# How do you implicitly differentiate -y^2=e^(y)-y/x ?

##### 1 Answer
Apr 18, 2016

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

#### Explanation:

As such functions cannot be explicitly written in the form $y = f \left(x\right)$, we differentiate them implicitly w.r.t $x$ or $y$, but when we differentiate w.r.t. $y$, we multiply by $\frac{\mathrm{dy}}{\mathrm{dx}}$ too (using chain rule).

Hence, differential of $- {y}^{2} = {e}^{y} - \frac{y}{x}$ is

$- 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{x \cdot \frac{\mathrm{dy}}{\mathrm{dx}} - y \cdot 1}{x} ^ 2$ or

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

$\frac{1}{x} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} - 2 y \frac{\mathrm{dy}}{\mathrm{dx}} - {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y}{x} ^ 2$ or

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

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