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: 13

Answer

$\dfrac{dy}{dx}=\dfrac{\cos{x}-\tan{y}-1}{x\sec^2{y}}.$

Work Step by Step

$\sin{x}=x+x\tan{y}\rightarrow$ $\dfrac{d}{dx}(\sin{x})=\dfrac{d}{dx}(x)+\dfrac{d}{dx}(x\tan{y})\rightarrow$ $\cos{x}=1+(\tan{y}+\dfrac{dy}{dx}(x\sec^2{y}))\rightarrow$ $\cos{x}-\tan{y}-1=\dfrac{dy}{dx}(x\sec^2{y})\rightarrow$ $\dfrac{dy}{dx}=\dfrac{\cos{x}-\tan{y}-1}{x\sec^2{y}}.$
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