for the initial-value problemdydx

= x2, with initial condition y (−3) = +1.

Solution:There are two steps to follow:

(i) Solve the differential equation by integrating both sides:

dydx

= x2

⇒ dy = x2dx

⇒ dy = x2dx

⇒ y = 13x3 + c

or y (x) = 13x3 + c. This is the general solution for the DE.

(ii) We find the constant c by substituting the initial condition y (−3) = 1 :

1 = 13(−3)3 + c

⇒ 1 = −9+ c⇒ c = 10.

Therefore, the particular solution to the given initial-value problem is

y (x) = 13x3 + 10.

123 MAT1512/1

5. Find a solution for the differential equationdydx

= 8x3 + 1 with initial conditon y (1) = −1.

Solution:

(i)dydx

= 8x3 + 1 (separate dx from dy and integrate)

⇒ dy = 8x3 + 1 dx⇒ dy = 8x3 + 1 dx

⇒ y = 8x4

4+ x + c

or y (x) = 2x4 + x + c·

(ii) Evaluate the value of c:

y (x) = 2x4 + x + c⇒ −1 = 2 (1)4 + 1+ c (initial condition y (1) = −1)⇒ −1 = 2+ 1+ c⇒ c = −1− 2− 1⇒ c = −4.

The particular solution to the given initial-value problem is

y = 2x4 + x − 4.

See examples 1.1, 1.2 and 1.3 on pages 567 to 570 in Smith.

Other initial-value problems:

6. Solve the initial-value problemdydx

= 2x3y − 1 , y (1) = 0.

Solution:

(3y − 1) dy = 2x dx

⇒ 32y2 − y = 2x2

2+ c.

Now substitute the given initial values and obtain:32(0)2 − 0 = (1)2 + c

⇒ 0 = 1+ c∴ c = −1.

Thus32y2 − y = x2 − 1

or 3y2 − 2y = 2x2 − 2 is the particular solution.

124

7. Solve the initial-value problemdydx

= x2y2, y (0) = b.

Solution:

Note that y (x) = 0 is the solution if b = 0Assume y = 0 and separate the variables:

dydx

= x2y2, y (0) = b

⇒ dyy2

= x2dx

⇒ y−2dy = x2dx

⇒ y−2dy = x2dx

⇒ y−1

−1 = x3

3+ c

⇒ −1y

= 13x3 + c.

Substitute the initial values:

−1y

= 13x3 + c

⇒ −1b

= 13(0)3 + c

⇒ −1b

= 0+ c

∴ c = −1b.

Therefore

−1y

= x3

3− 1b

⇒ −1y

= bx3 − 33b

⇒ −1 (3b) = y bx3 − 3⇒ 3b = −y bx3 − 3 = y −bx3 + 3

∴ y = 3b3− bx3 .

The solution includes y (x) = 0.

Note: In initial-value problems (IVPs), you are given initial conditions (in other words, the x and y values), and youuse them in order to evaluate a particular value of a constant c.

125 MAT1512/1

More examples:

8. Solving a logistic growth problem: Given a maximum sustainable population of M = 1 000 and a growthrate of k = 0.007, find an expression for the population at any time t , given an initial population of y(0) = 350and assuming logistic growth.

Solution:

Using the solution for a logistic equation, we have kM = 7 and

y = AMekMt

1+ AekMt , where

A = constant,M = carrying capacity,k = growth rate andt = time.

From the initial conditions, we have 350 = y (0) = 1 000A1+ A

and solving for A, we obtain A = 3565

which gives the solution of the IVP as y = 35 000e7t

65+ 35e7t .

9. Investment strategies for making a million: Money is invested at 8% interest, compounded continuously. Ifdeposits are made at a rate of R2 000 per year, find the size of the initial investment needed to reach R1million in 20 years.

Solution:

The constant of integration c is obtained by setting t = 0 and taking A (0) = x,that is 12.5 ln |0.08x + 2000| = cso that 12.5 ln |0.08A + 2000| = t + 12.5 ln |0.08x + 2000| .Now find the value of x such that A (20) = 1 000 000.12.5 ln |0.08 (1 000 000)+ 2000| = 20+ 12.5 ln |0.08x + 2000|or12.5 ln |82 000|− 20

12.5= ln |0.08x + 2000|

Solve for x by taking the exponential of both sides:

e(12.5 ln 82 000−20)/12.5 = 0.08x + 2000⇒ x = e

ln 82000−1.6 − 20000.08

≈ 181, 943.93.

126

II. Growth and decay

y (t) = Aekt is the general solution for the differential equation y (t) = dydt= ky (t) .

For k > 0 the equation reflects an exponential growth law and for k < 0, the equation reflects an exponentialdecay law.

k > 0 is the growth curve k < 0 is the decay curve

Exponential growth or decay means that:

y (t) = dydt= ky (t) ... (1)

Since equation (1) involves the derivative of an unknown function, we call it a differential equation (DE). The aimis to solve this DE. that is, to find the function y(t).

Recall how to solve separable differential equations. We have

dydt= ky (t) ... (2)

This is a separable DE, that isdyy (t)

= kdt.

Integrating both sides of equation (2) with respect to t , we obtain

1y (t)

dy = kdt

⇒ ln |y| + c1 = kt + c2

or ln |y| = kt + C (where C = c2 − c1).Since y (t) > 0, we have ln y (t) = kt + Cand taking exponentials of both sides , we get eln y(t) = ekt+C ,that is, y (t) = ekt .eC .Therefore y (t) = Aekt (where A = eC ).

127 MAT1512/1

Examples:

10. A bacterial culture contains 100 cells at a certain point in time. Sixty minutes later, there are 450 cells.Assuming exponential growth, determine the number of cells present at time t and find the doubling time.

Solution:

Exponential growth means that y (t) = ky (t) and after solving this equation, we have y (t) = Aekt .At t = 0, y (0) = 100 (initial conditions), we have 100 = y (0) = Ae0 = A.Then y (t) = 100ekt .

At t = 60 minutes, we have 450 = y (60) = 100e60k .Solve for k by taking natural logarithms of both sides and obtain

ln 4.5 = ln e60k = 60k,so that k = ln 4.5

60.

Therefore, at any time t we have y (t) = 100ekt = 100 exp ln 4.560

t .

At double the time, y (t) = 2y (0) = 200,so that 200 = y (t) = 100 exp ln 4.5

60t .

