   Chapter 3.11, Problem 42E ### 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

# Find the derivative. Simplify where possible.42. y = x tanh − 1 x + ln 1 − x 2

To determine

To find: The derivative of the function.

Explanation

Given:

The function y=xtanh1x+ln1x2.

Derivative rules used:

(1) The chain rule: dydx=dydududx.

(2) The derivative of the inverse hyperbolic tangent function is, ddx(tanh1x)=11x2.

Calculation:

The derivative of the function y=xtanh1x+ln1x2 is computed as follows,

dydx=ddx(xtanh1x+ln1x2)=ddx(xtanh1x)ddx(ln1x2)=[xddx(tanh1x)+tanh1xddx(x)]ddx(ln1x2)=[x(11x2)+tanh1x(1)]ddx(ln1x2)          [ddx(tanh1x)=11x2]

Let u=1x2 and apply the chain rule,

dydx=[x(11x2)+tanh1x]ddx(lnu)=[x(11x2)+tanh1x]ddu(lnu)dudx=[x(11x2)

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