Chapter 3_4.pptx

Chapter 3.4 1

Chapters 3.4

In Sections 3.2 and 3.3 we developed derivative expressions for simple power

functions, exponential, logarithmic, sine and cosine functions:

• _, • ,

• _, • _, • _, • _,

Chapter 3.4 3

We also found that the derivative of sums of functions is equal to the sum of their

derivatives:

,

We now seek to find derivatives for more complex functions.

Before we do so, it will be helpful to note that certain functions require arguments.

Functions such as , , , and are

meaningless without an argument. What do we want when we just say “sine”. We want the sine of what?

These functions can be written as , , , and , where refers to the argument of the

Chapter 3.4 5

Examples:

What are the arguments of the terms in the following: 1.

2. 3.

Example: Find for

We can distribute the perfect square and then find the derivative term by term:

� ′ ( �)= ��

�� =18 x −12

We now seek to find derivatives for more complex

functions by taking advantage of a concept called the

Chapter 3.4 7

Now consider rewriting as:

� (� )=�2 where

�

=

3

�

−

2

��

�� =2t

��

�� =3

The product of and is ��

��

��

�� =(2�) (3)=6 t but

�

=

3

�

−

2

�� ��

��

�� =6(3 �−2   )=18 x −12=

�� ��

This result agrees with the derivative, , we calculated on the previous slide.

This last result is known as the

chain rule

.

That is, if f(t) is a function of t, and

t(x) is a function of x, then

��

��

=

��

��

��

��

The chain rule is a powerful tool which can

be used to determine the derivatives of

Chapter 3.4 9

Example: where k is a constant. Find . Note that kx is the argument of the exponential function.

�

(

�

)

=

�

�

=

��

��

��

=

�

��

��

=

�

��

��

=

��

��

��

��

=

(

�

)

�

��

��

=

� �

If

,

Problem 1:

Find for

Problem 2:

Find for

Chapter 3.4 11

Problem 1 Solved:

Example: The logistic model is where L, A, and B are constants. Find .

� (� )= �

� =��

−1

where

�

=

1

+

����

(

−

��

)

��

�� = �(−1) �

−2 =− � �2 �� �� = �(− �)��� (−�� ) �� �� = �� �� �� �� =

(

Chapter 3.4 13 ��

�� = ��

��

��

�� =

(

− �

�2

)

�� �� =

��� ��� (−�� )

(

)

Recall that an exponential function with a base of e and the natural log function are inverse functions. That is

�

=

�

Consider the exponential function. Find .

� (�)=�� where

�

₌

�

recall, if ,

� ′ ( �)=��(�)���(�)�

� ′ ( �)=��(�)��

Chapter 3.4 15

If

,

Problem 3:

Find for

Problem 4:

Problem 3 Solved:

Problem 4 Solved:

Chapter 3.4 17

Example: Find for the sine model

a, b, c, and d are constants.

where

�

₌

��

₊

�

��

��

=

����

(

�

)

��

��

=

�

��

��

=

��

��

��

��

=

(

����

(

�

)

)

�

��

��

=

� ����

(

��

+

�

)

Derivatives of the models

Linear:

�

_′

(

�

)

=

�

Quadratic:

�

_′

(

�

)

=

2

��

+

�

Cubic:

�

_′

₍

�

₎

₌

₃

��

₊

₂

��

₊

�

Exponential:

_�

_′

₍

_�

₎

=

� ��

(

�

)

(

�

)

Logarithmic: � ′ ₍ �₎₌ �

�

Logistic: � _{′ (}�₎₌ ������(−��)

(

)

Chapter 3.4 19

We can avoid the use of the chain rule in taking derivatives of certain functions by summarizing chain rule results.

Example: ; find

Using the chain rule:

Generalizing the last example where, below, , refers to the derivative of the argument:

Example: ; find

Using the chain rule:

Chapter 3.4 21

Similar results may be gotten by considering

examples involving , , and .

In the following refers to the argument of a function, and refers to the derivative of an argument.

Summary of some derivative rules using Chain Rule

Ex:

Ex:

Ex:

Problem 5:

A. . Find .

Note:

B.

