Calculus 8th Edition

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

Chapter 6 - Inverse Functions - 6.3* The Natural Exponential Function - 6.3* Exercises - Page 453: 50


$y'=\frac{e^{-2x}(1-2x)}{2\sqrt (1+xe^{-2x})}$

Work Step by Step

$y= \sqrt (1+xe^{-2x})$ Let $t=(1+xe^{-2x})$ This implies $\frac{dt}{dx}=e^{-2x}(1-2x)$ ...(1) Now, $y'=\frac{d}{dt}\sqrt (t)\frac{dt}{dx}$ $=\frac{1}{2\sqrt t}\frac{dt}{dx}$ From equation (1), we get Hence, $y'=\frac{e^{-2x}(1-2x)}{2\sqrt (1+xe^{-2x})}$
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