# How do you find dy/dx if x+y=e^(xy)?

Feb 6, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{{y}^{2} + x y - 1}{{x}^{2} + x y - 1}$

#### Explanation:

Differentiate both sides of the equation with respect to $x$:

$\frac{d}{\mathrm{dx}} \left(x + y\right) = \frac{d}{\mathrm{dx}} {e}^{x y}$

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

substitute ${e}^{x y}$ from the original equation:

$1 + y ' = \left(y + x y '\right) \left(x + y\right)$

$1 + y ' = x y + {x}^{2} y ' + {y}^{2} + x y y '$

Solve for $y '$

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

$y ' = - \frac{{y}^{2} + x y - 1}{{x}^{2} + x y - 1}$