Solve for it and get ln 2 = ln 4.560

t .

11. A cup of coffee is 180◦F when poured. After 2 minutes in a room at 70◦F, the coffee has cooled to 165◦F.Find the temperature at time t and find the time at which the coffee will have cooled to 120◦F.

Solution:

The differential equation is y (t) = k y (t)+ 70 .Solving this differential equation, we obtain y (t) = Aekt + 70.Initial conditions are : 180 = y (0) = Ae0 + 70 = A + 70.This gives y (0) = 180,so that A = 110 and y (t) = 110ekt + 70.We use the second measured temperature (165◦F) to solve for k :165 = y (2) = 110e2k + 70.Solve for k by subtracting 70 from both sides and dividing by 110 to obtain

e2k = 165− 70110

= 95110

.

Taking natural logarithms of both sides, we get 2k = ln 95110

and, therefore k = 12 ln

95110

.

When the temperature T = 120, we have 120 = y (t) = 110ekt + 70.Solve for t and obtain t = 1

kln511.

128

12. Compound interest and an example of decay. Suppose the value of a R10 000 asset decreases continuouslyat a constant rate of 24% per year. Find its worth after

(a) 10 years

(b) 20 years

Solution:

The value v(t) of any quantity that is changing at a constant rate r satisfies the equation: v = rv .Here r = −0.24, so that v (t) = Ae−0,24t .Since the initial asset value is R10 000, we have v (t) = 10 000e−0.24t .We now have 10 000 = v (0) = Ae0 = A.At time t = 10, the asset value is R10 000−0.24(10) ≈ R907.18 andat time t = 20, the asset value is R10 000−0.24(20) ≈ R82.30.

More worked examples:

13. A bacterial culture starts with 1000 bacteria and after 2 hours there are 2500 bacteria. Assuming that theculture grows at a rate proportional to its size, find the population after 6 hours.

14. The earth’s atmospheric pressure p is often modeled by assuming that the rate dp/dh at which p changeswith altitude h above sea level is proportional to p. Suppose that the pressure at sea level is 1 013 millibars(about 14.7 pounds per square inch) and that the pressure at an altitude of 20 km is 90 millibars.

(a) Solve the initial value problem –Differential equation: dp/dh = kp (k is a constant).Initial condition: p = p0 when h = 0,Express p in terms of h. Determine the values of p0 and k from the given altitude-and pressure data.

(b) What is the atmospheric pressure at h = 50 km?(c) At what altitude does the pressure equal 900 millibars?

15. In some chemical reactions, the rate at which the amount of a substance changes with time is proportional tothe amount present. For the change of δ–gluconolactone into gluconic acid, for example,

dydt= −0.6y

when t is measured in hours. If there are 100 grams of δ–gluconolactone present when t = 0, how manygrams will be left after the first hour?

16. The intensity L (x) of light x feet beneath the surface of the ocean satisfies the differential equation

dLdx

= −kL .

As a diver, you know from experience that diving to 18 ft in the Caribbean Sea cuts the intensity in half. Youcannot work without artificial light when the intensity falls below one-tenth of the surface value. Up to abouthow deep can you expect to work without artificial light?

129 MAT1512/1

17. If R1000 is invested for 4 years at 12% interest compounded continuously, what is the value of the investmentat the end of the 4 years?

18. You have just placed A0 rands in a bank account that pays 14% interest, compounded continuously.

(a) How much money will you have in the account in 6 years’ time?

(b) How long will it take for your money to double and to triple?

19. Suppose an object takes 40 min to cool down from 30◦C to 24◦C in a room that is kept at 20◦C.

(a) What would the temperature of the object be 15 min after it has been 30◦C?

(b) How long will it take for the object to cool down to 21◦C?

20. Suppose that a cup of soup cooled down from 90◦C to 60◦C within 10 minutes in a room of which thetemperature was 20◦C. Use Newton’s Law of Cooling to answer the following questions.

(a) How much longer would it take the soup to cool to 35◦C?

(b) Instead of being left to stand in the room, the cup of 90◦C soup is put in a freezer in which the tempera-ture is −15◦C. How long will it take the soup to cool down from 90◦C to 35◦C?

Solutions:

13. Let y (t) be the number of bacteria after t hours. Then y0 = y (0) = 1000 and y (2) = 2500. Since we areassuming

dydt= ky, (Law of Exponential Change on page 380 in Smith),

y (t) = y0ekt = 1000ekt⇒ y (2) = 1000e2k = 2500.

Therefore e2k = 2.5 and 2k = ln 2.5.Thus k = ln 2.5

2.

Substituting the value of k = 12 ln 2.5 back into the expression for y (t), we have

y (t) = 1000eln 2, 52

t

= 1000 eln 2.5t2

= 1000 (2.5)t2 (since eln 2.5 = 2.5).

Therefore the population after 6 hours is

y (6) = 1000 (2.5)3 = 15 625.

130

14. (a) Differential equation:dpdh

= kp (k is a constant).Initial condition: p = p0 when h = 0.We separate the variables by dividing the differential equation by p to have

1pdpdh

= k

⇒ ln |p| = kh + c (by integrating the RHS will respect to h)⇒ |p| = ekh + c (by exponentiating)⇒ |p| = ecekh

⇒ p = ±ecekh⇒ p = Aekh where A = ±ec.

But the initial condition p = p0 when h = 0. Therefore p0 = Aek0 = A and thus

p = p0ekh

Then

p (0) = p0 = 1013 and p (20) = 90 = 1013e20k .

Thus

e20k = 901013

⇒ 20k = ln901013

⇒ k = 120ln901013

.

Therefore p (h) is given by

p (h) = 1013e ln 901013

h20= 1013 90

1013

h20

.

(b) If h = 50 km, then

p (50) = 1013eln901013 · 52

= 1013 eln901013

52

= 1013901013

52

(since eln901013 = 90

1013)

= 902.5

10131.5.

131 MAT1512/1

(c) If p = 900, then

900 = 1013901013

h20

⇒ 901013

h20

= 9001013

⇒ h20ln

901013

= ln9001013

⇒ h = 20 · 1ln 90

1013· ln 900

1013.

15. We have the following initial-value problem:

Differential equation:dydt= −0, 6y.

Initial condition: y0 = 100 when t = 0.We use the Radioactive decay equation on page 382 in Smith:

y = y0e−kt .

Since y0 = 100 and k = −0, 6 we havey = 100e−0,6t .

After the first hour the number of grams will be

y = 100e(−0,6)(1)

= 100e−0,6.

16. We have the following initial-value problem:

Differential equation:dLdx

= −kLInitial condition: L = L0 when x = 0. We are given that L = 1

2 L0 when x = 18. We want to find x whenL = 1

10L0.

We use equation (1.4) on page 567 in Smith:

L (x) = L0e−kx

where

L (x) = y (t) ,

L0 = A.

Thus12L0 = L0e−18k

∴ e−18k = 12

⇒ e18k = 2

⇒ 18k = ln 2

⇒ k = 118ln 2.

132

When L = 110L0 we have

110L0 = L0e−kx

⇒ e−kx = 110

⇒−kx = ln110

⇒−kx = − ln 10⇒ x = 1

kln 10

= 18ln 2

ln 10 (since k = 118ln 2)

17. We use the Continuous compound interest formula on page 381 in Smith:

A (t) = A0ert .

It is given that A0 =R1000 and r = 12% = 0, 12. So the value of the investment in 4 years will be

A (4) = 1000e(0,12)4

= 1000e0,48.

18. We use the continuous compound interest formula:

A (t) = A0ert .

(a) We are given that r = 14% = 0, 14. In 6 years the money will be

A (6) = A0e(0,14)6

= A0e0,84.

(b) We want to find t when A = 2A0, and when A = 3A0. We have A (t) = A0e0,14t .When A = 2A0 we have

2A0 = A0e0,14t

⇒ e0,14t = 2

⇒ 0, 14t = ln 2

⇒ t = ln 20, 14

= 10014ln 2.

When A = 3A0 we have

3A0 = A0e0,14t

⇒ e0,14t = 3

⇒ 0, 14t = ln 3

⇒ t = ln 30, 14

= 10014ln 3.

133 MAT1512/1

19. We use Newton’s Law of Cooling on page 569 in Smith:

T = Ts + (T0 − Ts) e−ktwhere T = y (t) , Ts = Tα

and T0 − Ts = A

where T is the temperature at any given time t , Ts is the surrounding temperature and T0 is the initial temper-ature (in other words, T0 is the value of T at t = 0).Now, we are given that Ts = 20◦C and T0 = 30◦C.Then the object’s temperature after t minutes is

T = 20+ (30− 20) e−kt= 20+ 10e−kt .

(a) We first find k, by using the information that T = 24 when t = 40 :24 = 20+ 10e−40k

⇒ 10e−40k = 4

⇒ e−40k = 410

⇒−40k = ln410

⇒ k = − 140ln

410

.

Now the object’s temperature after time t is

T = 20+ 10e 140 ln 410 ·t.

We now find T when t = 15 :T = 20+ 10e 140 ln 4

10 ·15

= 20+ 10 eln410

1540

= 20+ 10 410

1540

.

(b) We now find the time t when T = 21◦C:21 = 20+ 10 eln

410

t40

⇒ 10410

t40

= 1

⇒ 410

t40

= 110

⇒ t40ln

410

= ln110

⇒ t = 40ln 4

10· ln 1

10

= −40 ln 10ln (0, 4)

.

134

20. As in the solution of the question above,

T = Ts + (T0 − Ts) e−kt .

It is given that Ts = 20◦C and T0 = 90◦C. The soup’s temperature after t minutes is

T = 20+ (90− 20) e−kt= 20+ 70e−kt .

(a) We find k by using the information that T = 60◦C when t = 10 minutes:

60 = 20+ 70e−10k⇒ 40 = 70e−10k

⇒ e−10k = 47

⇒−10k = ln47

⇒ k = − 110ln

47.

The soup’s temperature at time t is

T = 20+ 70e 110 ln

47 ·t

We now find the time t when T = 35◦C:

35 = 20+ 70 eln47

t10

⇒ 15 = 7047

t10

⇒ 47

t10

= 1570

⇒ t10ln

47

= ln1570

⇒ t = 101

ln 47ln1570.

(b) With T = −15◦C and T0 = 90◦C, we have

T = −15+ (90− (−15)) e−kt= −15+ 105e−kt .

135 MAT1512/1

So we find t when T = 35◦C:

35 = −15+ 105 eln47

t10

⇒ 50 = 10547

t10

⇒ 47

t10

= 50105

⇒ t10ln

47

= ln50105

⇒ t = 101

ln 47ln

50105

.

III. Partial derivatives

Functions with more than one variable appear more often in science than functions with one variable. The reasonfor this is that in real life the mathematical application in respect of any dynamic phenomenon depends on morethan one variable. One example of a dynamic phenomenon is a moving train, where the motion will depend onvariables such as time, distance, friction, mass and so forth. The applied mathematician uses functions with severalvariables, their derivatives and their integrals to study continuum mechanics, probability, statistics, fluid mechanicsand electricity – to mention but a few examples.

If f is a function of two independent variables, we usually call the independent variables x and y and picture thedomain of f (x, y) as a region in the XY -plain. If f is a function of three independent variables, we call the variablesx , y, and z and picture the domain of f (x, y, z) as a region in space.

Now, to talk about the derivative of f doesn’t make sense if we do not specify with respect to which independentvariable the derivative is required. When we keep all but one of the independent variables of a function constant anddifferentiate with respect to that one variable, we get a partial derivative. We would like to encourage you to workthrough examples 3.1–3.6 in section 12.3 and to study section 12.5 of chapter 12 in Smith carefully, in conjunctionwith the examples in the study guide.

(a) The partial derivatives technique

This technique is used when functions have more than one variable.

Consider a function f of two variables, x and y. Suppose we let only x vary, while keeping y fixed and y = bwhere b is a constant.

Then this is a function of a single variable x : g (x) = f (x, b).

If g has a derivative at a, this is called the partial derivative of f with respect to x at (a, b), denoted by

fx (a, b) = g (x) , that is g (a) = limh→0

g (a + h)− g (a)h

or fx (a, b) = limh→0

f (a + h, b)− f (a, b)h

Similarly, the partial derivative of f with respect to y at (a, b), denoted by fy (a, b) is obtained by keeping x fixed(that is x = a) and finding the derivative at b of the function G (y) = f (a, y).

136

So fx (a, b) = limh→0

f (a + h, b)− f (a, b)h

and fy (a, b) = limh→0

f (a, b+ h)− f (a, b)h

.

If we now let the point (a, b) vary, fx and fy become functions of two variables.

