# How do you use implicit differentiation to find dy/dx given 2x-y+y^2=0?

Nov 13, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2}{2 y - 1}$

#### Explanation:

Given-

$2 x - y + {y}^{2} = 0$

$2 - \frac{\mathrm{dy}}{\mathrm{dx}} + 2 y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

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

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

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2}{2 y - 1}$