How do you implicitly differentiate  x/y = - x^2 + y^2 -4?

Nov 4, 2016

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

Explanation:

$\frac{x}{y} = - {x}^{2} + {y}^{2} - 4$

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

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

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

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 + 2 x y}{y} \cdot {y}^{2} / \left(2 {y}^{3} + x\right)$

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