DefinitionIf f is a function of two variables, its partial derivatives are the functions fx and fy defined as:

fx (x, y) = limh→0

f (x + h, y)− f (x, y)h

and

fy (x, y) = limh→0

f (x, y + h)− f (x, y)h

(b) Notations for partial derivatives

If z = f (x, y) , we write:

fx (x, y) = fx = ∂ f∂x= ∂

∂xf (x, y) = ∂z

∂x= f1 = D1 f = Dx f

fy (x, y) = fy = ∂ f∂y= ∂

∂yf (x, y) = ∂z

∂y= f2 = D2 f = Dy f

Note: f1 or D1 f indicates differentiation with respect to the first variable and f2 or D2 f indicates differentiationwith respect to the second variable.

(c) Rules for finding the partial derivatives of z = f (x, y)

1. To find fx , regard y as a constant and differentiate with respect to x .

2. To find fy , regard x as a constant and differentiate with respect to y.

Examples:

21. If f (x, y) = x3 + x2y3 − 2y2, find fx (2, 1) and fy (2, 1).

22. Find∂z∂xand

∂z∂yif z is defined implicitly as a function of x and y by the equation x3 + y3 + z3 + 6xyz = 1.

Solutions:

21. If f (x, y) = x3 + x2y3 − 2y2,then holding y constant and differentiating with respect to x , we get

fx (x, y) = 3x2 + 2xy3 − 0= 3x2 + 2xy3

137 MAT1512/1

and

fx (2, 1) = 3 (2)2 + 2 (2) 13= 12+ 4= 16.

Holding x constant and differentiating with respect to y, we get

fy (x, y) = 0+ 3x2y2 − 4y= 3x2y2 − 4y

and

fy (2, 1) = 3 22 12 − 4 (1)= 12− 4= 8.

22. Given that x3 + y3 + z3 + 6xyz = 1.

Then, holding y constant and differentiating implicitly with respect to x , we get

3x2 + 0+ 3z2 ∂z∂x+ 6yz + 6xy ∂z

∂x= 0.

Solving this equation for∂z∂x, we get

∂z∂x

= −3x2 − 6yz3z2 + 6yz = −3 x

2 + 2yz3 z2 + 2yz

= − x2 + 2yzz2 + 2yz .

Similarly, holding x constant and implicit differentiation with respect to y gives

0+ 3y2 + 3z2 ∂z∂y+ 6xz + 6xy ∂z

∂y= 0

⇒ 3z2 + 6xy ∂z∂y

= −3y2 − 6xz

⇒ ∂z∂y

= −3y2 − 6xz3z2 + 6xy

= −3 y2 + 2xz3 z2 + 2xy

= − y2 + 2xzz2 + 2xy .

138

(d) Partial differentiation

If z = f (x, y) where f is a differentiable function of x and y where x = x (s, t) and y = y (s, t) , and both havepartial derivatives, then the chain rule below holds:

∂z∂s

= ∂z∂x∂x∂s+ ∂z∂z∂y∂sand

∂z∂t

= ∂z∂x∂x∂t+ ∂z∂y∂y∂t.

Example:

23. Suppose f (x, y) = exy, x (u, v) = 3u sin v and y (u, v) = 4v2u.For g (u, v) = f (x (u, v) , y (u, v)) , find the parial derivatives

∂g∂u

and∂g∂v.

Solution:

f (x, y) = exy, x (u, v) = 3u sin v, y (u, v) = 4uv2 and

g (u, v) = f (x (u, v) , y (u, v)) .

Then∂g∂u= ∂ f∂x∂x∂u+ ∂ f∂y∂y∂u

and∂g∂v= ∂ f∂x∂x∂v+ ∂ f∂y∂y∂v.

So

∂ f∂x

= exy · y = yexy,∂ f∂y

= exy · x = xexy,∂x∂u

= 3 sin v,

∂y∂u

= 4v2,

∂x∂u

= 3u (− cos u (1)) = −3u cos v and∂y∂u

= 8uv.

Therefore

∂g∂u

= yexy (3 sin v)+ xexy 4v2

= 4uv2e3u sin v(4uv2) (3 sin v)+ 3u sin ve3u sin v(4uv2) 4v2

= 12uv2e12u2v2 sin v sin v + 12uv2e12u2u2 sin v sin v

= 24uv2e12u2v2 sin v sin v and

139 MAT1512/1

∂g∂v

= yexy (−3u cos v)+ xexy (8uv) = −3yuexy cos v + 8xuvexy

= −3 4uv2 ue3u sin v(4uv2) cos v + 8 (3u sin v) uve3u sin v(4uv2)= −12u2v2e12u2v2 sin v cos v + 24u2ve12u2v2 sin v sin v= 24uv2e12u2v2 sin v sin v.

(e) Functions of more than two variables

Partial derivatives can also be defined for functions of three or more variables . For example, if f is a function ofthree variables, x , y and z, then its partial derivative with respect to x is defined as:

fx (x, y, z) = limh→0

f (x + h, y, z)− f (x, y, z)h

and is found by, for example, regarding y and z as constants and differentiating f (x, y, z) with respect to x .

Example:

24. Find fx , fy and fz if f (x, y, z) = exy ln z.

Solution:If f (x, y, z) = exy ln z,then to compute the partial derivative with respect to x , we treat y and z as constants and obtain

fx = ∂

∂xexy ln z

= exy (y) ln z

= yexy ln z.

To compute the partial derivative with respect to y, we treat x and z as constants and obtain

fy = ∂

∂yexy ln z

= exy (x) ln z

= xexy ln z.

To compute the partical derivative with respect to z, we treat x and y as constants and obtain

fz = ∂

∂zexy ln z

= exy · 1z(1)

= exy

z.

140

(f) Higher–order derivatives

If f is a function of two variables, then its partial derivatives fx and fy are also functions of two variables, so wecan consider their partial derivatives ( fx)x , ( fx)y , fy x and fy y , which are called the second partial derivativesof f .

If z = f (x, y) , we use the following notation:

( fx)x = fxx = f11 = ∂

∂x∂ f∂x

= ∂2 f∂x2

= ∂2z∂x2

,

( fx)y = fxy = f12 = ∂

∂y∂ f∂x

= ∂2 f∂y∂x

= ∂2z∂y∂x

,

fy x = fyx = f21 = ∂

∂x∂ f∂y

= ∂2 f∂x∂y

= ∂2z∂x∂y

and

fy y = fyy = f22 = ∂

∂y∂ f∂y

= ∂2 f∂y2

= ∂2z∂y2

.

