# How do you use implicit differentiation to find dy/dx given ln(x^2+y^2-xy)=100?

Nov 13, 2016

${x}^{2} + {y}^{2} - x y = {e}^{100}$

We use the product rule, power rule, implicit differentiation and the rule $\left({e}^{f \left(x\right)}\right) ' = f ' \left(x\right) \times {e}^{f \left(x\right)}$

$2 x + 2 y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) - \left(y + x \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\right) = 0 \left({e}^{100}\right)$

$2 x + 2 y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) - y - x \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$

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

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

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

Hopefully this helps!