chapter 3 introduction to the derivative sections 3.5, 3.6, 4.1 and 4.2

Chapter 3Introduction to the

Derivative

Sections 3.5, 3.6, 4.1 and 4.2

Introduction to the Derivative

Average Rate of Change

The Derivative

change in

change in

f f

x x

( ) ( )f b f a

b a

The average rate of change of f (x) over the interval [a , b] is

The change of f (x) over the interval [a,b] is

( ) ( )f f b f a

Difference Quotient

Average Rate of Change

Average Rate of Change

, ( )P a f a

, ( )Q b f b

secant line has slope mS

( ) ( )S PQ

f f b f am m

x b a

Is equal to the slope of the secant line through the points (a , f (a)) and (b , f (b)) on the graph of f (x)

x b a

( ) ( )f f b f a

The average rate of change of f over the interval [a , b] can be written in two different ways:

( ) ( ) ( ) ( )f f b f a f a h f a

x b a h

Average Rate of Change as h0

We now look at the behavior of the average rate of change of f (x) as b a h gets smaller and smaller, that is, we will let h tend to 0 (h0) and look for a geometric interpretation of the result.For this, we consider an example of values of

and their corresponding secant lines.

( ) ( )S PQ

f f a h f am m

x h

Average Rate of Change as h0

2.57h

Average Rate of Change as h0Secant line tends to become the Tangent line

2h

Average Rate of Change as h0Secant line tends to become the Tangent line

1.5h

Average Rate of Change as h0Secant line tends to become the Tangent line

1h

Average Rate of Change as h0Secant line tends to become the Tangent line

0.5h

Average Rate of Change as h0Secant line tends to become the Tangent line

0.2h

Average Rate of Change as h0Secant line tends to become the Tangent line

Zoom in

0.1h

Average Rate of Change as h0Secant line tends to become the Tangent line

0.1h

Average Rate of Change as h0Secant line tends to become the Tangent line

0.05h

Average Rate of Change as h0Secant line tends to become the Tangent line

( ) ( ) smallT S

f a h f am m h

h

We observe that as h approaches zero,

1. The secant line through P and Q approaches the tangent line to the point P on the graph of f.

2. and consequently, the slope mS of the secant line approaches the slope mT of the tangent line to the point P on the graph of f.

Average Rate of Change as h0

That is, the limiting value, as h gets increasingly smaller, of the difference quotient

is the slope mT of the tangent line to the graph of the function f (x) at x = a.

( ) ( )S

f a h f am

h

Instantaneous Rate of Change of f at x = a

(slope of secant line)

This is usually written as

( ) ( )as 0S T

f a h f am m h

h

Instantaneous Rate of Change of f at x = a

tends to, orapproaches

That is, mS approaches mT as h tends to 0.

More briefly using symbols by writing

0

( ) ( )limTh

f a h f am

h

Instantaneous Rate of Change of f at x = a

and read0

limh

as “the limit as h approaches 0 of ”

that is, the limit symbol indicates that the value of

can be made arbitrarily close to mT by taking h to be a sufficiently small number.

( ) ( )S

f a h f am

h

Instantaneous Rate of Change of f at x = a

Definition: The instantaneous rate of change of f (x) at x = a is defined to be the slope of the tangent line to the graph of the function f (x) at x = a. That is,

0

( ) ( )lim "instantaneous rate of change at "Th

f a h f am a

h

Instantaneous Rate of Change of f at x = a

Remark: The slope mT gives a precise indication of how fast the graph of f (x) is increasing or decreasing at x = a.

Consider the function f (x) – x2/4 + 9/4, whose graph is given below, at the points a – 3 and a – 1

A Concrete Example

At which of these two points is the function increasing faster?

Intuition says at x – 3 because we notice the graph is steeper at the point x – 3. Why?

A Concrete Example

Because our brain makes no distinction between the graph of f and the tangent line to the graph at the point in question.

To make this more obvious we zoom in

Consider the function f (x) – x2/4 + 9/4, whose graph is given below, at the points a – 3 and a – 1

A Concrete Example

We see how the tangent line basically coincides with the graph of f near the point of contact. What is the slope of each line? 1.5 at -3Tm x

0.5 at -1Tm x

Consider the function f (x) – x2/4 + 9/4, whose graph is given below, at the points a – 3 and a – 1

A Concrete Example

That is, at the points of contact, the function is increasing as fast as its corresponding tangent line.

