# How do you implicitly differentiate -1=1/(xy)-1-3y^2?

Aug 30, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y}{x \left(6 x {y}^{3} + 1\right)}$

#### Explanation:

we rely on the chain rule which states

$\frac{d}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dx}} \cdot \frac{d}{\mathrm{dy}}$

$\frac{d}{\mathrm{dx}} = - \frac{1}{{x}^{2} y}$
$\frac{d}{\mathrm{dy}} = - \frac{1}{{y}^{2} x} - 6 y$

now we plug into our formula and solve
$- \frac{1}{{x}^{2} y} = \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) \left(- \frac{1}{{y}^{2} x} - 6 y\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{{x}^{2} y} / \left(\frac{1}{{y}^{2} x} + 6 y\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y}{x \left(6 x {y}^{3} + 1\right)}$