Answer
$y' = \frac{2x + ysinx}{cosx - 2y}$
Work Step by Step
$ycosx = x^2 + y^2$
differentiate both sides
$-ysinx + y'cosx = 2x + 2yy'$
$y'cosx - 2yy' = 2x + ysinx$
$y'(cosx-2y) = 2x + ysinx$
$y' = \frac{2x + ysinx}{cosx - 2y}$
Calculus: Early Transcendentals 9th Edition
by Stewart, James
Published by
Cengage Learning
ISBN 10:
1337613924
ISBN 13:
978-1-33761-392-7
Chapter 3 - Section 3.5 - Implicit Differentiation - 3.5 Exercises - Page 214: 15
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