Consider the function f (x) – x2/4 + 9/4, whose graph is given below, at the points a – 3 and a – 1

A Concrete Example

We now verify the claims by explicitly computing for this function, the limit,

0

( ) ( )limTh

f a h f am

h

Consider the function f (x) – x2/4 + 9/4, whose graph is given below, at the points a – 3 and a – 1

For the function f (x) – x2/4 + 9/4 at any point a we have

A Concrete Example

2 9( )

4 4

af a

2 2 2( ) 9 9( )

4 4 4 2 4 4

a h a ah hf a h

2

( ) ( ) 2 42 4

ah hf a h f a a h

h h

A Concrete Example

( ) ( )

2 4S

f a h f a a hm

h

Thus, for the function f (x) – x2/4 + 9/4 at any point a we have

Therefore, as h tends to 0, mS approaches – a/2 mT .

0 0

( ) ( )lim lim

2 4 2Th h

f a h f a a h am

h

A Concrete ExampleFor the function f (x) – x2/4 + 9/4 at any point a we have

2T

am

Thus,

1.5 at -3Tm a

0.5 at -1Tm a

A Concrete ExampleLet us try some other values of a for the same function.

0 at 0Tm a

0.5 at 1Tm a

1 at 2Tm a

1.5 at 3Tm a

2T

am

Remark

2T

am

The example clearly indicates that the slope mT of the tangent line is itself a function of the point a we choose.

We need a better notation reflecting the fact that this object is a function of a

The DerivativeThe instantaneous rate of change mT at a is also called the derivative of f at x = a and it is denoted by f '(a). That is,

0

( ) ( )( ) lim

h

f a h f af a

h

f '(a) is read “f prime of a”

mT = f '(a) = “slope of the tangent line to f (x) at a”

= “instantaneous rate of change of f at a”

The DerivativeThe instantaneous rate of change mT at a is also called the derivative of f at x = a and it is denoted by f '(a). That is,

0

( ) ( )( ) lim

h

f a h f af a

h

In our previous example, the derivative of the function f (x) – (x2 – 9)/4 at any point a is given by

( )2

af a

Rates of ChangeAverage rate of change of f over the interval [a, a+h] is

( ) ( )f f a h f a

x h

Instantaneous rate of change of f at x=a is

Slope of the secant line through the points (a , f (a)) and (a , f (a+h))

0

( ) ( )( ) lim

h

f a h f af a

h

Slope of the tangent line at the point (a , f (a))

Tangent Line and Secant Line

, ( )a f a

, ( )a h f a h

secant line

tangent line at a

Equation of Tangent Line at x = a

, ( )a f a

tangent line at a

( ) ( )( )y f a f a x a

Using the point-slope form of the line

Finding the derivative of the function f is called differentiating f.

If f '(a) exists, then we say that f is differentiable at x = a.

For some functions f , f '(a) may not exist. In this case we say that the function f is not differentiable at x = a.

Terminology

( ) ( )( )

f a h f af a

h

Recall that f '(a) is the limiting value of the expression

as we make h increasingly smaller. Therefore, we can approximate the numerical value of the derivative using small values of h. h = 0.001 often works. In this case,

Quick Approximation of the Derivative

( ) ( )f a h f a

h

Demand: The demand for an old brand of TV is given by

where p is the price per TV set, in dollars, and q is the number of TV sets that can be sold at price p . Find q (190) and estimate q '(190) . Interpret your answers.

100,000

10q

p

Quick Approximation - Example

Solution

100,000 100,000(190) 500

190 10 200q

(190 0.001) (190)

(190)0.001

q qq

499.9975 500(190) 2.5

0.001q

TV sets(190) 2.5

$q

Geometric Interpretation

a 190

At a 190, q(p) decreases as fast as the tangent line does, that is, at the ratemT q '(190) 2.5 TVs/$

What does this means?

It means that at the price of p $190, the demand will decrease by 2.5 TV sets per dollar we increase the price.

If we now set the price at $200, how many TV sets do we expect to sell?

Geometric Interpretation

p 200

At a 190, the equation of the tangent line is

y 2.5( p-190 )+500Thus, when we set the price at p $200, the line shows that we expect to sell y 475 TVs

The actual number we expect to sell according to the demand function is

q (200) 476.2 476 TVswhich is very close to the prediction given by the tangent line.

Now Recall The Example The slope mT f '(a) of the tangent line depends on the point a we choose on the graph of f .

(0) 0Tm f

(1) 0.5Tm f

(3) 1.5Tm f

( )2T

am f a

(2) 1Tm f

Chapter 4Techniques of

Differentiation with Applications

Sections 4.1 and 4.2

Techniques of Differentiation

Derivatives of Powers, Sums and Constant asMultiples

Marginal Analysis

The Derivative as a Function

0

( ) ( )( ) lim

h

f x h f xf x

h

If f is a function, its derivative function is the function whose value at x is the derivative of f at x. That is,

Notice that all we have done is substituted x for a in the definition of f '(a). This way, when we are done with the algebra, the answer will be given in terms of x.

