How do you use implicit differentiation to find dy/dx given x^2-2y^2+x+3y-4=0?

Mar 13, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x + 1}{2 y - 3}$

Explanation:

Differentiate both sides of the equation with respect to $x$:

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

$2 x - 2 y \frac{\mathrm{dy}}{\mathrm{dx}} + 1 + 3 \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$2 x + 1 = \left(2 y - 3\right) \frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x + 1}{2 y - 3}$