11
Techniques of Differentiation
with Applications
Copyright © Cengage Learning. All rights reserved.
11.4
The Chain Rule
Copyright © Cengage Learning. All rights reserved.
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.
3
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.
4
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
u(x) = 3x + 1
= f(u(x))
f(x) = x0.5
5
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)).
6
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.
7
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
Chain Rule
In words The derivative of f(quantity) is the derivative of f,
evaluated at that quantity, times the derivative of the
quantity.
8
The Chain Rule
Quick Example
Take f(u) = u2. Then
Because f '(u) = 2u
The derivative of a quantity squared is two times the
quantity, times the derivative of the quantity.
9
The Chain Rule
The following table gives more examples.
10
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
Generalized Power Rule with n = 3
But we could have done this using the product rule instead:
which gives us the same result.
11
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.
12
Example 1 – Using the Chain Rule
Compute the following derivatives.
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.
13
Example 1 – Solution
cont’d
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.
By the formula,
Generalized Power Rule
14
Example 1 – Solution
cont’d
Now substitute for u:
= 3(2x2 + x)2(4x + 1)
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).
15
Example 1 – Solution
cont’d
In symbols,
as we obtained above.
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.
16
Example 1 – Solution
cont’d
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.
In symbols,
17
Example 1 – Solution
cont’d
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,
18
Applications
19
Example 4 – Marginal Product
Precision Manufacturers is informed by a consultant that its
annual profit 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
Each worker contributes 100 lasers per year.
Use the chain rule to find the marginal product
20
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.
21
Example 4 – Solution
cont’d
By the chain rule,
Chain Rule
Notice how the “quantities” dq appear to cancel.
Now we compute
and
22
Example 4 – Solution
Substituting into the equation for
cont’d
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:
23
Applications
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:
24
Applications
Quick Example
If y = u3, where u = 4x + 1, then
25
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)
one has
Quick Example
In the demand equation q = –0.2p – 8, we have
Therefore,
26
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) one has
The terms dt appear to cancel.
Quick Example
If x = 3 – 0.2t and y = 6 + 6t, then
27
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.
28
Applications
For the first formula, use the second formula with y playing
the role of t:
29