chapter 4 differentiation

LFSE012 Sci/Eng Mathematics A Differentiation

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CHAPTER 4 DIFFERENTIATION

4.1 INTRODUCTION

Differential calculus involves the study of finding the rate of change of one quantity with respect

to another.

We can represent the motion of a car with a distance –time graph.

When a car travels at a constant speed (50

km.h–1

) , it has a straight line-graph. Its rate of

change of distance with respect to time is

constant. The gradient of the graph represents

the speed of the car.

dist (km)

200

150

100

50

0

0 1 2 3 4 time (hour)

When the car’s rate of travel with respect to

time is not constant, the graph is not linear.

The gradient of the graph is constantly

changing. The gradient of the curve at each

point gives the speed of the car at that point.

dist (km)

200

150

100

50

0

0 1 2 3 4 time (hour)

The process of differentiation is used to find the rate of change of one quantity with respect to

another, especially when the rate of change is not constant.

4.1.1 FUNCTION NOTATION

Mathematical equations can be written in Cartesian form or function notation.

A function is a rule for which each x-value has a unique y-value.

Function notation is often used in differential calculus.

Cartesian form Function notation

2

2

2

2

2

When = 4, = 4 =16

When = 0, = 0 =0

When = , =

When = , =

y x

x y

x y

x h y h

x x h y x h

2

2

2

2

2

4 4 16

0 0 0

( )

( )

( )

( )

( )

f x x

f

f

f h h

f x h x h

4.1.2 LIMITS

We will consider what happens to the value of a function ( )f x as the value of x approaches

some value. Such a value is called a limiting value. The notation we use is:

lim x a

f x

This is read as “ the limit , as x approaches a, of ( )f x ”


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Example 1

Evaluate each of the following limits (where they exist) :

2

4 3

9a lim 2 (b) lim

3x x

xx

x

( )

Solution:

(a) The function ( )f x = 2 x is a continuous function. So the limit can be evaluated by

substitution.

4

lim 2 =2 4=8 x

x

(b) The function ( )f x = 2 9

3

x

x

is not a continuous function. It does not exist at the point

where x = 3 (we cannot evaluate the expression 0

0, because we cannot divide by 0)

However, if 3x : 2 3 39

3

x xx

x

3x 3x

So 2

3 3

9lim lim 3 6

3x x

xx

x

*The limit exists at x = 3, but the function does not exist.

Theorems of Limits

* In the following, and are different functions of lim and lim( ) ( ) . ( ) , ( )x a x a

f x g x x f x L g x M

1. If ( )f x = c , lim x a

f x c

*If the function is independent of x, the limit is a

constant.

e.g. If ( )f x = 5 , 2

lim 5x

f x

2. lim limx a x a

f x g x L M

( ) ( )

*We can find the limit of a polynomial term by

term.

e.g. If ( )f x = x2 + 3x ,

2

1 1 1lim lim lim 3 1 3 4x x x

f x x x

3. lim lim x a x a

f x g x L M

( ) . ( ) .

*We can split the limit of a product into 2 limits.

e.g. If ( )f x = (x+3) (2x-1) ,

4 4 4lim lim 3 lim 2 1

7 7 49

.x x x

f x x x

4. limlim 0

lim

x a

x a

x a

f xf x LM

g x g x M

( )( )

,( ) ( )

*We can split the limit of a quotient into 2 limits.

e.g. If 3

2 1( )

xf x

x

,

4

4

4

lim 3lim

lim 2 1

7 1

7

x

x

x

xf x

x


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Exercise 4.1

Evaluate each of the following:

1. 4

lim 6x

x

2. 2

1lim 3x

x x

3. 0

lim 5x

4.

2

3 2lim

2x

x x

x

5. 3

0

3lim x

x x

x

6. 2

3

25lim

5x

x

x

7.

2

0

5 25lim h

h

h

8. 2

0

5 4lim h

x h xh

h

9. 2

0lim , where

( ) ( )( )

h

f x h f xf x x

h

4.2 DERIVATIVE OF A FUNCTION

4.2.1 GRADIENT OF A STRAIGHT LINE

The gradient of a straight line is given by:

m = rise

run

For points 2 11 1 2 2

2 1

P and Q m, , ,y y

x y x yx x

The gradient is constant.

y

y = ( )f x

Q(x2,y2)

P(x1,y1)

x

4.2.2 GRADIENT OF A CURVE

The gradient of a curve is constant changing. We find the gradient at any point by drawing a

tangent line at that point.

How do we find the gradient at a point mathematically?

On the curve ( )y f x :

P is the point ,x f x

Q is the point ,x h f x h

We draw a straight line (called a secant)

to join P and Q.

The gradient of this straight line is :

rise

run

f x h f xkm

h h

y y = ( )f x

Q(x+h,f(x+h))

P(x,f(x)) h

x

k

As Q gets closer to P :

h gets smaller and

the gradient of the secant line gets closer to the tangent line at point P.