Example: Given the function

0 0

( )lim lim 1h h

x h x h

h h

( ) , find ( )f x x f x

0

( ) ( )( ) lim

h

f x h f xf x

h

Thus, if ( ) , then ( ) 1f x x f x

The Derivative as a Function

Thus, if ( ) , then ( ) 1f x x f x

Geometric Verification At any point on the graph of f , the tangent line agrees with the graph of which already is a straight line of slope 1

2 2 2

0

2limh

x xh h x

h

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

2 2

0

( )limh

x h x

h

2

0

2limh

xh h

h

0lim(2 )h

x h

The Derivative as a Function

Example: Given the function

0

( ) ( )( ) lim

h

f x h f xf x

h

2Thus, if ( ) , then ( ) 2f x x f x x

Geometric Verification

2Thus, if ( ) , then ( ) 2f x x f x x

x y=x2

-3

-2

-1

0

1

2

3

x y'=2x

-3

-2

-1

0

1

2

3

( )f x ( )f x

3 2 2 3 3

0

3 3limh

x x h xh h x

h

3( ) , find ( )f x x f x

3 3

0

( )limh

x h x

h

2 2

0lim 3 3h

x xh h

2 2 3

0

3 3limh

x h xh h

h

Example: Given the function

0

( ) ( )( ) lim

h

f x h f xf x

h

3 2Thus, if ( ) , then ( ) 3f x x f x x

The Derivative as a Function

The Power Rule

3.7( )g x x

11 ( )f x x

x

Example: 2.7( ) 3.7g x x

22

1 ( )f x x

x

1/ 33

1 ( )f x x

x 4 / 31

( )3

f x x

1

If ( ) , where is any constant, then

( )

n

n

f x x n

f x nx

Example:

Example:

0limh

x h x x h x

h x h x

( ) , find ( )f x x f x

0limh

x h x

h

0

1limh x h x

0

limh

x h x

h x h x

Example: Given the function

0

( ) ( )( ) lim

h

f x h f xf x

h

1Thus, if ( ) , then ( )

2f x x f x

x

Verification when f (x) x1/2

Leibniz’ d Notation For Derivatives

Average rate of change f

x

0limx

f df

x dx

Instantaneous rate of change

df

dx( ).f xmeans the same as

Differential Notation: Differentiation

d

dxmeans “the derivative with respect to x”.

The derivative of y f (x) with respect to x is written

( )d dy

ydx dx

5 4If ( ) , then 5 .dy

y f x x xdx

or as ( )d df

f xdx dx

Example:

55 4( )

If ( ) , then 5 .d x

y f x x xdx

Differential Notation: Differentiation

5If ( ) , then find ( 2)y f x x f Example:

5 4If ( ) , then find ( ) 5 and ( 2) 80y f x x f x x f

To find f '(2), the derivative of y f (x) at x 2, we first need to find the function f '(x).

In Leibniz’ notation f '(2) is denoted by the symbol

54 4

22 2

( )5 5( 2) 80

xx x

df d xx

dx dx

Differentiation Rules

Example: If ( ) 14 then ( ) 0f x f x

1) 0 a constantd

c cdx

1

2) If ( ) , where is any constant, then

( )

n

n

f x x n

f x nx

Example: 2 / 3 1/ 32If ( ) then ( )

3f x x f x x

3) ( ) ( ) a constantd d

cf x c f x cdx dx

8 8 7 7( ) 3 3 3 8 24f x x x x x

Differentiation Rules

Example: 8If ( ) 3 then f x x

8 8 7 7

or

3 3 3 8 24df d d

x x x xdx dx dx

4) ( ) ( ) d d d

f x g x f x g xdx dx dx

2 12 11( ) 7 4 14 12f x x x x x

Differentiation Rules

Example: 2 12If ( ) 7 4 then f x x x

2 12 11

or

7 4 14 12df d d d

x x x xdx dx dx dx

Example: Find the derivative of 2

3( )

2f x

x

2 23 3( )

2 2f x x x

First write f (x) as constant times a power 23( )

2f x x

2 2 33 3 3( ) 2

2 2 2f x x x x

2 2 3 33

3 3 3 3( ) 2 3

2 2 2f x x x x x

x

More Examples

33

1 5( ) 4 7

4f x x

x x

1 1/ 3 31( ) 5 4 7

4f x x x x

First write f (x) as

1 1/ 3 31( ) 5 4 7

4f x x x x

2 4 / 3 21 1( ) 1 5 4 3 0

4 3f x x x x

1 1/ 3 31( ) 5 4 7

4f x x x x

2 4 / 3 21 5( ) 12

4 3f x x x x

Example: Find the derivative of

Functions Not Differentiable at a Point

a) Vertical tangent x = a b) Cusp x = a

c) f (a) = Undefined

Example: The Absolute Value of x.

