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

Dec 5, 2015

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

#### Explanation:

We assume that y is a function of x, ie $y = f \left(x\right)$ and then differentiate each side of the equation with respect to x, then re-arrange and solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$.

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

$\therefore 3 {x}^{2} - 3 x \frac{\mathrm{dy}}{\mathrm{dx}} - 3 y + 6 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$. (Used product rule and power rule)

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} \left(- 3 x + 6 {y}^{2}\right) = 3 y - 3 {x}^{2}$. (Took out common factor)

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

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y - {x}^{2}}{2 {y}^{2} - x}$ (Divide all by 3).