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

Oct 12, 2016

Use implicit differentiation. (Please see the explanation)

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

#### Explanation:

The process is called implicit differentiation.

Multiply both sides by -y:

$- 2 x y + {y}^{2} = - {y}^{3} + {x}^{2}$

Move all of the terms containing y to the left side:

${y}^{3} + {y}^{2} - 2 x y = {x}^{2}$

Differentiate remembering that $\frac{\mathrm{df} \left(y\right)}{\mathrm{dx}} = \left(\frac{\mathrm{df} \left(y\right)}{\mathrm{dy}}\right) \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$:

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

Note: Differentiation of the term $- 2 x y$ required the use of the product rule $\left(g h\right) ' = g ' h + g h '$ where $g = - 2 x , g ' = - 2 , h = y \mathmr{and} h ' = \frac{\mathrm{dy}}{\mathrm{dx}}$

Move all of the terms NOT containing $\frac{\mathrm{dy}}{\mathrm{dx}}$ to the right:

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

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

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

Divide both sides by $\left(3 {y}^{2} + 2 y - 2 x\right)$:

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