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

Nov 6, 2016

$\setminus \frac{2 x {y}^{2} + y - 3 {x}^{2} - 2}{3 {y}^{2} - 4 y - 2 y {x}^{2} + 1 - x} = \setminus \frac{\mathrm{dy}}{\mathrm{dx}}$

#### Explanation:

First, you remove the fraction by multiplying by $x - {y}^{2}$
$- 2 \left(x - {y}^{2}\right) = y - y \left(x - {y}^{2}\right) + {x}^{2} \left(x - {y}^{2}\right)$
expand
$- 2 x + 2 {y}^{2} = y - y x + {y}^{3} + {x}^{3} - {x}^{2} {y}^{2}$
Next, I like to move all terms onto one side and put them in a good order:
$0 = {y}^{3} - 2 {y}^{2} - {x}^{2} {y}^{2} + y - y x + {x}^{3} + 2 x$
Next, take the derivative:
$0 = \setminus \frac{\mathrm{dd}}{x} \left[{y}^{3}\right] - \setminus \frac{\mathrm{dd}}{x} \left[2 {y}^{2}\right] - \setminus \frac{\mathrm{dd}}{x} \left[{x}^{2} {y}^{2}\right] + \setminus \frac{\mathrm{dd}}{x} \left[y\right] - \setminus \frac{\mathrm{dd}}{x} \left[y x\right] + \setminus \frac{\mathrm{dd}}{x} \left[{x}^{3}\right] + \setminus \frac{\mathrm{dd}}{x} \left[2 x\right]$
Use the product rule, chain rule, and power rule:
$0 = 3 {y}^{2} \setminus \frac{\mathrm{dy}}{\mathrm{dx}} - 4 y \setminus \frac{\mathrm{dy}}{\mathrm{dx}} - \left({x}^{2} \setminus \cdot 2 y \setminus \frac{\mathrm{dy}}{\mathrm{dx}} + 2 x {y}^{2}\right) + \setminus \frac{\mathrm{dy}}{\mathrm{dx}} - \left(y + x \setminus \frac{\mathrm{dy}}{\mathrm{dx}}\right) + 3 {x}^{2} + 2$

group all $\setminus \frac{\mathrm{dy}}{\mathrm{dx}}$ terms
$0 = 3 {y}^{2} \setminus \frac{\mathrm{dy}}{\mathrm{dx}} - 4 y \setminus \frac{\mathrm{dy}}{\mathrm{dx}} - 2 y {x}^{2} \setminus \frac{\mathrm{dy}}{\mathrm{dx}} + \setminus \frac{\mathrm{dy}}{\mathrm{dx}} - x \setminus \frac{\mathrm{dy}}{\mathrm{dx}} - 2 x {y}^{2} - y + 3 {x}^{2} + 2$

factor $\setminus \frac{\mathrm{dy}}{\mathrm{dx}}$
$0 = \setminus \frac{\mathrm{dy}}{\mathrm{dx}} \left(3 {y}^{2} - 4 y - 2 y {x}^{2} + 1 - x\right) - 2 x {y}^{2} - y + 3 {x}^{2} + 2$

subtract non $\setminus \frac{\mathrm{dy}}{\mathrm{dx}}$ terms
$2 x {y}^{2} + y - 3 {x}^{2} - 2 = \setminus \frac{\mathrm{dy}}{\mathrm{dx}} \left(3 {y}^{2} - 4 y - 2 y {x}^{2} + 1 - x\right)$

divide both sides by$3 {y}^{2} - 4 y - 2 y {x}^{2} + 1 - x$
$\setminus \frac{2 x {y}^{2} + y - 3 {x}^{2} - 2}{3 {y}^{2} - 4 y - 2 y {x}^{2} + 1 - x} = \setminus \frac{\mathrm{dy}}{\mathrm{dx}}$