# How do you use implicit differentiation to find dy/dx given x^2-y^3=9?

##### 2 Answers
Nov 12, 2016

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

#### Explanation:

${x}^{2} - {y}^{3} = 9$

$2 x - 3 {y}^{2} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$

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

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

The answer is:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x}{3 {y}^{2}}$

Nov 12, 2016

Please see the explanation.

#### Explanation:

First term:

$d \frac{{x}^{2}}{\mathrm{dx}} = 2 x$

Second term:

$\frac{d \left(- {y}^{3}\right)}{\mathrm{dx}}$ = -3y^2(dy/dx)#

Last term:

$\frac{d \left(9\right)}{\mathrm{dx}} = 0$

Put each term into its corresponding locaton:

$2 x - 3 {y}^{2} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$

Solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$:

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

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