Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.5 Exercises: 15

Answer

$\dfrac{dy}{dx}=\dfrac{y\cos{xy}}{1-x\cos{xy}}.$

Work Step by Step

$\dfrac{d}{dx}(y)=\dfrac{d}{dx}(\sin{xy})\rightarrow$ Using the Chain Rule with $u=xy\rightarrow \dfrac{du}{dx}=y+x\dfrac{dy}{dx}\rightarrow$ $\dfrac{d}{dx}(\sin{xy})=(y+x\dfrac{dy}{dx})\cos{xy}\rightarrow$ $\dfrac{dy}{dx}=(y+x\dfrac{dy}{dx})\cos{xy}\rightarrow$ $\dfrac{dy}{dx}(1-x\cos{xy})=y\cos{xy}\rightarrow$ $\dfrac{dy}{dx}=\dfrac{y\cos{xy}}{1-x\cos{xy}}.$
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