# What is the implicit derivative of 1= xy-(xy-1)^2+2y ?

Dec 17, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x {y}^{2} - 3 y}{3 x - 2 {x}^{2} y + 2}$

#### Explanation:

Differentiate both sides of the equation with respect to $x$, bearing in mind that $y$ is a function of $x$. Then solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$.

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

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

$\therefore 0 = x \frac{\mathrm{dy}}{\mathrm{dx}} + y - 2 {x}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} - 2 x {y}^{2} + 2 x \frac{\mathrm{dy}}{\mathrm{dx}} + 2 y + 2 \frac{\mathrm{dy}}{\mathrm{dx}}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x {y}^{2} - 3 y}{3 x - 2 {x}^{2} y + 2}$