# How do you find dy/dx by implicit differentiation of x^3+y^3=8?

Feb 13, 2017

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

#### Explanation:

We have:

${x}^{3} + {y}^{3} = 8$

Differentiate wrt $x$; then Apply the chain rule (implicit differentiation):

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