# How do you differentiate 2x^2-3xy+3y^2=2 at (1,1)?

May 2, 2015

To do this you differentiate implicitly

$2 {x}^{2} - 3 x y + 3 {y}^{2} = 2$ becomes,

$4 x - 3 y - 3 x \frac{\mathrm{dy}}{\mathrm{dx}} + 6 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\implies \left(6 y - 3 x\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 3 y - 4 x$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 y - 4 x}{6 y - 3 x}$

At point $\left(1 , 1\right)$ $y = 1 \mathmr{and} x = 1$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 - 4}{6 - 3} = - \frac{1}{3}$