Calculus 8th Edition

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

Chapter 6 - Inverse Functions - 6.2* The Natural Logarithmic Functions - 6.2* Exercises: 50

Answer

$y'=\frac{y(xycosx-1)}{x(1-ysinx)}$

Work Step by Step

$\frac{d}{dx}(lnxy)=\frac{d}{dx}(ysinx)$ $\frac{1}{xy}[x.y'+y]=ycosx+sinxy'$ Hence, $y'=\frac{y(xycosx-1)}{x(1-ysinx)}$
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