# How do you find dy/dx by implicit differentiation of x^2y+y^2x=-2?

Dec 6, 2016

The answer is $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2 x y + {y}^{2}}{{x}^{2} + 2 x y}$

#### Explanation:

We use the product rule for differentiation

$\left(u v\right) ' = u ' v + u v '$

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

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

$\left(- 2\right) ' = 0$

Putting it all together

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

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

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