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So….

0

Gradient of tangent line at P = limh

f x h f x

h

This expression is called the derivative of the function ( )f x and is written as ( )f x

0 = lim

h

f x h f xf x

h

Example 1

If 2( )f x x x , find ( )f x

Solution: 2( )f x x x and

2 2 22( )f x h x h x h x xh h x h

0

2 2 2

0

2

0

= lim

2 lim

lim

h

h

h

f x h f xf x

h

x xh h x h x x

h

x

22xh h x 2h x x

2

0

0

2 lim

lim

h

h

h

xh h h

h

h

2 1x h

h

0

lim 2 1 2 1h

x h x

This process of finding the derivative of a function is called differentiation from first principles.

4.2.3 CARTESIAN NOTATION

On the curve ( )y f x :

P is the point ,x y

Q is the point ,x x y y

means a small change in ;

means a small change in

*

* .

x x

y y

The derivative can be written as dy

dx

y y = ( )f x

P(x,y)) x

x

y

Q ,x x y y

0 0 = lim lim

x x

f x x f xdy yf x

dx x x


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Example 1

Differentiate 25y x from first principles.

Solution:

2 2

22 2

5 5

5 2 5

5

y f x x f x

x x x

x x x x x

2x 2

2 5x x x 2x

2

0

2

0

0

2

lim

2 lim

lim 2 2

x

x

x

x x x

f x x f xdy

dx x

x x x

x

x x x

Exercise 4.2

1. Find from first principles, the derivatives of the following:

(a) 2( )f x x

(b) 2 4 3y x x

(c) 4 3( )f x x x

2. For the function 3 4( )f x x x find

(i) f x h f x

h

(ii)

3 3f h f

h

(iii)

0

3 3limh

f h f

h

The technique used in finding derivatives from first principles is very time-consuming. There are

rules which enable us to write down the derivatives of certain functions. These are called

standard derivatives.

4.3 DERIVATIVE OF ncx

2 2

3 2 3 2

We have observed that the derivative of is 2 this can be written as 2

the derivative of is 3 3 and so on.

:

dx x x x

dx

dx x x x

dx

From this we can make a general rule:

1If then ,n nf x cx f x ncx

where c is any constant, and n is any real number.


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Example 1

Find the derivatives given the following functions:

(a) 61

2( )f x x (b) 4y x

Solution:

6

5 5

1(a)

2

1 6 3

2

( )

( )

f x x

f x x x

12

1 12 2

1

(b) 4 4

1 2 4 2

2

y x x

dyx x

dx x

4.3.1 DERIVATIVE OF A CONSTANT

If then 0,f x c f x

4.3.2 DERIVATIVE OF A POLYNOMIAL

If a function is made up of the sum of a number of terms, the expression is called a polynomial.

e.g. 3 24 5 3 4f x x x x

To find the derivative of a polynomial, differentiate each term in order.

3 2

3 2

2

e.g. If 4 5 3 4

d d d d 4 5 3 4

dx dx dx dx

12 10 3

f x x x x

f x x x x

x x

Example 1

Find the derivatives of the following polynomials:

(a) 4 3 23 2 7 7( )f x x x x x (b) 3 35y x x

x

Solution:

4 3 2

3 2 1 0

3 2

3 2 7 7

3 4 2 3 7 2 1 0

12 6 14 1

( ) ( )a f x x x x x

f x x x x x

x x x

1

31 1 12

2 2 2 2

12

3 1 12 2 2

31 12 2 2

3 3

1 1 1

3 3(b) 5 = 5 5 3

3 1 1 5 3

2 2 2

3 5 3

2 2 2

3 5 3

2 2 2

y x x x x x x xx x

dyx x x

dx

x x x

xx x x


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Exercise 4.3

For questions 1 –10, find the derivatives of the given functions.

1. 4 3( )f x x 2. 24 5 3y x x

3. 4 3 21 1 1

4 3 2( )f x x x x x 4. 24 5 3y x x

5. 1 12 22 3( )f x x x

6.

3

25( )f x x

x

7. 2 1 3y x x x 8. 2y ax bx c

9. 3 24 3 2

0( ) ,x x

f x xx

10.

32( )f x x

11. If 23 7 3y x x , find the value of y for which dy

0dx

12. For 3 2 3 , findf x x x

(i) 0 (ii) the value of for which =0 f x f x

13. Find the gradient of the curve 2 4 3y x x at the y-intercept.

14. Find the gradient of the curve 2 3 10f x x x at the points where the graph cuts the

x-axis.

15. For the function : 2

1 2f x x x find :

(a) 0( )f (b) the value(s) of x for which 0( )f x

(c) ( )f x (d) 0( )f (e) the value(s) of x for which 0( )f x

4.4 THE PRODUCT AND QUOTIENT RULES

4.4.1 DERIVATIVE OF A PRODUCT

When two functions are multiplied together, the derivative of the product can be found using the

product rule. The proof of this rule is in appendix A.

