# How do you implicitly differentiate 2= xy-ysin^2x-cos^2xy ?

Dec 30, 2016

$y ' = - \frac{y + y \sin 2 x + y \sin 2 x y}{x - {\sin}^{2} x - x \sin 2 x y}$

#### Explanation:

Differentiate the equation with respect to $x$:

$0 = \frac{d}{\mathrm{dx}} \left(x y - y {\sin}^{2} x - {\cos}^{2} x y\right)$

$0 = y + x y ' + y ' {\sin}^{2} x + 2 y \sin x \cos x + 2 \cos x y \sin x y \left(y + x y '\right)$

$0 = y + x y ' + y ' {\sin}^{2} x + y \sin 2 x + y \sin 2 x y + x y ' \sin 2 x y$

$- y ' \left(x - {\sin}^{2} x - x \sin 2 x y\right) = y + y \sin 2 x + y \sin 2 x y$

$y ' = - \frac{y + y \sin 2 x + y \sin 2 x y}{x - {\sin}^{2} x - x \sin 2 x y}$