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

Nov 16, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \left({y}^{2} - x\right)}{y \left(3 y - 4 x\right)}$

#### Explanation:

As we will differentiate implicitly we may as well multiply through by $y , y \ne 0$ to get:

$2 x y = {y}^{2} - {x}^{2} / y$
$\therefore 2 x {y}^{2} = {y}^{3} - {x}^{2}$

We now differentiate wrt $x$, and apply the product rule:

$\left(2 x\right) \left(\frac{d}{\mathrm{dx}} {y}^{2}\right) + \left(\frac{d}{\mathrm{dx}} 2 x\right) \left({y}^{2}\right) = \frac{d}{\mathrm{dx}} {y}^{3} - 2 x$
$\therefore \left(2 x\right) \left(2 y \frac{\mathrm{dy}}{\mathrm{dx}}\right) + \left(2\right) \left({y}^{2}\right) = 3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} - 2 x$

$\therefore 4 x y \frac{\mathrm{dy}}{\mathrm{dx}} + 2 {y}^{2} = 3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} - 2 x$
$\therefore 3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} - 4 x y \frac{\mathrm{dy}}{\mathrm{dx}} = 2 {y}^{2} - 2 x$
$\therefore \left(3 {y}^{2} - 4 x y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 2 \left({y}^{2} - x\right)$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \left({y}^{2} - x\right)}{y \left(3 y - 4 x\right)}$