Note: The notation fxy or∂2 f∂y∂x

means we first differentiate with respect to x and then with respect to y, and in

computing fyx , the order of differentiation is reversed.

Example:

25. Find the second partial derivatives of f (x, y) = x3 + x2y3 − 2y2.

Solution:Given f (x, y) = x3 + x2y3 − 2y.In example 21 we found thatfx (x, y) = 3x2 + 2xy3 andfy (x, y) = 3x2y2 − 4y.Thereforefxx (x, y) = ∂

∂x3x2 + 2xy3 = 6x + 2y3,

fxy (x, y) = ∂

∂y3x2 + 2xy3 = 0+ 6xy2 = 6xy2,

fyx (x, y) = ∂

∂x3x2y2 − 4y = 6xy2 − 0 = 6xy2 and

fyy (x, y) = ∂

∂y3x2y2 − 4y = 6x2y − 4.

Now work through the rest of the examples provided.

Worked examples:

26. Determine ∂ f/∂x and ∂ f/∂y if:

(a) f (x, y) = sin(x + y) and

141 MAT1512/1

(b) f (x, y) = xx2 + y2 .

27. Use the chain rule for partial derivatives to determine the value of dw/dt when t = 0 if w = x2 + y2,x = cos t + sin t and y = cos t − sin t.

28. Use the chain rule for partial derivatives to find dw/dt when w = − sin (xy)x = 1+ t and y = t2.

29. Suppose

w = 1xy+ 1y, x = t2 + 1 and y = t4 + 3.

Use the chain rule for partial derivatives to find dw/dt.

30. Assume that the equationyx

43 + y 43 = 144

defines y as a differentiable function of x . Use partial derivatives to find the value of dy/dx at the point(−8, 8).

31. Determine ∂z/∂u when u = 0 and v = 1, if z = sin(xy)+ x sin y, x = u2 + v2 and y = uv.32. Let g be a function of x and y, and suppose that x and y are both functions of u and v . Which one of the

following is ∂g/∂v?

(a)∂g∂u· ∂u∂x+ ∂g∂u· ∂u∂y

(b)dgdx· ∂x∂v+ dgdy· ∂y∂v

(c)∂g∂x· ∂x∂v+ ∂g∂y· ∂y∂v

(d)dgdx· dxdv+ dgdy· dydv

33. Determine ∂2 f/∂y2 if f (x, y) = tan(xy)− cot x .34. Suppose that the equation

y = sin(xy)defines y as a differentiable function of x . Use partial derivatives to find dy/dx .

35. Suppose w = (−2x2 + 2y2 − 2)2, x = 2u − 2v + 2 and y = −u + 4v + 2. Find the value of ∂w/∂v whenu = 0 and v = −1.

36. Use the chain rule for partial derivatives to find dw/dt when w = − sin(xy), x = 1+ t and y = t2.37. Assume that the equation

xy + y2 − 3x − 3 = 0defines y as a differentiable function of x . Find the value of dy/dx at the point (−1, 1) by using partialderivatives.

142

38. Use the chain rule for partial derivatives to find ∂w/∂r when r = π and s = 0, if w = sin(2x − y),x = r + sin s and y = rs.

39. Suppose that the equation7y4 + x3y + x = 4

defines y as a differentiable function of x . Find the value of dy/dx at the point (4, 0).

40. The dimensions a, b and c of a rectangular solid vary with time (t). At the instant in question

a = 13 cm, b = 9 cm, c = 5 cm, dadt= dcdt= 2 cm/sec

anddbdt= −5 cm/sec.

a

b

c

diagonal

(a) How fast is the volume V changing at the given instant? Is V increasing or decreasing?

(b) How fast is the surface area S changing at that moment? Is the area increasing or decreasing?

(c) How fast is the length D of the diagonal changing at that instant? Does the length increase or decrease?

Solutions:

26. (a)

f (x, y) = sin(x + y)⇒ ∂ f∂x

= cos(x + y) and∂ f∂y= cos(x + y).

(b)

f (x, y) = xx2 + y2

⇒ ∂ f∂x

= 1. x2 + y2 − x (2x + 0)x2 + y2 2

= y2 − x2x2 + y2 2

.

143 MAT1512/1

Also,

f (x, y) = xx2 + y2 = x(x

2 + y2)−1

⇒ ∂ f∂y

= x(−(x2 + y2)−2).(0+ 2y)

= − 2xy

x2 + y2 2.

27.

w = x2 + y2, x = cos t + sin t and y = cos t − sin t⇒ dw

dt= ∂w

∂xdxdt+ ∂w∂ydydt

= (2x + 0)(− sin t + cos t)+ (0+ 2y)(− sin t − cos t)= 2x (cos t − sin t)− 2y (sin t + cos t)

by substituting for x and y

= 2(cos t + sin t)(cos t − sin t)− 2(cos t − sin t)(cos t + sin t)= 0.

Hence,dwdt t=0

= 0.

28.w = − sin (xy)

⇒ ∂w

∂x= − cos (xy) · y

and∂w

∂y= − cos (xy) · x .

Also,

x = 1+ t⇒ dx

dt= 1,

and

y = t2

⇒ dydt

= 2t.

We havedwdt

= ∂w

∂x· dxdt+ ∂w∂y

· dydt

= − cos (xy) · y (1)− cos (xy) · x · 2t= − cos ((1+ y) (+2)) · y2 − cos (1+ t) t2 (1+ t) · 2t .

144

29.

w = 1xy+ 1y

⇒ ∂w

∂x= − 1

yx2

and∂w

∂y= 1xy2

− 1y2.

Also,

x = t2 + 1

⇒ dxdt

= 2t

and

y = t4 + 3

⇒ dydt

= 4t3.

Hence,

dwdt

= ∂w


· dydt

= − 1yx2

· 2t + − 1xy2

− 1y2

· 4t3

= − 2t

t4 + 3 t2 + 1 2 −4t3

t2 + 1 t4 + 3 2 −4t3

t4 + 3 2 .

30. LetF(x, y) = yx 43 + y 43 − 144.

Thendydx

= −FxFy

= −43 yx

13

x4/3 + 43 y

13,

so

dydx (x,y)=(−8,8)

= −43 · 8 · (−2)16+ 4

3 · 2= 8

7.

