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: 35



Work Step by Step

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