Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.2 Exponential Functions and Their Derivatives - 6.2 Exercises: 49

Answer

$y'=sin(\frac{1-e^{2x}}{1+e^{2x}})[\frac{4e^{2x}}{1+e^{2x})^{2}}]$

Work Step by Step

$y'=\frac{d}{dx}[cos(\frac{1-e^{2x}}{1+e^{2x}})]$ $=-sin(\frac{1-e^{2x}}{1+e^{2x}})\frac{d}{dx}(\frac{1-e^{2x}}{1+e^{2x}})$ $=-sin(\frac{1-e^{2x}}{1+e^{2x}})[\frac{-4e^{2x}}{1+e^{2x})^{2}}]$ Hence, $y'=sin(\frac{1-e^{2x}}{1+e^{2x}})[\frac{4e^{2x}}{1+e^{2x})^{2}}]$
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