# How do you find the derivative of  x^2 + xy + y^2 =7?

Oct 12, 2016

$\frac{d}{\mathrm{dx}} \left({x}^{2} + x y + {y}^{2}\right) = \frac{d}{\mathrm{dx}} \left(7\right)$

$\frac{d}{\mathrm{dx}} \left({x}^{2}\right) + \frac{d}{\mathrm{dx}} \left(x y\right) + \frac{d}{\mathrm{dx}} \left({y}^{2}\right) = \frac{d}{\mathrm{dx}} \left(7\right)$

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

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

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

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

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

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

Hopefully this helps!