145 MAT1512/1

31.

∂z∂u

= ∂z∂x· ∂x∂u+ ∂z∂y· ∂y∂u

= (y cos(xy)+ sin y)2u + (x cos(xy)+ x cos y)v.

If u = 0 and v = 1, then x = u2 + v2 = 1 and y = uv = 0. Then∂z∂u

= (0 · cos 0+ sin 0) · 0+ (1 · cos 0+ 1 · cos 0) · 1= 2.

32.∂g∂v= ∂g∂x· ∂x∂v+ ∂g∂y· ∂y∂v.

33.

f (x, y) = tan(xy)− cot x⇒ ∂ f∂y

= sec2(xy).x

⇒ ∂2y∂y2

= 2 sec(xy). sec(xy) tan(xy).x .x

= 2x2. sec2(xy). tan(xy).

34.

y = sin(xy)

⇒ y − sin(xy) = 0.

LetF(x, y) = y − sin(xy).

Then

dydx

= −FxFy

= − − cos (xy) · y1− cos (xy) · x

= y cos (xy)1− x cos (xy) .

35.

∂w

∂v= ∂w

∂x· ∂x∂v+ ∂w∂y

· ∂y∂v

= 2(−2x2 + 2y2 − 2)(−4x)(−2)+ 2(−2x2 + 2y2 − 2)(4y)(4).

If u = 0 and v = −1, thenx = 0+ 2+ 2 = 4

146

and

y = 0− 4+ 2 = −2

so that

∂w

∂v= 2 (−32+ 8− 2) (−16) (−2)+ 2 (−32+ 8− 2) (−8) (4)= 0.

36.

w = − sin(xy)⇒ ∂w

∂x= − cos(xy).y

and∂w

∂y= − cos(xy).x .

Also,

x = 1+ t⇒ dxdt

= 1

and

y = t2

⇒ dydt

= 2t.

We have

dwdt

= ∂w


· dydt

= − cos(xy) · y · 1− cos(xy) · x · 2t= − cos((1+ t)t2) · t2 − cos((1+ t)t2) · (1+ t)2t.

37. Put

F(x, y) = xy + y2 − 3x − 3.

Thendydx

= −FxFy= − y − 3

x + 2y .

Hence,dydx (x,y)=(−1,1)

= − 1− 3−1+ 2 = 2.

147 MAT1512/1

38.∂w

∂r= ∂w

∂x· ∂x∂r+ ∂w∂y

· ∂y∂r

= 2 · cos(2x − y) · 1+ (− cos(2x − y)) · s.

If r = π and s = 0, then x = π and y = 0, so that∂w

∂r= 2 · cos(2π) · 1− cos(2π).0 = 2.

39. LetF(x, y) = 7y4 + x3y + x − 4.

Thendydx

= −FxFy

= − 3x2y + 1

28y3 + x3 .

Hencedydx (x,y)=(4,0)

= − 164.

40. (a) Denote derivatives with respect to time by a dot. If

V = abc,

then

V = ∂V∂aa + ∂V

∂bb + ∂V

∂cc

= abc + abc+ abc.

Substitute the given values:

V = (2)(9)(5)+ (13)(−5)(5)+ (13)(9)(2)= 90− 325+ 234= −1.

Therefore the volume decreases at 1 cm3/sec.

(b) Differentiate with respect to time. If

S = 2ac + 2bc + 2ab,

then

S = ∂S∂aa + ∂S

∂bb + ∂S

∂cc

= (2c+ 2b)a + (2c + 2a)b + (2a + 2b)c.

148

Substitute the given values:

S = (2× 5+ 2× 9)(2)+ (2× 5+ 2× 13)(−5)+ (2× 13+ 2× 9)(2)= −36.

The area decreases at a rate of 36 cm2/sec.

(c) According to the Theorem of Pythagoras

D = (a2 + b2 + c2) 12 .

Differentiate with respect to time:

D = 12(a2 + b2 + c2)− 12 (2aa + 2bb + 2cc)

= aa + bb + cc√a2 + b2 + c2 .

Thus D (13, 9, 5) = (13) (2)+ (9) (−5)+ (5) (2)(13)2 + (9)2 + (5)2

(from the given values)

= 26− 45+ 10√169+ 81+ 25

= −9√275

= −95√11.

Therefore the diagonal decreases at 9/ 5√11 cm/sec.

149 MAT1512/1

Key points

In this chapter we have introduced you to specific solutions of simple, first-order differential equations. You shouldnow be comfortable with solving some basic growth and decay problems. We have also introduced the notion ofpartial derivatives, where we have seen how to approach scientific problems involving functions of several variables,by changing one of the variables at a time, and keeping the remaining variable(s) fixed.

At this stage you should be able to:

• solve first-order differential equations with initial values• solve basic real-life problems involving exponential growth and decay• determine the partial derivatives of functions of several variables

Continue practising solving problems until you have mastered the basic techniques! Go through the section “Foryour review ” at the end of each chapter to consolidate what you have learnt. Also use other textbooks and theinternet.

150

APPENDIX

The purpose of the Appendix is to help you to make the transition from high school to the more advanced first-yearcalculus module MAT1512.

I. Sequences and summation (sigma) notationII. Mathematical inductionIII. The Binomial Theorem

I. Sequences and summation notation

A sequence is a list of numbers written in a specific order. Sequences have many applications.

To read: Smith and Minton, section 4.2, pp 354–359

Outcomes:After studying this section you should:

(a) know the definition of a sequence.

(b) know what is meant by

• a recursive sequence• a partial sum of a sequence• sigma notation

(c) be able to

• find the terms of a sequence

151 MAT1512/1

• find the nth term of a sequence• find the partial sum of a sequence• use sigma notation

Examples:

I. 1.1Write the first four terms of the sequence given by the formula in each case.

(a) c j = 3 110

j−1

Solution:

c1 = 3110

1−1= 3 1

10

0

= 3,

c2 = 3110

2−1= 3 1

10

1

= 310,

c3 = 3110

3−1= 3 1

10

2

= 3100

c4 = 3110

4−1= 3 1

10

3

= 31000

.

(b) an = (−1)n+1n + 3

Solution:

a1 = (−1)1+11+ 3 = 1

4, a2 = (−1)2+1

2+ 3 = −15

a3 = (−1)3+13+ 3 = 1

6, a4 = (−1)4+1

4+ 3 = −17.

