Calculus 8th Edition

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

Chapter 2 - Derivatives - Review - Exercises: 29

Answer

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

Work Step by Step

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