# What is the derivative of x^(2/3)+y^(2/3)=5 at the given point of (8,1)?

## What is the derivative of ${x}^{\frac{2}{3}} + {y}^{\frac{2}{3}} = 5$ at the given point of $\left(8 , 1\right)$?

##### 1 Answer
Oct 17, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{2}$ at $\left(x , y\right) = \left(8 , 1\right)$

#### Explanation:

First, let's find $\frac{\mathrm{dy}}{\mathrm{dx}}$ using implicit differentiation:

$\frac{d}{\mathrm{dx}} \left({x}^{\frac{2}{3}} + {y}^{\frac{2}{3}}\right) = \frac{d}{\mathrm{dx}} 5$

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

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

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

Now, we evaluate $\frac{\mathrm{dy}}{\mathrm{dx}}$ at our given point of $\left(x , y\right) = \left(8 , 1\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} {|}_{\left(x , y\right) = \left(8 , 1\right)} = - {\left(\frac{8}{1}\right)}^{- \frac{1}{3}}$

$= - {8}^{- \frac{1}{3}}$

$= - \frac{1}{2}$