# How do you find (dy)/(dx) given x^(2/3)+y^(2/3)=pi^(2/3)?

Jan 13, 2017

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

#### Explanation:

${x}^{\frac{2}{3}} + {y}^{\frac{2}{3}} = {\pi}^{\frac{2}{3}}$

differentiate wrt $x$

$\frac{d}{\mathrm{dx}} \left({x}^{\frac{2}{3}} + {y}^{\frac{2}{3}}\right) = \frac{d}{\mathrm{dx}} \left({\pi}^{\frac{2}{3}}\right)$

the RHS $= 0$ since $\pi$ is a constant.

using the power rule, and chain rule on the $y$ term

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

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

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

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