(c) xk = kk + 1 −

k + 1k

Solution:

x1 = 11+ 1 −

1+ 11

= 12− 2 = −3

2,

x2 = 22+ 1 −

2+ 12

= 23− 32= −5

6,

x3 = 33+ 1 −

3+ 13

= 34− 43= − 7

12,

x4 = 44+ 1 −

4+ 14

= 45− 54= − 9

20.

152

I. 1.2 Find the ninth and tenth terms of: 0, 4, 0, . . . ,2n + (−2)n

n, . . .

Solution:

an = 2n + (−2)nn

Then a9 = 29 + (−2)99

= 29 − 299

= 09= 0 and

a10 = 210 + (−2)1010

= 210 + 21010

= 2 · 210

10= 2048

10= 1024

5.

I. 1.3 Find the first six terms of the sequence defined by: a1 = 6 and an = 3an−1

.

Solution:

a1 = 6, a2 = 3a2−1

= 36= 12,

a3 = 3a3−1

= 3a2= 3

12= 6,

a4 = 3a4−1

= 3a3= 36= 12,

a5 = 3a5−1

= 3a4= 3

12= 6,

a6 = 3a6−1

= 3a5= 36= 12.

I. 1.4 Find the sum of the first five terms of the sequence given by the formula in each case.

(a) ak = (−1)k 1kSolution:

5

k=1ak = a1 + a2 + a3 + a4 + a5

= (−1)1 11+ (−1)2 1

2+ (−1)3 1

3+ (−1)4 1

4+ (−1)5 1

5

= −1+ 12− 13+ 14− 15

= −4760.

153 MAT1512/1

(b) bi = i3

Solution:5

i=1i3 = b1 + b2 + b3 + b4 + b5

= 13 + 23 + 33 + 43 + 53= 1+ 8+ 27+ 64+ 125= 225.

(c) Find8

n=0xn where xn = 1

2n.

Solution:8

n=0xn = x0 + x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8

= 120+ 121+ 122+ 123+ 124+ 125+ 126+ 127+ 128

= 1+ 12+ 14+ 18+ 116+ 132+ 164+ 1128

+ 1256

= 511256

.

I. 1.5 Evaluate each of the following:

(a)6

k=1(5k)

Solution:6

k=1(5k) = 5.1+ 5.2+ 5.3+ 5.4+ 5.5+ 5.6

= 5+ 10+ 15+ 20+ 25+ 30= 105.

(b)7

k=1(−1)k

Solution:7

k=1(−1)k = (−1)1 + (−1)2 + (−1)3 + (−1)4 + (−1)5 + (−1)6 + (−1)7

= −1+ 1− 1+ 1− 1+ 1− 1= −1.

154

(c)3

n=1

n + 1n

− nn + 1

Solution:

3

n=1

n + 1n

− nn + 1 = 1+ 1

1− 11+ 1 + 2+ 1

2− 22+ 1 + 3+ 1

3− 33+ 1

= 2− 12+ 32− 23+ 43− 34

= 3512.

I. 1.6 Rewrite each series using sigma notation.

(a) 4+ 8+ 12+ 16+ 20+ 24

Solution:

4+ 8+ 12+ 16+ 20+ 24 = 4 · 1+ 4 · 2+ 4 · 3+ 4 · 4+ 4 · 5+ 4 · 6

=6

k=1(4k).

(b) −4− 2+ 0+ 2+ 4+ 6+ 8

Solution:

−4− 2+ 0+ 2+ 4+ 6+ 8 = 2(−2)+ 2(−1)+ 2 · 0+ 2 · 1+ 2 · 2+ 2 · 3+ 2 · 4

=4

k=−2(2k).

155 MAT1512/1

II. Mathematical induction

We consider here a special kind of proof method called mathematical induction.

To read: Smith and Minton, section 4.2, pp 359–360

Outcomes:After studying this section you should

(a) know what the principle of mathematical induction is

(b) be able to prove using mathematical induction that a statement P (n) is true for all natural numbers n

Examples:

II. 2.1 Use mathematical induction to prove the following statement for all positive integers n :

1+ 4+ 7+ . . .+ (3n − 2) = n(3n − 1)2

.

Solution:Let P (n) be the statement

1+ 4+ 7+ . . .+ (3n − 2) = n(3n − 1)2

(∗)for any positive integer n.

We first show that P (1) is true.

LHS of (∗) = 3 · 1− 2= 1.

RHS of (∗) = 1(3 · 1− 1)2

= 22

= 1.

Thus P (1) is true.

Suppose that P (k) is true for any k, that is, we assume

1+ 4+ 7+ . . .+ (3k − 2) = k(3k − 1)2

.

156

We want to prove that P (k + 1) is true, in other words:

1+ 4+ 7+ . . .+ (3k − 2)+ (3(k + 1)− 2)

= (k + 1) (3(k + 1)− 1)2

= (k + 1)(3k + 2)2

.

Now

1+ 4+ 7+ . . .+ (3k − 2)+ (3(k + 1)− 2)

= k(3k − 1)2

+ (3k + 1)

= 12[k(3k − 1)+ 2(3k + 1)]

= 123k2 − k + 6k + 2

= 12(3k2 + 5k + 2)

= (k + 1)(3k + 2)2

.

Since both conditions of the principle of mathematical induction have been satisfied, it follows that

1+ 4+ 7+ . . .+ (3n − 2) = n(3n − 1)2

is true for all integers n ≥ 1.

II. 2.2 Prove that

1 · 2+ 2 · 3+ 3 · 4+ . . .+ n(n + 1) = n(n + 1)(n + 2)3

for all n ∈ N.

Solution:We must verify that the equation is true for n = 1.The LHS is 1(1+ 1) = 2.The RHS is

1(1+ 1)(1+ 2)3

= 2.So the equation is true for n = 1.

Suppose that the equation is true for n = k, that is

1 · 2+ 2 · 3+ 3 · 4+ . . .+ k(k + 1) = k(k + 1)(k + 2)3

(∗)

157 MAT1512/1

We now use the assumption (∗) to deduce that the equation is true for n = k + 1, in other words:

1 · 2+ 2 · 3+ 3 · 4+ . . .+ k(k + 1)+ (k + 1) [(k + 1)+ 1]

= (k + 1) [(k + 1)+ 1] [(k + 1)+ 2]3

.

