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: 43

Answer

$$y'=\sinh^{-1}({\frac{x}{3}})$$

Work Step by Step

$y'=\frac{d}{dx}(x\sinh^{-1}(\frac{x}{3})-\sqrt{9+x^2})$ $=\frac{d}{dx}x\sinh^{-1}(\frac{x}{3})-\frac{d}{dx}\sqrt{9+x^2}$ Using the product rule: $y'=(1\times\sinh^{-1}(\frac{x}{3})+x\times\frac{d}{dx}\sinh^{-1}(\frac{x}{3})-\frac{d}{dx}\sqrt{9+x^2}$ Using the chain rule: $y'=(\sinh^{-1}(\frac{x}{3})+x\times\frac{d\sinh^{-1}(\frac{x}{3})}{d\frac{x}{3}}\times \frac{d\frac{x}{3}}{dx})-\frac{d\sqrt{9+x^2}}{d9+x^2} \times\frac{d9+x^2}{dx}$ $=\sinh^{-1}({\frac{x}{3}})+x\times\frac{1}{\sqrt{1+(\frac{x}{3})^2}} \times\frac{1}{3} -\frac{1}{2\sqrt{9+x^2}} \times2x$ $=\sinh^{-1}({\frac{x}{3}})+\frac{x}{3\sqrt{\frac{x^2+9}{9}}}-\frac{x}{\sqrt{9+x^2}}$ $=\sinh^{-1}({\frac{x}{3}})+\frac{x}{3\times\frac{1}{3}\sqrt{{x^2+9}}}-\frac{x}{\sqrt{x^2+9}}$ $=\sinh^{-1}({\frac{x}{3}})+\frac{x}{\sqrt{{x^2+9}}}-\frac{x}{\sqrt{x^2+9}}$ $=\sinh^{-1}({\frac{x}{3}})$
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