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: 55


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

Work Step by Step

Differentiate $e^{x/y}=x-y$ with respect to $x$. Apply chain rule. $e^{x/y}\frac{d}{dx}(\frac{x}{y})=1-\frac{dy}{dx}$ Apply quotient rule. $\frac{dy}{dx}=e^{x/y}\frac{d}{dx}(\frac{x}{y})+1$ $y'=e^{x/y}[\frac{1}{y}-\frac{x}{y^{2}}\frac{dy}{dx}]+1$ Hence, $y'=\frac {y({y-e^{x/y})}}{{y^{2}-xe^{x/y}}}$
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