Note:

C.

. Find .

Chapter 3.4 23

Problem 5 Solved:

A. .

B. .

C. .

We now pursue a method to take derivatives of complex functions by introducing the

generalized power rule.

Once we become acquainted with the

generalized power rule, we can totally bypass using the chain rule to take derivatives and

greatly simplify the process.

Chapter 3.4 25

How might we find for the function:

Summarizing the problem:

Now consider the quantity in parentheses in the original

problem as an argument, arg. Then a summary of the

Chapter 3.4 27

This example illustrates the use of the

generalized power rule. If

then

Problem 6: Find the derivatives of the following. A.

B.

Chapter 3.4 29

A.

B.

With the generalized power rule, and the

use of arguments, we can find derivatives of fairly complex functions without needing to use the chain rule.

However, the chain rule has other important applications, as we will see in the following problem.

Chapter 3.4 31

Problem 7:

A company’s weekly cost ($) to produce g gallons of milk can be modeled by:

Projected weekly production over the next 13 years has been estimated, and is

displayed in the table and scatterplot below.

1 5.8

4 8.4

7 14.1

10 22.9

13 31.7

1 5.8

4 8.4

7 14.1

10 22.9

Chapter 3.4 33

1 5.8

4 8.4

7 14.1

10 22.9

13 31.7

1 5.8

4 8.4

7 14.1

10 22.9

A. Find a model for weekly milk production, in thousands of gallons, as a function of time, g(t), in years. Two different functions may provide appropriate models.

Determine both models, reporting model parameters with 3 decimal digits.

Chapter 3.4 35

Using quadratic model for weekly production:

��

��

=

0.246

�

+

0.488

�

(

�

)

=

0.123

�

+

0.488

�

+

4.923

�

(

�

)

₌

4.940

(

1.158

)

Using exponential model for weekly production:

��

��

=

4.940

��

(

1.158

)

(

1.158

)

Problem 7A Solved (cont):

Chapter 3.4 37

B. By the chain rule, . What are the units for ?

For the C(g) function:

in units of $

g in units of gallons

For the g(t) function:

t in units of years

g in units of thousands of gallons

Units of :

Chapter 3.4 39

C. Find the rate of change of cost as a function of time, , using the quadratic

model.

Recall, by the chain rule

Write such that g does not appear in the expression. That is, express as a function of t only.

Using quadratic model for weekly production: �� �� = �� �� �� �� =

(

0.123�2+0.488�+4.923  

)

�� �� =

74.95 � =

74.95

0.123 �2+0.488�+4.923  

�(� )=0.123�2+0.488� +4.923 ��_�� =0.246 � +0.488

�� 74.95( 0.246�+0.488)

�

(

�

)

=

3250.23

+

74.95

��

(

�

)

�� =

74.95

� Problem 7C Solved:

Chapter 3.4 41

D. Find the rate of change of cost as a

function of time, , using the exponential

model.

Recall, by the chain rule

Write such that g does not appear in the expression. That is, express as a function of t only.

Using exponential model for weekly production:

�

(

�

)

=

3250.23

+

74.95

��

(

�

)

�� =

74.95

� Problem 7D Solved:

Recall:

�(� )=4.940

(

)

(

)

�� �� = �� �� �� �� =

(

4.940 (1.158� )

)

(

� )

₎

4.940

(

)

Chapter 3.4 43

E. Estimate the rate of change of costs as a function of time 5 and 10 years from now using both models. What are the units and what does this rate of change

Using quadratic model:

��

��

|

74.95

₍

₎

0.123 (5)2+0.488 (5)+4.923   =12.3

$ h�

��

��

�� =

74.95( 0.246�+0.488)

0.123�2+0.488�+4.923  

��

��

|

74.95

₍

₎

0.123(10 )2+0.488 (10)+4.923   =10.0

$ h� ��

Chapter 3.4 45

Using exponential model: ��

�� =74.95 ��(1.158 )

��

��

|

=

��

��

|

=

74.95

��

(

1.158

)

=

11.0

$ h

�

��