# How do you implicitly differentiate 2=e^ysinx-x^2y ?

Jan 23, 2017

$y ' = \frac{2 x y - {e}^{y} \cos x}{2 + {x}^{2} \left(y - 1\right)}$

#### Explanation:

Differentiate both sides of the equation with respect to $x$, keeping in mind that:

d/dx(f(y(x)) = (df)/(dy)(dy)/(dx) = f'(y)y'

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

Solve for $y '$:

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

SUbstitute ${e}^{\sin} y = 2 + {x}^{2} y$ from the original equation:

$y ' = \frac{2 x y - {e}^{y} \cos x}{2 + {x}^{2} y - {x}^{2}} = \frac{2 x y - {e}^{y} \cos x}{2 + {x}^{2} \left(y - 1\right)}$