# How do you differentiate (y+x)^2+2y-x=0?

Dec 11, 2016

If ${\left(y + x\right)}^{2} + 2 y - x = 0$:

${y}^{2} + 2 x y + {x}^{2} + 2 y - x = 0$

$2 y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) + 2 \left(y + x \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\right) + 2 x + 2 \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) - 1 = 0$

$2 y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) + 2 y + 2 x \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) + 2 x + 2 \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) - 1 = 0$

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

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

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

Hopefully this helps!