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Sullivan 04 apcalc4e 45342 ch02 166 233 5pp August 7, 2023 12:54

Section 2.4 • Differentiating the Product and the Quotient of Two Functions; Higher-Order Derivatives 207

EXAMPLE 3 Differentiating the Quotient of Two Functions

2
x + 1
′
Find y if y = .
2x − 3
Solution
2
The function y is the quotient of f (x) = x + 1 and g(x) = 2x − 3. Using the Quotient
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Rule, we have
d 2 2 d

(x + 1) (2x − 3) − (x + 1) (2x − 3)
2
d x + 1 dx dx
′
y = =
dx 2x − 3 (2x − 3) 2
2
2
2
2
(2x)(2x − 3) − (x + 1)(2) 4x − 6x − 2x − 2 2x − 6x − 2
= = =
(2x − 3) 2 (2x − 3) 2 (2x − 3) 2
3
provided x 6= .
2
R
NOW WORK Problem 23 and AP Practice Problems 1, 2, 3, 7, and 8.
COROLLARY Derivative of the Reciprocal of a Function
If a function g is differentiable, then
IN WORDS The derivative of the reciprocal of
d
a function is the negative of the derivative of g(x)
′
d 1 dx g (x)
the denominator divided by the square of the = − 2 = − 2 (1)
dx g(x) [g(x)] [g(x)]
denominator. That is,
1 g ′
′
= − . provided g(x) 6= 0.
g g 2
The proof of the corollary is left as an exercise. (See Problem 98.)
EXAMPLE 4 Differentiating the Reciprocal of a Function

d 2
d 1 dx (x + x) 2x + 1
(a) = − = −
2
2
2
dx x + x ↑ (x + x) 2 (x + x) 2
Use (1).
d x
d −x d 1 dx e e x 1 −x
(b) e = = − = − = − = −e
x 2
dx dx e x ↑ (e ) e 2x e x
Use (1).
NOW WORK Problem 25.
Notice that the derivative of the reciprocal of a function f is not the reciprocal of
the derivative. That is,
d 1 1
6=
dx f (x) f (x)
′
The rule for the derivative of the reciprocal of a function allows us to extend the
Simple Power Rule to all integers. Here is the proof.

Proof Suppose n is a negative integer and x 6= 0. Then m = −n is a positive
integer, and

d m
d n d 1 dx x mx m − 1 m − 1 − 2m −m − 1 n − 1
x = = − = − = −mx = −mx = nx
m 2
dx dx x m ↑ (x ) ↑ x 2m ↑
Use (1). Simple Power Rule Substitute n = −m.
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