# How do you differentiate xy^2-x^3y=6?

Mar 9, 2015

You have an Implicit Function in which $y$ is function of $x$ so you have to derive it as well.

You get:
$\left(1 \cdot {y}^{2} + 2 x y \frac{\mathrm{dy}}{\mathrm{dx}}\right) - \left(3 {x}^{2} y + {x}^{3} \frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$ using the Product Rule.

Then:
${y}^{2} + 2 x y \frac{\mathrm{dy}}{\mathrm{dx}} - 3 {x}^{2} y - {x}^{3} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

Collecting $\frac{\mathrm{dy}}{\mathrm{dx}}$:
$\frac{\mathrm{dy}}{\mathrm{dx}} \left[2 x y - {x}^{3}\right] = 3 {x}^{2} y - {y}^{2}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y \left(3 {x}^{2} - y\right)}{x \left(2 y - {x}^{2}\right)}$

Hope it helps