# How do you find the derivative of x^3+xy^2=y^3+yx^2?

Nov 3, 2016

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

#### Explanation:

Differentiating both sides of the given equation w.r.t x, it would result in

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

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

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