   Chapter 3.2, Problem 64E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# (a) If g is differentiable, the Reciprocal Rule says that d d x [ 1 g ( x ) ] = − g ' ( x ) [ g ( x ) ] 2 Use the Quotient Rule to prove the Reciprocal Rule.(b) Use the Reciprocal Rule to differentiate the function in Exercise 16.(c) Use the Reciprocal Rule to verify that the Power Rule is valid for negative integers, that is, d d x ( x − n ) = − n x − n − 1 for all positive integers n.

(a)

To determine

To show: The reciprocal rule ddx[1g(x)]=g(x)[g(x)]2.

Explanation

Given:

The function g(x) is differentiable.

Derivative rules:

(1) Quotient Rule: If f1(x) and f2(x) are both differentiable, then

ddx[f1(x)f2(x)]=f2(x)ddx[f1(x)]f1(x)ddx[f2(x)][f2x]2

(2) Power rule: ddx(xn)=nxn1

(3) Sum rule: ddx(f+g)=ddx(f)+ddx(g)

(4) Difference rule: ddx(fg)=ddx(f)</

(b)

To determine

To find: Differentiation of the function by using the reciprocal rule.

(c)

To determine

To verify: The power rule is valid for negative integer. That is, ddx(xn)=nxn1

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Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th 