# How do you implicitly differentiate -y=y-sqrt(x^3y-y) ?

Sep 14, 2016

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

#### Explanation:

implicit differentiation is derived from the following
$\frac{d}{\mathrm{dx}} = \frac{d}{\mathrm{dy}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

so we need to only solve
$\frac{d}{\mathrm{dy}} = 2 - \frac{1}{2 \sqrt{{x}^{3} y - y}} \left({x}^{3} - 1\right) = \frac{4 \sqrt{{x}^{3} y - y}}{2 \sqrt{{x}^{3} y - y}} \left({x}^{3} - 1\right)$

$\frac{d}{\mathrm{dx}} = - \frac{3 y {x}^{2}}{2 \sqrt{{x}^{3} y - y}}$

now lets divide and solve

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \sqrt{{x}^{3} y - y}}{\left(4 \sqrt{{x}^{3} y - y}\right) \left({x}^{3} - 1\right)} \cdot \frac{3 y {x}^{2}}{2 \sqrt{{x}^{3} y - y}}$

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