Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.5 - Implicit Differentiation - 3.5 Exercises: 11

Answer

$y' = \frac{2x + ysinx}{cosx - 2y}$

Work Step by Step

$ycosx = x^2 + y^2$ differentiate both sides $-ysinx + y'cosx = 2x + 2yy'$ $y'cosx - 2yy' = 2x + ysinx$ $y'(cosx-2y) = 2x + ysinx$ $y' = \frac{2x + ysinx}{cosx - 2y}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After 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.