University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.11 - Linearization and Differentials - Exercises - Page 199: 38

Answer

${dy}=e^{(\tan^{-1}{\sqrt{(x^2+1)}})}\times{\frac{x}{(2+x^2)(\sqrt{(1+x^2)})}}{dx}$

Work Step by Step

We are given: $y=e^{(\tan^{-1}{\sqrt{(x^2+1)}})}$ on differentiating both sides: ${\frac{dy}{dx}}={\frac{d(e^{(\tan^{-1}{\sqrt{(x^2+1)}}})}{dx}}$ on applying the chain rule: ${\frac{dy}{dx}}=e^{(\tan^{-1}{\sqrt{(x^2+1)}})}\times{\frac{1}{(1+(\sqrt{(1+x^2)})^2)}}{\frac{d({({\sqrt{(x^2+1)}})})}{dx}}$ ${\frac{dy}{dx}}=e^{(\tan^{-1}{\sqrt{(x^2+1)}})}\times{\frac{1}{(2+x^2)}}{\frac{2x}{2\sqrt{(x^2+1)}}}$ ${dy}=e^{(\tan^{-1}{\sqrt{(x^2+1)}})}\times{\frac{x}{(2+x^2)(\sqrt{(1+x^2)})}}{dx}$ The final answer is: ${dy}=e^{(\tan^{-1}{\sqrt{(x^2+1)}})}\times{\frac{x}{(2+x^2)(\sqrt{(1+x^2)})}}{dx}$
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