# How do you differentiate x^3+y^3=3xy?

Jan 7, 2017

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

#### Explanation:

differentiate $\textcolor{b l u e}{\text{implicitly with respect to x}}$

The $\textcolor{b l u e}{\text{product rule}}$ has to be used on the right side.

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

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

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

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