   Chapter 3.6, Problem 12E ### 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

# Differentiate the function. h ( x ) = ln ( x + x 2 − 1 )

To determine

To find: The derivative of h(x).

Explanation

Given:

The function, h(x)=ln(x+x21).

Result used: Chain Rule

If h is differentiable at x and g is differentiable at h(x), then the composite function F=gh defined by F(x)=g(h(x)) is differentiable at x and F is given by the product

F(x)=g(h(x))h(x) (1)

Calculation:

Obtain the derivative h(x).

h(x)=ddx(h(x))=ddx(ln(x+x21))

Let k(x)=x+x21 and f(u)=lnu  where u=h(x).

Apply the chain rule as shown in equation (1),

h(x)=f(k(x))k(x) (2)

The derivative f(k(x)) is computed as follows,

f(k(x))=f(u)=ddu(f(u))=ddu(lnu)=1u

Substitute u=x+x21 in the above equation,

f(k(x))=1x+x21

Thus, the derivative f(k(x)) is f(k(x))=1x+x21

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