# How do you implicitly differentiate xy^2-x/y=-1?

Dec 22, 2016

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

#### Explanation:

Firstly multiply everything by $y$ to get rid of the awkward denominator.

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

now differentiate both sides wrt $x$ remembering that any $f \left(y\right)$ will be differentiated wrt $y$ then multiplied by $\frac{\mathrm{dy}}{\mathrm{dx}}$

Also the first term will need the use of the product rule.

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

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

now rearrange for$\text{ } \frac{\mathrm{dy}}{\mathrm{dx}}$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} \left[3 x {y}^{2} + 1\right] = 1 - {y}^{3}$

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