The product rule states:

If where and

then

y uv u u x v v x

dy du dvv u

dx dx dx

In function notation: If then f x u x v x f x v x u x u x v x


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Example 1

Find if 5 3 2 1dy

y x xdx

Solution:

Let 5 3 5 and let 2 1 2

2 1 5 5 3 2

10 5 10 6

20 1

du dvu x v x

dx dx

dy du dvv u x x

dx dx dx

x x

x

Example 2

Find 2 if 2 3 2 5dy

y x x xdx

Solution

2

2

2 2

2

Let 2 3 4 3 and let 2 5 2

2 5 4 3 2 3 2

8 6 20 15 4 6

12 32 15

du dvu x x x v x

dx dx

dy du dvv u x x x x

dx dx dx

x x x x x

x x

Example 3 y = (2x

2 + 6x) (2x

3 + 5x

2)

4.4.2 DERIVATIVE OF A QUOTIENT

When one function is divided by another, the derivative of the quotient can be found using the

quotient rule. The proof of this rule is in appendix C.

The quotient rule states:

2

If where and

then

u

y u u x v v xv

du dvv u

dy dx dx

dx v

In function notation:

2If then

u x v x u x u x v xf x f x

v x v


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Example 1

Find 3

if 4 1

dy xy

dx x

Solution:

2 2

2 2

Let 3 1 and let 4 1 4

4 1 1 3 4

4 1

4 1 4 12 13

4 1 4 1

du dvu x v x

dx dx

du dvv u x xdy dx dx

dx v x

x x

x x

Example 2

Find

34 if

2 6

dy xy

dx x

Solution:

3 2

2 3

22

3 2 3

2

3 2

2

Let 4 12 and let 2 6 2

2 6 12 4 2

2 6

24 72 8

2 6

416 72

2 6

du dvu x x v x

dx dx

du dvv u x x xdy dx dx

dx v x

x x x

x

x x

x

3 24 18

4

x x

2

2 2

2 2 9

3 3

x x

x x

Exercise 4.4

Find the derivative of each of the following functions.

1. 7 4 3y x x 2. 2 42 4f x x x

3. 3 2

2 3

xy

x

4. 1²y x x

5. 3

4

xy

x

6.

2

4

2

4

xy

x

7. 2 3f x x x 8. 2 3

1 2f xx x

9. 2 42 5y x x 10. 2 1x

yx


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4.5 THE CHAIN RULE

Some functions are composed of two functions, one inside the other. For example:

Example 1

2

4 4 is subtraced from ; then the result is squared.f x x x ( ) .

This can be written as 2 where 4( ) ,f x u u x

Example 2

2 4 4 is added to the square of ; then the square root of the result is taken.y x x .

This can be written as 2 where 4,y u u x

Functions like these are called composite functions.

The derivative of a composite function can be found by means of the chain rule. The proof of this

rule is in appendix B.

The chain rule states:

If where theny y u u u x

dy dy du

dx du dx

In function notation: If where then ' f x f u u u x f x f u u x

Note: To solve problems, first decide on an appropriate substitution.

Example 3

Use the chain rule to find the derivative of the function 5

3 2y x

Solution:

5 4

44 4

Let 3 2 =3

So =5

5 3 15 15 3 2. .

duu x

dx

dyy u u

du

dy dy duu u x

dx du dx

Example 4

Find the derivative of the function 2 3( )f x x x

Solution:

1 12 2

12

12

2Let 3 2 3

1So

2

1 2 3 2 32 3

2 22

( )

( ) .

u x x u x x

f u u u f u u

x xf x f u u x u x

uu


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Exercise 4.5A

In each of the following examples, find the derivative of the given function.

1. 3

3 2y x x 2. 6

33 1( )f x x

3. 8

42 4y x 4. 4

26 3( )f x x x

5. 5 4y x 6.

3

2

2

3 1y

x x

7. 4 32y x x 8. 15210 3( )f x x x

9.

5

3 2

4

2 3 15y

x x x

10. 3 2 5y x

4.5.1 DERIVATIVES INVOLVING CHAIN RULE AND OTHER RULES

Example 1

Find the derivative of the function

2

2 3

3 1

xy

x

Solution:

2

Let 2 3 and let 3 1

2 2(3 1) (3 1)

6(3 1)

u x v x

du dv dx x

dx dx dx

x

du dv u

dy dx

dx

2

42

3 1 (2) (2 3)6(3 1)

3 1

(3 1)

vx x xdx

v x

x

4

2(3 1) 6(2 3)

3 1

x x

x

3

3

3

6 2 12 18

3 1

2(8 3 )

3 1

x x

x

x

x

Exercise 4.5B

Find the derivatives of the following functions:

1. 3

5 7 5y x x 2. 5

32 7 3 5y x x

3.

