Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Review - Exercises: 15

Answer

$\dfrac {2xy-\cos y}{1- x\sin y-x^2}$

Work Step by Step

$y+x\cos y=x^{2}y\Rightarrow \dfrac {d}{dx}\left( y+x\cos y\right) =\dfrac {d}{dx}\left( x^{2}y\right) \Rightarrow \dfrac {dy}{dx}+\dfrac {d}{dx}\left( x\cos y\right) =2xy+x^{2}\dfrac {dy}{dx}\Rightarrow \dfrac {dy}{dx}+\cos y-x\sin y\dfrac {dy}{dx}=2xy+x^{2}\dfrac {dy}{dx}\Rightarrow \dfrac {dy}{dx}=\dfrac {2xy-\cos y}{1- x\sin y-x^2}$
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