# What is the derivative of (x^(2/3 ))(8-x)?

Feb 28, 2016

$\left[\left({x}^{\frac{2}{3}}\right) \left(8 - x\right)\right] ' = \frac{16 - 5 x}{3 \sqrt[3]{x}}$

#### Explanation:

Recall the following:

$1$. Product Rule: $\left[f \left(x\right) g \left(x\right)\right] ' = f \left(x\right) g ' \left(x\right) + f ' \left(x\right) g \left(x\right)$
$2$. Difference Rule: $\left[f \left(x\right) - g \left(x\right)\right] ' = f ' \left(x\right) - g ' \left(x\right)$
$3$. Constant Rule: $c ' = 0$
$4$. Power Rule: $\left({x}^{n}\right) ' = n {x}^{n - 1}$

Finding the Derivative
$1$. Start by rewriting the equation using the product rule.

$\left[\left({x}^{\frac{2}{3}}\right) \left(8 - x\right)\right] '$

$= \left({x}^{\frac{2}{3}}\right) \left(\textcolor{red}{8 - x}\right) ' + \left(\textcolor{p u r p \le}{{x}^{\frac{2}{3}}}\right) ' \left(8 - x\right)$

$2$. Use the difference rule and constant rule for $\left(\textcolor{red}{8 - x}\right)$. Recall that the derivative of a constant is always $\textcolor{\mathmr{and} a n \ge}{0}$ and the derivative of $x$ is always $\textcolor{b l u e}{1}$.

$= \left({x}^{\frac{2}{3}}\right) \left(\textcolor{red}{8 ' - x '}\right) + \left(\textcolor{p u r p \le}{{x}^{\frac{2}{3}}}\right) ' \left(8 - x\right)$

$= \left({x}^{\frac{2}{3}}\right) \left(\textcolor{\mathmr{and} a n \ge}{0} - \textcolor{b l u e}{1}\right) + \left(\textcolor{p u r p \le}{{x}^{\frac{2}{3}}}\right) ' \left(8 - x\right)$

$3$. Use the power rule for $\textcolor{p u r p \le}{{x}^{\frac{2}{3}}}$.

$= \left({x}^{\frac{2}{3}}\right) \left(0 - 1\right) + \left(\frac{2}{3} {x}^{- \frac{1}{3}}\right) \left(8 - x\right)$

$4$. Simplify.

$= - {x}^{\frac{2}{3}} + \left(\frac{2}{3 {x}^{\frac{1}{3}}}\right) \left(8 - x\right)$

$= \left(\frac{2}{3 \sqrt[3]{x}}\right) \left(8 - x\right) - {x}^{\frac{2}{3}}$

$= \frac{16 - 2 x}{3 \sqrt[3]{x}} - {x}^{\frac{2}{3}}$

$= \frac{16 - 2 x}{3 \sqrt[3]{x}} - \frac{{x}^{\frac{2}{3}} \left(3 \sqrt[3]{x}\right)}{3 \sqrt[3]{x}}$

$= \frac{16 - 2 x - {x}^{\frac{2}{3}} \left(3 \sqrt[3]{x}\right)}{3 \sqrt[3]{x}}$

$= \frac{16 - 2 x - {x}^{\frac{2}{3}} \left(3 {x}^{\frac{1}{3}}\right)}{3 \sqrt[3]{x}}$

$= \frac{16 - 2 x - 3 x}{3 \sqrt[3]{x}}$

$= \textcolor{g r e e n}{\frac{16 - 5 x}{3 \sqrt[3]{x}}}$

$\therefore$, the derivative is $\frac{16 - 5 x}{3 \sqrt[3]{x}}$.