Now,

LHS = 1 · 2+ 2 · 3+ 3 · 4+ . . .+ k(k + 1)+ (k + 1) [(k + 1)+ 1]

= k(k + 1)(k + 2)3

+ (k + 1) [(k + 1)+ 1]

= (k + 1) k(k + 2)3

+ (k + 1)+ 1

= (k + 1) k2 + 2k + 3k + 63

= (k + 1) k2 + 5k + 63

= (k + 1)(k + 2)(k + 3)3

= (k + 1) [(k + 1)+ 1] [(k + 1)+ 2]3

= RHS.

Since the equation is true for n = 1, and if it is true for n = k, then it is true for n = k + 1. We conclude bymathematical induction that the equation is true for all integers n ≥ 1.

II. 2.3 Prove that

3+ 32 + 33 + . . .+ 3n = 3n+1 − 32

for all integers n ≥ 1.Solution:We must verify that the equation is true for n = 1.The LHS is 31 = 3.The RHS is

31+1 − 32

= 3So the equation is true for n = 1.

Suppose that the equation is true for n = k, that is

3+ 32 + 33 + . . . 3k = 3k+1 − 32

(∗)

158

We now use the assumption (∗) to deduce that the equation is true for n = k + 1, in other words:

3+ 32 + 33 + . . .+ 3k + 3k+1 = 3(k+1)+1 − 3

2.

Now,

LHS = 3+ 32 + 33 + . . .+ 3k + 3k+1

= 3k+1 − 32

+ 3k+1

= 3k+1 − 3+ 2 · 3k+12

= 3k+1(1+ 2)− 32

= 3k+1 · 3− 32

= 3(k+1)+1 − 32

= RHS.

Since the equation is true for n = 1, and if it is true for n = k, then it is true for n = k + 1. We conclude bymathematical induction that the equation is true for all integers n ≥ 1.

II. 2.4 Use mathematical induction to prove that

34

n

<34for n ≥ 2.

Solution:Our starting point is n = 2, so we first verify that the inequality is true for n = 2.

LHS is34

2

= 34· 34<34=RHS since 0 < 3

4< 1.

So the inequality is true for n = 2.

Suppose that the inequality is true for n = k, that is34

k

<34

(∗)

We now use the assumption (∗) to deduce that the inequality is true for n = k + 1, in other words34

k+1<34.

159 MAT1512/1

Now,

LHS = 34

k+1

= 34

k

· 34

<34· 34

by (∗)

<34

by the case n = 2

= RHS.

Therefore if the inequality is true for n = k, it is true for n = k + 1. By mathematical induction we conclude that itis true for all positive integers n ≥ 2.

III. The Binomial Theorem

To read: Smith and Minton, section 2.3, pp 171–172 and section 8.8, p 691. It is not necessary to learn the proof ofthe Binomial Theorem (3.4) on p 171, but you may know it as another example of “proving by using mathematicalinduction”.

Outcomes: After studying this section you should

(a) know what is meant by

• binomial expansion• binomial coefficients• Pascal’s triangle

160

(b) be able to

• expand a polynomial using Pascal’s Triangle• expand a polynomial using the Binomial Theorem• find the general term of a binomial expansion• find a particular term in a binomial expansion

Examples:

III. 3.1 Expand and simplify:

(a) (3x − 2)4

Solution:

We use the Binomial Theorem with a = 3x and b = −2,

(3x)4 + 4(3x)3(−2)+ 4 · 31 · 2 (3x)

2(−2)2 + 4 · 3 · 21 · 2 · 3(3x) · (−2)

3 + (−2)4

= 81x4 − 216x3 + 216x2 − 96x + 16.

(b)x2+ 2y

5

Solution:

We use the Binomial Theorem with a = x2and b = 2

y,

x2+ 2y

5

= x2

5 + 51x2

4 2y+ 5 · 41 · 2

x2

3 2y

2

+ 5 · 4 · 31 · 2 · 3

x2

2 2y

3

+5 · 4 · 3 · 21 · 2 · 3 · 4

x2

2y

4

+ 2y

5

= 132x5 + 5

8x4

y+ 5x

3

y2+ 20 x

2

y3+ 40 x

y4+ 32y5.

161 MAT1512/1

(c) 2a − 1a2

6

Solution:

2a − 1a2

6

= (2a)6 + 61(2a)5 − 1

a2+ 6 · 51 · 2(2a)

4 − 1a2

2

+ 6 · 5 · 41 · 2 · 3(2a)

3 − 1a2

3

+6 · 5 · 4 · 31 · 2 · 3 · 4(2a)

2 − 1a2

4

+ 6 · 5 · 4 · 3 · 21 · 2 · 3 · 4 · 5(2a) − 1

a25

+ − 1a2

6

= 64a6 − 192a3 + 240− 160 1a3+ 60 1

a6− 12 1

a9+ 1a12.

III. 3.2 Find the x5 y3 term in the expansion of (2x − y)8.

Solution:The general term in the binomial expansion is

8r(2x)8−r (−y)r = 8

r28−r x8−r (−1)r yr .

We need 8− r = 5, that is, r = 3.The required term is

8328−3 x8−3(−1)3 y3 = −8 · 7 · 6

1 · 2 · 3 25 x5 y3

= −1792 x5 y3.

III. 3.3 Find the a8 b4 term in the expansiona2+ b

2

3

10

.


10r

a2

10−r b2

3

r

= 10r2r−10 a10−r 3−r b2r .

We need 10− r = 8, that is, r = 2.The required term is

10222−10 a10−2 3−2b2· 2 = 10 · 9

1 · 2 · 2−8 a83−2 b4 = 5

256a8 b4.

162

III. 3.4 Find the term that contains y10 in the expansion of (x − 2y2)8.


8rx8−r (−2y2)r = 8

rx8−r (−2)r y2r .

We need 2r = 10, that is, r = 5.The required term is

85x8−5 (−2)5 y2·5 = 8 · 7 · 6

1 · 2 · 3 x3(−2)5 · y10

= −1792 x3 y10.

Extra reading:

Stewart, J, Redlin, L, and Watson, S. Precalculus, 5th edition (or 4th or 3rd edition).Chapter 11: Sections 11.1, 11.2, 11.3, 11.5 and 11.6 (or chapter 10.6 in the 4th and3rd editions)

mat 1512 (calculus a)

Documents

b functions

logarithmic functions

derivatives of exponential

e functions

c examples

simplification of functions

logarithmic dierentiation

partial derivatives