# How do you find the first, second derivative for 3x^(2/3)-x^2?

##### 1 Answer
Aug 4, 2015

You can use the power rule.

#### Explanation:

This function's first and second derivates can be found by using the power rule

$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left({x}^{n}\right) = n \cdot {x}^{n - 1}}$

The first derivative will be equal to

$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = \left[\frac{d}{\mathrm{dx}} \left(3 {x}^{\frac{2}{3}}\right)\right] - \frac{d}{\mathrm{dx}} \left({x}^{2}\right)$

${f}^{'} = \textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} \cdot \frac{2}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}} \cdot {x}^{\frac{2}{3} - 1} - 2 {x}^{2 - 1}$

${f}^{'} = \textcolor{g r e e n}{2 \cdot {x}^{- \frac{1}{3}} - 2 x}$

The second derivative will be equal to

$\frac{d}{\mathrm{dx}} \left({f}^{'} \left(x\right)\right) = \left[\frac{d}{\mathrm{dx}} \left(2 {x}^{- \frac{1}{3}}\right)\right] - \frac{d}{\mathrm{dx}} \left(2 x\right)$

${f}^{' '} = 2 \cdot \left(- \frac{1}{3}\right) \cdot {x}^{- \frac{1}{3} - 1} - 2$

${f}^{' '} = \textcolor{g r e e n}{- \frac{2}{3} \cdot {x}^{- \frac{4}{3}} - 2}$