# How do you find x^2y- xy^3 = 2-2x^3 implicitly?

Aug 7, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{y}^{3} - 6 {x}^{2} - 2 x y}{{x}^{2} - 3 x {y}^{2}}$

#### Explanation:

Differentiate term by term (using the product rule):

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

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

$\frac{d \left(2\right)}{\mathrm{dx}} = 0$

$\frac{d \left(2 {x}^{3}\right)}{\mathrm{dx}} = 6 {x}^{2}$

This gives:
$2 x y + {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} - \left({y}^{3} + 3 x {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}\right) = - 6 {x}^{2}$

Rearranging terms to make $\frac{\mathrm{dy}}{\mathrm{dx}}$ the subject we have:

$\left({x}^{2} - 3 x {y}^{2}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = {y}^{3} - 6 {x}^{2} - 2 x y$

=>

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{y}^{3} - 6 {x}^{2} - 2 x y}{{x}^{2} - 3 x {y}^{2}}$