How do you implicitly differentiate  x^(2/3) + y^(2/3) = 4?

Jan 13, 2016

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

Explanation:

Differentiating with respect to x :

$\frac{2}{3} {x}^{- \frac{1}{3}} + \frac{2}{3} {y}^{- \frac{1}{3}} . \frac{d}{\mathrm{dx}} \left(y\right) = 0$

$\Rightarrow \frac{2}{3} {x}^{- \frac{1}{3}} + \frac{2}{3} {y}^{- \frac{1}{3}} . \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

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

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

now cancelling $\frac{2}{3}$ and writing with positive indices gives :

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