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

Jun 21, 2016

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

#### Explanation:

start by making life easier for yourself

this is $- 2 y = {e}^{{x}^{2} - {y}^{2}}$

and then further simplification $\ln \left(- 2 y\right) = {x}^{2} - {y}^{2}$

and then we differentiate to get

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

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

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