# How do you find the derivative of (x^(2/3 ))(8-x)?

Jun 29, 2015

$\frac{d \left(8 {x}^{\frac{2}{3}} - {x}^{- \frac{1}{3}}\right)}{\mathrm{dx}} = \frac{1}{3 \sqrt[3]{x}} \cdot \left(16 + \frac{1}{x}\right)$

#### Explanation:

$\left({x}^{\frac{2}{3}}\right) \left(8 - x\right)$
$\textcolor{w h i t e}{\text{XXXX}}$$= 8 {x}^{\frac{2}{3}} - {x}^{- \frac{1}{3}}$

$\frac{d \left(8 {x}^{\frac{2}{3}} - {x}^{- \frac{1}{3}}\right)}{\mathrm{dx}} = \textcolor{red}{\frac{d \left(8 {x}^{\frac{2}{3}}\right)}{\mathrm{dx}}} - \textcolor{b l u e}{\frac{d \left({x}^{- \frac{1}{3}}\right)}{\mathrm{dx}}}$

$\textcolor{w h i t e}{\text{XXXX}}$$= \textcolor{red}{\frac{2}{3} \cdot 8 {x}^{-} \left(\frac{1}{3}\right)} - \textcolor{b l u e}{\left(- \frac{1}{3}\right) {x}^{- \frac{4}{3}}}$

$\textcolor{w h i t e}{\text{XXXX}}$$= \frac{16}{3 \sqrt[3]{x}} + \frac{1}{3 {\left(\sqrt[3]{x}\right)}^{4}}$

$\textcolor{w h i t e}{\text{XXXX}}$$= \frac{16}{3 \sqrt[3]{x}} + \frac{1}{3 \left(x \cdot \sqrt[3]{x}\right)}$

$\textcolor{w h i t e}{\text{XXXX}}$$= \frac{1}{3 \sqrt[3]{x}} \cdot \left(16 + \frac{1}{x}\right)$