## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

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

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.