## Calculus: Early Transcendentals 8th Edition

$y' = \frac{1+y-e^{y}cos(x)}{e^{y}sin(x)-x}$
Start with equation: $e^{y}sin(x) = x+xy$. Differentiate Both Sides: $e^{y}y'sin(x)+e^{y}cos(x)=1+xy'+y$. Solve for y': $y'(e^{y}sin(x)-x) = 1+y-e^{y}cos(x)$. $y' = \frac{1+y-e^{y}cos(x)}{e^{y}sin(x)-x}$