# What is the implicit derivative of 1=xy^2-y^2+x^2 ?

Feb 1, 2017

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

#### Explanation:

Assuming $y$ is a function of $x$ and not the opposite:

Differentiate both sides with respect to $x$.

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

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

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

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

(If $f$ is a function of $y$, then $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{dy}} \frac{\mathrm{dy}}{\mathrm{dx}}$)