University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Practice Exercises - Page 202: 64



Work Step by Step

We are given that: $y=(1+x^2)e^{\tan^{-1}{x}}$ $\frac{dy}{dx}=\frac{d((1+x^2))}{dx}e^{\tan^{-1}{x}}+(1+x^2)\frac{d(e^{\tan^{-1}{x}})}{dx}$ $\frac{dy}{dx}={(2x)}e^{\tan^{-1}{x}}+(1+x^2)(e^{\tan^{-1}{x}})\frac{d({\tan^{-1}{x}})}{dx}$ $\frac{dy}{dx}={(2x)}e^{\tan^{-1}{x}}+(1+x^2)(e^{\tan^{-1}{x}})\frac{{{1}}}{1+x^2}$ $\frac{dy}{dx}={(2x)}e^{\tan^{-1}{x}}+(e^{\tan^{-1}{x}})$
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