# How do you use implicit differentiation to find dy/dx given 4x^2-2xy+3y^2=8?

Jan 26, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y - 4 x}{3 y - x}$

#### Explanation:

Differentiate both sides of the equation with respect to $x$, keeping in mind that:

$\frac{d}{\mathrm{dx}} f \left(y \left(x\right)\right) = f ' \left(y \left(x\right)\right) \cdot y ' \left(x\right)$

$\frac{d}{\mathrm{dx}} \left(4 {x}^{2} - 2 x y + 3 {y}^{2}\right) = 0$

$8 x - 2 y - 2 x y ' + 6 y y ' = 0$

Solve now for $y '$:

$2 y ' \left(3 y - x\right) = 2 y - 8 x$

$y ' = \frac{y - 4 x}{3 y - x}$