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, $y'=\frac{y(xycosx-1)}{x(1-ysinx)}$
Calculus 8th Edition
by Stewart, James
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 - Page 446: 50
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
