# How do you find (dy)/(dx) given x^3+y^3=2xy?

Aug 18, 2016

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

#### Explanation:

${x}^{3} + {y}^{3} = 2 x y$

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

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

Here, by the Chain Rule, $\frac{d}{\mathrm{dx}} \left({y}^{3}\right) = \frac{d}{\mathrm{dy}} \left({y}^{3}\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 3 {y}^{2} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$,

&, by, the Product Rule, $\frac{d}{\mathrm{dx}} \left(x y\right) = x \cdot \frac{d}{\mathrm{dx}} \left(y\right) + y \cdot \frac{d}{\mathrm{dx}} \left(x\right) = x \frac{\mathrm{dy}}{\mathrm{dx}} + y \cdot 1$.

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

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

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