Calculus 8th Edition

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

Chapter 2 - Derivatives - Review - Exercises: 36

Answer

$\dfrac {dy}{dx}=\dfrac {\tan y}{1-xsec^2y}$

Work Step by Step

$x\tan y=y-1\Rightarrow \dfrac {d}{dx}\left( x\tan y\right) =\dfrac {d}{dx}\left( y-1\right) \Rightarrow \tan y+xsec^2y\times \dfrac {dy}{dx}=\dfrac {dy}{dx}$ $\Rightarrow \dfrac {dy}{dx}=\dfrac {\tan y}{1-xsec^2y}$
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