# How do you differentiate (y-1)^2+y-x=0?

Nov 27, 2016

${\left(y - 1\right)}^{2} + y - x = 0$

Implicitly differentiate (use the chain rule and power rule):
$2 \left(y - 1\right) \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) + \frac{\mathrm{dy}}{\mathrm{dx}} - 1 = 0$

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

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

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

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