# How do you find (dy)/(dx) given y(x+1)-y^2=x?

Oct 30, 2016

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

#### Explanation:

Distribute y:

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

Differentiate:

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

Move y to the right side:

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

Factor out $\frac{\mathrm{dy}}{\mathrm{dx}}$

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

Divide both sides by the leading coefficient:

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