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

Jul 15, 2015

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

#### Explanation:

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

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

Apply the product rule and the chain rule $\Rightarrow$

$2 x y ' + 2 y - 2 y . y ' = 0$

$2 y ' \left(x - y\right) = - 2 y$

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

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