32

5

xf x

x

4.

2

2 3

4

xy

x

5. 22 5y x x 6. 22 3 6y x x


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4.6 DERIVATIVE OF TRIGONOMETRIC FUNCTIONS

The derivation of the derivatives of sin, cos and tan can be found in appendix 1 at the end of the

notes.

DERIVATIVE OF sin x

sin cos d

x xdx

DERIVATIVE OF sin ax

sin

Let

sin cos

By chain rule cos cos : .

y ax

duu ax a

dx

dyy u u

du

dy dy duu a a ax

dx du dx

sin cos d

ax a axdx

DERIVATIVE OF cos x

cos sin d

x xdx

DERIVATIVE OF cos ax

The derivation of the derivative of cos ax can be found using the chain rule:

cos sin d

ax a axdx

DERIVATIVE OF tan x

2tan sec d

x xdx

DERIVATIVE OF tan ax

Similarly if we use the chain rule we obtain:

2tan sec d

ax a axdx


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Example 1

Find down the derivatives of the following functions:

2

tan 2(a) sin 3 (b) cos sin 3 (c) .

xy x f x x x y

x

Solution:

(a) sin 3

3 cos 3

(b) cos sin 3

cos sin 3 + sin 3 cos (product rule)

cos .3cos 3 sin 3 sin

3

y x

dyx

dx

f x x x

d df x x x x x

dx dx

x x x x

.

2

2 2

22

2 2

4

2

cos cos 3 sin sin 3

tan 2(c)

tan 2 tan 2

(quotient rule)

2 sec 2 2 tan 2

2 sec 2 ta

x x x x

xy

x

d dx x x x

dy dx dx

dx x

x x x x

x

x x x

.

4

2

3

n 2

2 sec 2 tan 2

x

x

x x x

x

Exercise 4.6A

Find the derivatives of the following functions

12

sin 2 . 4 tan

cos 3

1. 2

3. 4.

y x y x

y x f

2

2 12

2

6 sin3

cos 3 2 1 tan .

sin

3 cos 3. .

tan 2

5. 6

7 8

x x x

xf x x x x y

x

x xy y

x x


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The derivatives of composite functions involving sine, cosine and tangent can be found by using

substitution and the chain rule. Examples of composite functions are 3 2sin cos 2 5,x x x

3and tan x

Example 2

Find the derivatives of the following functions:

3 2 3a sin b cos 2 5 c tan y x y x x y x

Solution:

3

3 2

2 2 3

a sin

Let 3

sin cos

cos 3 3 cos . .

y x

duu x x

dxdy

y u udu

dy dy duu x x x

dx du dx

2

2

2

3

1 2

3 3 3

b cos 2 5

Let 2 5 2 2

cos sin

sin 2 2 2 1 sin 2 5

c tan

1 Let

3

y x x

duu x x x

dxdy

y u udu

dy dy duu x x x x

dx du dx

y x

duu x x x

dx

. .

2

2 32 2

2 2 33 3

3 2

tan sec

sec1 1 sec sec

3 3 3

dyy u u

du

xdy dy duu x x x

dx du dx x

. .

Exercise 4.6B


3

2

sin 3 5 . 3 tan 5

cos si

1. 2

3. 4.

y x y x

xy x f x

2

2

n 4cos

tan . sin

. 3 sin 2 . cos sin 4

5. 6

7 8

xx

f x x x yx

y x y x


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4.7 DERIVATIVE OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS

The derivation of the derivatives of and xln x e can be found in the appendix 2 at the end of the

notes.

DERIVATIVE OF ln x

1

d

ln xdx x

DERIVATIVE OF ex

x xde e

dx

DERIVATIVE OF eax

ax axde a e

dx

Example 1


33 2 1

2(a) (b) 3 (c) (d) sin xx xy e f x ln x y e y ln

Solution:

3 2

3 2

(a) (b) 3

3 Let 3 2

x

x

y e f x ln x

dy due u x x

dx dx

3

2

1

1 2 2

3

(c)

. .

x

dyy ln u

du u

dy dy du xx

dx du dx u x

y e

12

3 2 1 1 12 2 2

(d) sin x

Let 3 Let sin x cos

u u

y ln

du duu x x u x

dx dx

dyy e e

du

2 2 1 1

2 2

1

1 3 3 cos

. . . .u u

dyy ln u

du u

dy dy du dy dy due x x e x

dx du dx dx du dx u

1 1

2 2 1 1

2 21

2

cos cot x

sin x

x


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Exercise 4.7


2 1

4

5 2 .

1. 2

3. 4.

x

x x

y ln x y e

y e f x e

3 2 2

2

3

2 .

6. .

