Calculus 8th Edition

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

Chapter 2 - Derivatives - Review - Exercises - Page 197: 26

Answer

$$y' =\frac{y-2x\cos y}{2 \cos 2y -x^2\sin y-x}$$

Work Step by Step

Given $$ x^2 \cos y + \sin 2y =xy$$ differentiate with respect to $x$ , then \begin{align*} x^2( -\sin y)y'+2x\cos y +2 \cos 2y y'&=xy'+y\\ y'( 2 \cos 2y -x^2\sin y-x)&=y-2x\cos y\\ y'&=\frac{y-2x\cos y}{2 \cos 2y -x^2\sin y-x} \end{align*}
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