## Calculus: Early Transcendentals 8th Edition

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