5. 6

7 8

x

x lnx

e

f x ln x y x ln x xx

ln xy y x e

x

4.8 DERIVATIVE OF INVERSE TRIGONOMETRIC FUNCTIONS

The derivation of the derivatives of 1 1 1 Sin , Cos and Tanx x x can be found in the appendix 3

at the end of the notes.

DERIVATIVE OF -1

sin x

1

2

1sin for 1 < < 1

1

dx x

dx x

DERIVATIVE OF -1

cos x

1

2

1cos for 1 < < 1

1

dx x

dx x

DERIVATIVE OF -1

tan x

1

2

1tan

1

dx

dx x

Example 1


1 1 2 1(a) sin 5 (b) cos (c) tanxy x y x y e x

Solution:

1 1 2

2

(a) sin 5 (b) cos

Let 5 5 Let 2

y x y x

du duu x u x x

dx dx

1 1

2 2

2 2 2 4

1

1 1 sin cos

1 15 5 2 2

1 1 25 1 1

(c) tan x

dy dyy u y u

du duu udy dy du dy dy du x x

dx du dx dx du dxu x u x

y e x

1 1

1 1

2 2

(Product rule)

tan +tan

1 1 tan tan

1 1

x x

x x x

dy d de x x e

dx dx dx

e e x e xx x


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Exercises 4.8


1 1 12

1 3

sin 3 2 . tan

cos sin

1. 2

3. 4.

y x y x

y x f x x

1

11tan

. sin 2 1

5. 6

x

xf x y ln x

x

4.9 IMPLICIT FUNCTIONS

We have found derivatives for functions where y f x ; these functions are called explicit

functions ( is in terms of y x ).

However many functions do not have the x and y terms separated. These are called implicit

functions. Examples of implicit functions are : 2 23 5 8 ; 12 ; 6x y xy x y xy x

In some cases the (eg 3x+5y=12) the expressions can be easily separated.

In other equations ( eg. 2 2 6x y xy x ) this cannot be done.

All the above equations can be solved by a method called implicit differentiation.

Example 1

Find the derivative of 3 5 8x y by means of implicit differentiation

Solution:

Differentiate both sides of the equation with respect to x.

3 5 8

3 5 8

3 5 8

3 33 5 0

5 5

d dx y

dx dx

d d dx y

dx dx dx

d d dx y

dx dx dx

dy dy

dx dx

Example 2

Find the derivative of xy =12 by means of implicit differentiation

Solution:


12 Use the product rule for

0

0

:d d d

xy xydx dx dx

d dx y y x

dx dx

dyx y

dx

dy y

dx x


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Example 3

Find the derivative of 2 3 2 3x y y by means of implicit differentiation

Solution:

2 3

2 3 3

3

2

2

2 2

2 (3)

2 0 use chain rule for

2 (2 ) 0

2 3 2 0

3 2 2

2 2 or

3 2 2 3

d dx y y

dx dx

d d d dx y y y

dx dx dx dx

d dy d dyx y y

dy dx dy dx

dy dyx y

dx dx

dyy x

dx

dy x x

dx y y

Example 4

Find the derivative of 2 2 6x y xy x by means of implicit differentiation

Solution:


2 2

2 2 2 2

2 2 2 2 2

2 2 2

2

6

0 Use the product rule for and

0 Use chain rule for

2 1 1 0

d dx y xy x

dx dx

d d d d dx y xy x x y xy

dx dx dx dx dx

d d d d d dx y y x x y y x x y

dx dx dx dx dx dx

dy d dyx y x x y y

dx dy dx

dyx

d

2

2 2

2 2

2 2

2

2

2 2 1 0

2 2 1 0 Rearrange terms

+2 2 1

2 2 1

2 1

2

dyxy x y y

x dx

dy dyx xy xy y

dx dx

dy dyx xy xy y

dx dx

dyx xy xy y

dx

dy xy y

dx x xy


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Example 5

Differentiate xy a where a is a constant.

Solution:

ln ln

ln

Differentiate with respect to :

ln ln

1 ln ln

0 ln 1

ln

x

x

y a

y a

x a

x

d dy x a

dx dx

dy d dx a a x

y dx dx dx

x a

dyy

dx

lnx

a

a a

Exercise 4.9

Find the derivatives of the following using implicit differentiation

2

2 3 2

4 +3 7 2 5

6

x y x y

x y x

1. 2.

3. 4.2

2 2 2 2

2

16

1 3 4 12

3 4 2 3 2 4

y

y x x y

x y y xy x y

5. 6.

7. 8.

2 2 2

2

3

2 4 3 7 2

1

5 2

3

x x

x

xx y xy y

x

y y

y

9. 10.

11. 12.

13.

2 2 3 2 2

2

sin 10

2 = 3 4 2

2

sin

xy y

x yx y x y x y x xy y

x

xy

14.

15. 16.