( )f x x (0)f DNE

1 if 0( )

1 if 0

xf x

x

Example: The cube root of x.1/ 33( )f x x x (0)f DNE

1/ 3 2 / 31( )

3f x x x

2 / 3

1( )

3f x

x

2 / 3

1(0)

3 0f DNE

Introduction to the Derivative

Average Rate of Change

The Derivative

Derivatives of Powers, Sums and Constant asMultiples

Marginal Analysis

Cost Functions

A cost function specifies the total cost C as a function of the number of items x produced. Thus, C(x) is the cost of x items. The cost functions is made up of two parts:

C(x)= “variable costs” + “fixed costs”

The marginal cost function is the derivative C′(x) of C(x). It measures the instantaneous rate of change of cost with respect to x.

By definition the Marginal Cost Function is C′(x).

The units of marginal cost are units of cost (say, $) per item.

Interpretation: The marginal cost function C′(x) approximates the change in actual cost of producing an additional unit when x items are already being produced.

0

( ) ( )( ) lim

h

C x h C xC x

h

Interpretation of Marginal Cost

The Marginal Cost Function is C′(x). Thus, for small h we have the quick approximation given by

( ) ( )( )

C x h C xC x

h

If we let h=1 (one more item being produced ), then

( 1) ( )( )

1

C x C xC x

That is, for h=1, we have

( ) ( 1) ( )C x C x C x

Notice that C(x+1) - C(x) is the actual (exact) cost of producing one additional item when the production level is x.

In practice, h=1 is considered to be small in relation to production level x, which in general is a large number.

Interpretation of Marginal Cost

Average Cost Functions

Given a cost function C(x) , the average cost function given by:

( )( )

C xC x

x

Total cost of producing items( )

xC x

x

Example

The total monthly cost function C, in dollars, of producing x computer games is given by:

2( ) 0.001 20 3000 C x x x

a) Find the marginal cost function and use it to estimate the cost of producing the 1,001st game.

b) Find the actual cost of producing the 1,001st game.

The marginal cost function is given by:

( ) 0.002 20C x x

An estimate for the cost of producing the 1,001st game is

(1000) 0.002(1000) 20 18C

That is, it costs an additional $18 to produce one more game when the production level is at 1,000 games a month

Solution

The the actual cost of producing the 1,001st game

(1001) (1000) =C C

2

2

0.001(1001) 20(1001) 3000

0.001(1000) 20(1000) 3000

(1001) (1000) =22017.999-22000C C

(1001) (1000) = $17.999C C

Solution

Example Continued

a) Find the average cost per game if 1,000 games are manufactured.

b) Find the average cost function.

The total monthly cost function C, in dollars, of producing x computer games is given by:

2( ) 0.001 20 3000 C x x x

The average cost function:

20.001 20 3000x x

x

( )( )

C xC x

x

10.001 20 3000x x

(1000) 22,000(1000) 22 $ /

1000 1000

CC game

Solution

More Marginal Functions

The Marginal Profit Function measures the rate of change of the profit function. It approximates the profit from the sale of an additional unit.

The Marginal Revenue Function measures the rate of change of the revenue function. It approximates the revenue from the sale of an additional unit.

Given a revenue function, R(q), The Marginal Revenue Function is:

( )dR

R qdq

Given a profit function, P(q), The Marginal Profit Function is:

( )dP

P qdq

More Marginal Functions

The monthly demand for T-shirts is given by

0.05 25 0 400p q q

where p denotes the wholesale unit price in dollars and q denotes the quantity demanded. The monthly cost for these T-shirts is $8 per shirt.

a) Find the revenue and profit functions.

b) Find the marginal revenue and marginal profit functions.

Marginal Functions - Examples

2( ) 0.05 25 0.05 25 R q q q q q

2( ) 0.05 25 8P q q q q

a) Find the revenue and profit functions.

Revenue = qp

Profit = revenue – cost

2( ) 0.05 17P q q q

Solution

b) Find the marginal revenue and marginal profit functions.

Marginal revenue 2( ) 0.05 25R q q q

0.1 25q

0.1 17q

2( ) 0.05 17P q q q Marginal profit

Solution