Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.11 - Hyperbolic Functions - 3.11 Exercises: 42

Answer

$$y'=\tanh^{-1}x$$

Work Step by Step

$y'=\frac{d}{dx}(x\tanh^{-1}x+\ln{\sqrt{1-x^2}})$ $=\frac{d}{dx}x\tanh^{-1}x+\frac{d}{dx}\ln\sqrt{1-x^2}$ Using the product rule: $y'=(\tanh^{-1}x+x\frac{d}{dx}\tanh^{-1}x)+\frac{d}{dx}\ln{\sqrt{1-x^2}}$ $=\tanh^{-1}x+x\times\frac{1}{1-x^2}+\frac{d}{dx}\ln{\sqrt{1-x^2}}$ $=\tanh^{-1}x+\frac{x}{1-x^2}+\frac{d}{dx}\ln{\sqrt{1-x^2}}$ Using the chain rule: $y'=\tanh^{-1}x+\frac{x}{1-x^2}+\frac{d\ln{\sqrt{1-x^2}}}{d{\sqrt{1-x^2}}} \times\frac{d{\sqrt{1-x^2}}}{d1-x^2} \times\frac{d1-x^2}{dx}$ $=\tanh^{-1}x+\frac{x}{1-x^2}+\frac{1}{\sqrt{1-x^2}} \times\frac{1}{2\sqrt{1-x^2}} \times -2x$ $=\tanh^{-1}x+\frac{x}{1-x^2}+\frac{-x}{{1-x^2}}$ $=\tanh^{-1}x$
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