17. cos x y x ln y y ln x x 18.


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4.10 PARAMETRIC EQUATIONS

A function y f x has variables and x y .Sometimes x and y can both be expressed in terms of

another variable. This variable is called a parameter.

A curve given by , x x t y y t is called a parametrically defined curve and the functions

and x x t y y t are called the parametric equations for the curve.

Example 1 2

2 2

2 2

and

For square both sides :

So the parametric equations and represent the parabola

x t y t

x t

x t y

x t y t y x

:

Example 2

2 2 2 2

2 2 2 2

3cos and 3sin

Square both equations :

9cos ; 9sin Add:

9cos 9sin 9

The parametric equations 3cos and 3sin represent a circle of ra

x t y t

x t y t

x y t t

x t y t

dius 3 units.

4.10.1 DERIVATIVES OF PARAMETRIC EQUATIONS

The derivative of a function which is in parametric form can be found by applying the chain rule.

If where , :y f x x x t y y t

.dy dy dt dy dt

dx dt dx dx dt

Example 1

2Find the derivative for the parametric equations: dy

x t and y tdx

Solution:

2

1

2

22

1

Note : this derivative is in terms of the parameter . We can also express in terms of .

2

dxx t

dt

dyy t t

dt

dy dy dt tt

dx dx dt

dyt x

dx

dyx

dx


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Example 2

Find the derivative for the parametric equations: 4cos 4sin dy

x t and y tdx

Solution:

4 cos 4 sin

4 sin 4 cos

4 cos cot

4 sin

Note : this derivative is in terms of the parameter . We can also express in terms of .

4 cos 4 c

4 sin

dxx t t

dt

dyy t t

dt

dy dy dt tt

dx dx dt t

dyt x

dx

dy t

dx t

2 2 2 2

os 4 cos 4 cos

4 1 cos 16 16 cos 16 16 4 cos

t t t x

t t xt

Exercise 4.10

2 2

Find the derivative for the following pairs of parametric equations.

. 2 1 . sin

. sin 2 cos

1 ; 2 ;

3 ;

dy

dx

x t y t x t y t

x t y t

3 5

2 2

2

.

. 3 . 2 sin

. sin cos 1

4 ;

5 ; 6 ;

7 ;

t t

x t y t

x e y e x t y t

x t y t

2 3

3 2

. 1 1

. ; . 2 1 ; cos

8 ;

9 10 ,

x t y t

x t t y t t x t y t t

4.11 HIGHER ORDER DERIVATIVES

3 2 2

2

is the derivative of the function

In fact is itself a function which may also have its own derivative.

For example, if 2 3 then 6 6

This function 6 6 can also be diffe

.

, .

dyy f x

dx

dy

dx

dyy x x x x

dx

x x

2

rentiated.

6 6 12 6 This is called the derivative. . second d dy d

x x xdx dx dx

We could also find third derivatives, fourth order derivatives and so on. Such derivatives are

called higher order derivatives.


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There are a number of different notations which are used for higher order derivatives.

For the function :y f x

2 2

2 2

3 2

3 2

First derivative is : or or

Second derivative is : or or or

Third derivative is : or

dy df x f x

dx dx

d y d dy df x f x

dx dx dx dx

d y d d y

dx dx dx

3

3 or or

And so on.

df x f x

dx

We can find derivatives of higher. In fact the first and second derivatives will be most important

to us, as they are used in sketching curves.

Example 1

If 2

3

26 sin2 , find and

dy d yy x x

dx dx

Solution: 2

2

218 2 cos 2 36 4 sin 2

dy d yx x x x

dx dx

Example 2

If , find and f x x ln x f x f x

Solution:

12

121

2

2 2

2 2

1 1 1 1 2

2 22

For the second derivative, we use the quotient rule:

2 2 2 2 2 2 2

42

2 4 4

4 4

. .

xf x x

x x xx

d dx x x x x x x

dx dxf xxx

x x x

x x

Exercise 4.11

Find the second derivatives of each of the following:

2 2 5 4 2

32 2

4 7 . 7 7 2 11 3

5. 2 . 3 + .

1. 2 3.

4 5 6

7.

y x x y x x f x x x x

f x x y x x y xx

2 2

. tan 2 3

1. sin sin 3 + cos .

2

8 9.

10 11. 12

x x

x

y e e f x x f x ln x

y x y x x y x e


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4.12 STANDARD DERIVATIVES

We have covered a number of derivatives which you are required to recognise. These are called

standard derivatives.

You do not have to memorise these derivatives; a table of standard derivatives (as shown below)

will be provided.

You should try to become very familiar with these derivatives, as you must be able to recofise the

standard derivative, and the rule which is to be used.

