Math 135 Business Calculus Spring 2009 Class Notes 1.7 The Chain Rule THE EXTENDED POWER RULE According to the Power Rule, the derivative of the power function y = x k is given by d dx (x k )= kx k−1 . For a power function y = x k , we have a variable x raised to a power k. What is the derivative if, instead of the variable x raised to a power, we have a function raised to a power, such as d dx £ g(x) § k In particular, suppose we want the derivative of y = £ g(x) § 2 . We can compute this using the Product Rule: d dx £ g(x) § 2 = d dx £ g(x) · g(x) § = g 0 (x) · g(x)+ g(x) · g 0 (x) =2g(x) · g 0 (x) This resembles the result of differentiating y = x 2 . The exponent 2 has been pulled down in front of the function and the power has been reduced by 1. However, we also have a factor of g 0 (x) that appears in the derivative. This is a special case of the following general rule: THEOREM 7 The Extended Power Rule Suppose that g(x) is a differentiable function of x. Then, for any real number k, d dx £ g(x) § k = k £ g(x) § k−1 · d dx £ g(x) § = k £ g(x) § k−1 · g 0 (x). In this rule, we can think of g(x) as the “inside” function and the kth power as the “outside” function. The rule can then be viewed as follows. d dx £ g(x) § k = derivative of power function z }| { k £ g(x) § k−1 · g 0 (x) | {z } derivative of inside function EXAMPLE Differentiate f (x) = (7x 2 + x 3 ) 5 . 27
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Math 135 Business Calculus Spring 2009Class Notes1.7 The Chain Rule
THE EXTENDED POWER RULE
According to the Power Rule, the derivative of the power function y = xk is given byd
dx(xk) = kxk−1.
For a power function y = xk, we have a variable x raised to a power k. What is the derivative if,instead of the variable x raised to a power, we have a function raised to a power, such as
d
dx
£g(x)
§k
In particular, suppose we want the derivative of y =£g(x)
§2. We can compute this using the ProductRule:
d
dx
£g(x)
§2 =d
dx
£g(x) · g(x)
§
= g0(x) · g(x) + g(x) · g0(x)= 2g(x) · g0(x)
This resembles the result of differentiating y = x2. The exponent 2 has been pulled down in front ofthe function and the power has been reduced by 1. However, we also have a factor of g0(x) that appearsin the derivative. This is a special case of the following general rule:
THEOREM 7 The Extended Power RuleSuppose that g(x) is a differentiable function of x. Then, for any real number k,
d
dx
£g(x)
§k = k£g(x)
§k−1 · d
dx
£g(x)
§= k
£g(x)
§k−1 · g0(x).
In this rule, we can think of g(x) as the “inside” function and the kth power as the “outside”function. The rule can then be viewed as follows.
d
dx
£g(x)
§k =
derivative ofpower functionz }| {k£g(x)
§k−1 · g0(x)| {z }
derivative ofinside function
EXAMPLE Differentiate f(x) = (7x2 + x3)5.
27
28 Chapter 1 Differentiation
EXAMPLE Differentiate h(x) = (7x2 + x3)1/2.
EXAMPLE Differentiate f(x) = 4
rx + 3x− 2
.
EXAMPLE Differentiate f(x) = (3x− 5)4(7− x)10.
1.7 The Chain Rule 29
COMPOSITION OF FUNCTIONS AND THE CHAIN RULE
The Extended Power Rule is a special case of a more general differentiation rule that can be used tocompute the derivative of the composition of any two functions.
EXAMPLE The number of gallons of paint needed to paint a house depends upon the size of thehouse. A gallon of paint typically covers 250 square feet. This means the number of gallons of paint,n, is a function of the area, A, to be painted according to the function
n = g(A) =A
250.
Suppose a gallon of paint costs $27.50. Then the cost C of n gallons of paint is given by the function
C = f(n) = 27.5n.
Find a function for the cost C to paint an area of A square feet.
In the above example, the cost C of n gallons of paint is a function of n, C = f(n), where n is itselfa function of the area A to be painted, n = g(A). The cost to paint an area A is then a “function of afunction,” or a composite function. If the function giving C in terms of A is denoted h, so C = h(A),then we write
C = h(A) = (f ◦ g)(A).The function h is called the composition of the functions f and g.
DEFINITION OF COMPOSITION OF FUNCTIONS
Suppose f(x) and g(x) are two functions. The composition of f and g, denoted f ◦ g, is defined by
(f ◦ g)(x) = f°g(x)
¢.
If we represent the two functions f andg by two machines, we can visualize thecomposition of functions as shown in thefigure. The input to g is x. The outputg(x) of g is used as the input to f . Theoutput to f is then f
°g(x)
¢.
30 Chapter 1 Differentiation
EXAMPLE Let f(x) = x3 and g(x) =1x
. Find
a) (f ◦ g)(x)
b) (g ◦ f)(x)
Given a composition (f ◦ g)(x) = f°g(x)
¢, we can think of g as the “inside function” and f as the
“outside function.”
(f ◦ g)(x) =
outside functionz }| {f°
g(x)|{z}inside
function
¢
We sometimes have to “decompose” a complicated function and write it as the composition of twosimpler functions.
EXAMPLE Suppose h(x) =1√
7x + 2. Find functions f(x) and g(x) such that h(x) = (f ◦ g)(x).
1.7 The Chain Rule 31
The next theorem now tells us how to differentiate a composition of functions.
THEOREM 8 The Chain RuleThe derivative of the composition f ◦ g is given by
d
dx
£(f ◦ g)(x)
§=
d
dx
£f°g(x)
¢§= f 0
°g(x)
¢· g0(x) =
derivative ofouter functionz }| {
f 0°g(x)
¢· g0(x)
| {z }derivative ofinner function
In Leibniz’s notation, if y = f(u) and u = g(x), thendy
dx=
dy
du· du
dx.
This says that the derivative of the composition is the product of the derivatives of the outside andinside functions.
EXAMPLE Suppose f(u) = 2 +√
u and u = g(x) = x3 + 1. Find (f ◦ g)0(x)a) By using the Chain Rule.
b) By first finding (f ◦ g)(x) and then differentiating.