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

##### 1 Answer
Oct 23, 2016

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

#### Explanation:

So firstly differentiate all parts individually

$- 1 = {y}^{2} x - 2 x y - y {e}^{y}$
$- 1 \implies 0$
$- {y}^{2} x \implies - 2 y x \frac{\mathrm{dy}}{\mathrm{dx}} + {y}^{2}$
$- 2 x y \implies - 2 \frac{\mathrm{dy}}{\mathrm{dx}} - 2 y$
$- y {e}^{y} \implies - {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} - y {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}}$

(this is done using mostly the product rule i.e.
$y = f \left(x\right) g \left(x\right)$
$y ' = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$)

leaving us with,

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

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

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

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