Copyright © Cengage Learning. All rights reserved. 4 Techniques of Differentiation with...

Copyright © Cengage Learning. All rights reserved.

4 Techniques of Differentiationwith Applications

Copyright © Cengage Learning. All rights reserved.

4.4 The Chain Rule

3

The Chain Rule

We can now find the derivatives of expressions involving powers of x combined using addition, subtraction, multiplication, and division, but we still cannot take the derivative of an expression like (3x + 1)0.5.

For this we need one more rule.

The function h(x) = (3x + 1)0.5 is not a sum, difference, product, or quotient.

4

The Chain Rule

To find out what it is, we can use the calculation thought experiment and think about the last operation we would perform in calculating h(x).

1. Calculate 3x + 1.

2. Take the 0.5 power (square root) of the answer.

The last operation is “take the 0.5 power.” We do not yet have a rule for finding the derivative of the 0.5 power of a quantity other than x.

5

The Chain Rule

There is a way to build h(x) = (3x + 1)0.5 out of two simpler functions: u(x) = 3x + 1 (the function that corresponds to the first step in the calculation above) and f (x) = x0.5 (the function that corresponds to the second step):

h(x) = (3x + 1)0.5

= [u(x)]0.5

= f (u(x))

u(x) = 3x + 1

f (x) = x0.5

6

The Chain Rule

We say that h is the composite of f and u. We read f (u(x)) as “f of u of x.”

To compute h(1), say, we first compute 3 1 + 1 = 4 and then take the square root of 4, giving h(1) = 2.

To compute f (u(1)) we follow exactly the same steps: First compute u(1) = 4 and then f (u(1)) = f (4) = 2.

We always compute f (u(x)) numerically from the inside out: Given x, first compute u(x) and then f (u(x)).

7

The Chain Rule

Now, f and u are functions whose derivatives we know.

The chain rule allows us to use our knowledge of the derivatives of f and u to find the derivative of f (u(x)).

For the purposes of stating the rule, let us avoid some of the nested parentheses by abbreviating u(x) as u.

Thus, we write f (u) instead of f (u(x)) and remember that u is a function of x.

8

The Chain Rule

Chain Rule

If f is a differentiable function of u and u is a differentiable function of x, then the composite f (u) is a differentiable function of x, and

In words The derivative of f(quantity) is the derivative of f, evaluated at that quantity, times the derivative of the quantity.

Chain Rule

9

The Chain Rule

Quick Example

Take f (u) = u2. Then

The derivative of a quantity squared is two times the quantity, times the derivative of the quantity.

Because f '(u) = 2u

10

The Chain Rule

The following table gives more examples.

11

The Chain Rule

To motivate the chain rule, let us see why it is true in the special case when f (u) = u3, where the chain rule tells us that

But we could have done this using the product rule instead:

which gives us the same result.

Generalized Power Rule with n = 3

12

The Chain Rule

A similar argument works for f (u) = un where n = 2, 3, 4, …

We can then use the quotient rule and the chain rule for

positive powers to verify the generalized power rule for

negative powers as well.

13

Example 1 – Using the Chain Rule

Compute the following derivatives.

14

Example 1 – Solution

a. Using the calculation thought experiment, we see that the last operation we would perform in calculating (2x2 + x)3 is that of cubing.

Thus we think of (2x2 + x)3 as a quantity cubed.

There are two similar methods we can use to calculate its derivative.

Method 1: Using the formula We think of (2x2 + x)3 as u3, where u = 2x2 + x.

15

Example 1 – Solution

By the formula,

Now substitute for u:

= 3(2x2 + x)2(4x + 1)

cont’d

Generalized Power Rule

16

Example 1 – Solution

Method 2: Using the verbal form If we prefer to use the verbal form, we get:

The derivative of (2x2 + x) cubed is three times (2x2 + x) squared, times the derivative of (2x2 + x).

In symbols,

as we obtained above.

cont’d

17

Example 1 – Solution

b. First, the calculation thought experiment: If we were computing (x3 + x)100, the last operation we would

perform is raising a quantity to the power 100.

Thus we are dealing with a quantity raised to the power 100, and so we must again use the generalized power rule.

According to the verbal form of the generalized power rule, the derivative of a quantity raised to the power 100 is 100 times that quantity to the power 99, times the derivative of that quantity.

cont’d

18

Example 1 – Solution

In symbols,

cont’d

19

Example 1 – Solution

c. We first rewrite the expression as and then use the generalized power rule as in parts (a) and (b):

The derivative of a quantity raised to the 0.5 is 0.5 times the quantity raised to the –0.5, times the derivative of the quantity.

Thus,

cont’d

20

Example 1 – Solution

d. The calculation thought experiment tells us that |4x2 − x| is the absolute value of a quantity, so we use the generalized rule for absolute values (above):

The derivative of the absolute value of a quantity is the absolute value of the quantity divided by the quantity, times the derivative of the quantity.

cont’d

21

Example 1 – Solution

Thus,

cont’d

22

Applications

23

Example 4 – Marginal Product

A consultant determines that Precision Manufacturers’

annual profit (in dollars) is given by

P = –200,000 + 4,000q – 0.46q2 – 0.00001q3

where q is the number of surgical lasers it sells each year.

The consultant also informs Precision that the number of

surgical lasers it can manufacture each year depends on

the number n of assembly line workers it employs

according to the equation

q = 100n

Use the chain rule to find the marginal product

Each worker contributes 100 lasers per year.

24

Example 4 – Solution

We could calculate the marginal product by substituting the expression for q in the expression for P to obtain P as a function of n and then finding dP/dn.

Alternatively—and this will simplify the calculation—we can use the chain rule.

To see how the chain rule applies, notice that P is a function of q, where q in turn is given as a function of n.

25

Example 4 – Solution

By the chain rule,

Now we compute

and

cont’d

Chain Rule

Notice how the “quantities” dq appear to cancel.

26

Example 4 – Solution

Substituting into the equation for gives

Notice that the answer has q as a variable. We can express dP/dn as a function of n by substituting 100n for q:

cont’d

27

Chain Rule in Differential Notation

If y is a differentiable function of u, and u is a differentiable function of x, then

Notice how the units cancel:

Applications

The terms du cancel.

28

Applications

Quick Example

If y = u3, where u = 4x + 1, then

29

Applications

Manipulating Derivatives in Differential Notation

1. Suppose y is a function of x. Then, thinking of x as a function of y (as, for instance, when we can solve for x) we have

Quick Example

In the demand equation q = –0.2p – 8, we have

Therefore,

30

Applications

2. Suppose x and y are functions of t. Then, thinking of y as a function of x (as, for instance, when we can solve for t as a function of x, and hence obtain y as a function of x) we have

Quick Example

If x = 3 – 0.2t and y = 6 + 6t, then

The terms dt appear to cancel.

31

Applications

To see why the above formulas work, notice that the second formula,

can be written as

which is just the differential form of the chain rule.

32

Applications

For the first formula, use the second formula with y playing the role of t: