# How do you implicitly differentiate -1=-y^2x-2xy-xye^x ?

Apr 23, 2016

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

#### Explanation:

Given -

$- {y}^{2} x - 2 x y - x y {e}^{x} = - 1$

$\left[- {y}^{2} \left(1\right) + x \left(- 2\right) y \frac{\mathrm{dy}}{\mathrm{dx}}\right] - \left[2 x \frac{\mathrm{dy}}{\mathrm{dx}} \left(1\right) + 2 y\right] - \left[x y {e}^{x} + x {e}^{x} \frac{\mathrm{dy}}{\mathrm{dx}} \left(1\right) + y {e}^{x} \left(1\right)\right] = 0$

$- {y}^{2} - 2 x y \frac{\mathrm{dy}}{\mathrm{dx}} - 2 x \frac{\mathrm{dy}}{\mathrm{dx}} - 2 y - x y {e}^{x} - x {e}^{x} \frac{\mathrm{dy}}{\mathrm{dx}} - y {e}^{x} = 0$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} \left[- 2 x y - 2 x - x {e}^{x}\right] = {y}^{2} + 2 y + x y {e}^{x} + y {e}^{x}$

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