Standard Derivatives

or ( )y f x or dy

f xdx

c 0

nx n 1n x

ex ex axe axa e

ln x 1

x

sin x cos x

sin ax a cos ax

cos x – sin x

cos ax –a sin ax

tan x 2sec x

tan ax 2 seca ax

1sin x 2

1

1 x

1cos x 2

1

1 x

1tan x 2

1

1 x


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APPENDICES

APPENDIX A : PRODUCT RULE

For the product function where and

Let be a small increment in ; this produces small increments in and of

and respectively

,

, .

. . .

y uv u u x v v x

x x u v y

u v y

y y u u v v uv u v v u u v

So as 0 and 0

So

. . .

. . .

, , , ,

y u v v u u v

y v u vu v u

x x x x

y dy v dv u dux u

x dx x dx x dx

dy dv duu v

dx dx dx

APPENDIX B : QUOTIENT RULE

1 1 1

For the product function where and


and respectively

,

, .

uy u u x v v x

v

x x u v y

u v y

uy y u u v v u v v u v v

u

v v

u u u uvy

v v v

.v u uv

2

So as 0 and 0

So

. .

.. .

.

, , , ,

.

u v v u u v

v v v v v v

u vv u

y v u u v x x

x x v v v v v v

y dy u du v dvx v

x dx x dx x dx

du dvv u

dy dx dx

dx v


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APPENDIX C : CHAIN RULE

For the composite function where


and respectively

If 0 we can write

So as 0

.

,

. ,

y f u u u x

x x u y

u y

x

y y ux

x u x

0

and

So

,

, ,

.

u

y dy y dy u du

x dx u du x dx

dy dy du

dx dx dx

APPENDIX D : DERIVATIVES OF TRIGONOMETRIC FUNCTIONS

Derivatives for the trigonometric functions can be derived using the using the fundamental limits:

θ 0 θ 0

sin θ cos θ 1lim 1 lim 0

θ θ

Derivative of sin x

0 0

0

0

sin

+ sin sin cos +cos sin

sin sinlim lim

sin cos +cos sin sin lim

sin cos sin +c lim

x x

x

x

y x

y y x x x x x x

x x xdy y

dx x x

x x x x x

x

x x x

0

0 0

0 0

os sin

sin cos 1 +cos sin lim

sin cos 1 cos sin lim lim

cos 1 sin lim cos lim

x

x x

x x

x x

x

x x x x

x

x x x x

x x

xx x

x

sin

sin 0 cos 1

cos

x

x

x x

x

Derivative of cos x

cos

cos sin cos2 2 2

cos 1 sin 12

sin

.

. .

y x

d d dx x x x

dx dx dx

x x

x


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Derivative of tan x

The derivation of the derivative of tan x can be found using the quotient rule.

2

2 2

2 2

sin tan =

cos

sin tan =

cos

cos sin sin cos

= (Quotient rule)cos

cos cos sin sin cos sin = =

coscos

. .

. .

xx

x

d d xx

dx dx x

d dx x x x

dx dx

x

x x x x x x

xx

2

2

1 = sec

cosx

x

APPENDIX E : DERIVATIVE OF EXPONENTIAL AND LOGARITHMIC

FUNCTIONS

DERIVATIVE OF ln x

1

0

+ .

1

1

1 1

1 1

Let

1 1 1. 1 1

As 0 p 0.

lim

,

p

x

y ln x

y y ln x x ln x ln x

ln x x ln xy

x x

ln x x ln xx

x xln

x x

xln

x x

x xln

x x x

xp

x

yln p ln p

x x p x

x

dy y

dx

1

0

1

0

1 lim 1

1 lim 1

1 ***refer back to exponential functions

1

p

p

p

p

x

ln px

ln px

ln ex

x


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DERIVATIVE OF xe

Take natural logs of both sides

Differentiate both sides with respect to :

1

By the chain rule :

1

11

.

.

x

x

x

y e

ln y ln e x

x

d dln y x

dx dx

d dyln y

dy dx

dy

y dx

dyy e

dx

APPENDIX F : DERIVATIVE OF INVERSE TRIGONOMETRIC

FUNCTIONS

DERIVATIVE OF -1

sin x

2 2 2 2

2

sin

cos

1 cos 1 sin 1 cos 1 for 1 <x <1

cos

1 for 1 <x <1

1

x y

dxy

dy

dyy y x y x

dx y

dy

dx x

DERIVATIVE OF -1

cos x

2 2 2 2

2

cos

sin

1 sin 1 cos 1 sin 1 for 1 <x <1

sin

1 for 1 <x <1

1

x y

dxy

dy

dyy y x y x

dx y

dy

dx x

DERIVATIVE OF -1

tan x

2

2 2

2

2

2

tan

sec

1 And 1 + tan sec

sec

1

1 tan

1

1

x y

dxy

dy

dyy y

dx y

dy

dx y

x


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ANSWERS

Exercise 4.1

1. 24 2. –2 3. 5 4. 5 5. –3

6. –2 7. 10 8. 5x2 – 4x 9. 2x

Exercise 4.2

1. (a) 2x (b) 2x –4 (c) 2x –1

2. (i) 2 23 3 4x xh h (ii) 223 9h h (iii) 23

Exercise 4.3

1. –3 2. 8 5x 3. 3 2 1x x x 4. 212 5 1x

5. 2 3

2

x

x x

or

3

1 3

x 2 x 6.

2

2

5 6

2

x

x x

or

5

5 3

2 x x 7. 26 10 3x x

8. 2ax b 9. 3 2

2

8 3 2x x

x

10.

23 2x

11. 7 1

76 12

,x y 12. (i) 9 (ii) 3

4 13. 4

14. –7 , 7 15. (a) –2 (b) –1 , 2 (c) 23 3x (d) –3 (f) 1

Exercise 4.4

1. 25 6x 2. 4 22 3 4 4 x x x 3.

2

13

2 3

x 4. 23 1x

5.

2

2

2 6

4

x x

x 6.

2 4

24

2 4 4

4

x x x

x

7. 3 3

22 2

x

x

8. 3 2

12 1

x x 9. 2 42 5 4 3x x x 10.

23 1

2

x

x x

Exercise 4.5A

1. 2

3 23 2 3 2x x x 2. 5

2 354 1x x 3. 7

3 464 4x x

4. 5

224 6 1 3x x x

5. 5

2 5 4x 6.

4

2

6 2 3

3 1

x

x x

7. 2

4 3

2 3

2

x x

x x

8.

652

2 2 3

3

x

x x


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9.

2

63 2

20 3 4 3

2 3 15

x x

x x x

10.

2

3

2

3 2 5x

Exercise 4.5B

1. 2

2 10 23 5x x 2. 4

3 2 33 32 50 7 2 7x x x

3.

2

2

2 17 2

5

x x

x

4.

3

2 7

4

x

x

5.

2

2

4 5

2 5

x

x

6.

248 10 3

2 6

x x

x

Exercise 4.6A

2 12

2 2 21 1 12 2 2

2 cos2 . 2 sec 3 sin 3 2 9 cos3

3 2 1 sec 2 3 1 tan . cosec

6 tan 2 sec.

x x x x x

x x x x x x

x x x

1. 2 3. 4.

5. 6

7 2

2

2 6 sin3 cos 3 .

2tan 2

x x x x

x xx

8

Exercise 4.6B

2 2 3

2

2

1 3 cos 3 5 . 45 sec 5 sin

2

2 sin 4 2 cos 4 2 sec tan . 2

sin 4 2

x x x xx

x x x x x x xx

x x

1. 2 3.

4. 5. 6

2 2 cosec

. 6 cos 2 . cos sin sin 4

x

x y x x

7 8 .

Exercise 4.7

52 1 4

5

2 4

5 3 4 . 2 4

5 2 2

1 3 62 1. 2 . . 1

1. 2 3. 4. 5.

6 7 8

x x x x

x lnx

xe e e e

x x x

ln xxx e x x l

x x x n x

Exercise 4.8

1. 2

3

12 9 3 x x

2. 2

2

4 x 3.

2

6

3

1

x

x


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4. 1

2sin

1

xx

x

5. 2 1

2 2

2 1 2(1 ) tan

(2 1) (1 )

x x x

x x 6.

2

1

1 ln

x x

Exercise 4.9

2

2

4 2 3 . .

3 43

2 3 2 .

2 13

x x x xx

y y yy

xy y

xx

1. 2 3. 4. 5. 6

7. 8

2

2

2 23

22 2

3 8

2 2 3

1 2 1. . 5 ln 5 . 3 ln 2 2

2 1 2 1

3 ln 3

x x

x

y xy

x xy

xy x

y x y x

9.

10 / 11 12

13. 14.

3 2 2

4 2

2 2

3 2

1 4 4 .

cos 2 2 1

3 3 2 4 cos cos .

sin sin 2 4 2

.

y x y xy

x y x y x y

x y xy y x y y

x x yx x y x

y x y x ln y

x x y ln x

15

16 17.

18

Exercise 4.10

2cos 5. . . 2 tan . . 2

2 3

cos 3 2 1. . 2 sin . .

2 2

1 2 3 4 5

6 7 8 9

tt tt t e

t

t t tt

t t

2 cos sin .

61

10

t t t

t

Exercise 4.11

3 2

3

33

2

3

10 2 . 14 40 132 6 .

8 3 3 2. 2 .

4 94

2 sin . or 2sec t

cos

x x

x xx

x xor e e

x x x xx

xx

x

1. 2 3. 4

5 6 7.

8

2

2

4an . 2 cos 2

2 3

1 1 9 sin 3 cos . 4 2

4 2

x

x xx

x x x x e

9. 10

11. 12

chapter 4 differentiation

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