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

May 22, 2015

Use the power rule along with "linearity":

$f ' \left(x\right) = 2 {x}^{- \frac{1}{3}} - 2 x$ and $f ' ' \left(x\right) = - \frac{2}{3} {x}^{- \frac{4}{3}} - 2$.

The power rule says that $\frac{d}{\mathrm{dx}} \left({x}^{n}\right) = n {x}^{n - 1}$ for any fixed number $n$. Linearity of the derivative operator says that $\frac{d}{\mathrm{dx}} \left(a f \left(x\right) + b g \left(x\right)\right) = a f ' \left(x\right) + b g ' \left(x\right)$ for any fixed numbers $a$ and $b$ and any differentiable functions $f$ and $g$.

Linearity can be extended, by using mathematical induction, to any finite "linear combination" of differentiable functions:

$\frac{d}{\mathrm{dx}} \left({a}_{1} {f}_{1} \left(x\right) + {a}_{2} {f}_{2} \left(x\right) + \setminus \cdots + {a}_{n} {f}_{n} \left(x\right)\right) = {a}_{1} {f}_{1} ' \left(x\right) + {a}_{2} {f}_{2} ' \left(x\right) + \setminus \cdots + {a}_{n} {f}_{n} ' \left(x\right)$

This can also be written using summation (sigma) notation as:

$\frac{d}{\mathrm{dx}} \left(\setminus {\sum}_{k = 1}^{n} {a}_{k} {f}_{k} \left(x\right)\right) = \setminus {\sum}_{k = 1}^{n} {a}_{k} {f}_{k} ' \left(x\right)$

which can be thought of as a "distributive-like" property (even though the symbol $\frac{d}{\mathrm{dx}}$ is not a number that is being multiplied).

Furthermore, this is often extended to infinite series (sums), especially "power series", as long as $x$ is within the interior of the "interval of convergence":

$\frac{d}{\mathrm{dx}} \left(\setminus {\sum}_{k = 0}^{\setminus \infty} {a}_{k} {\left(x - c\right)}^{k}\right) = \setminus {\sum}_{k = 1}^{\setminus \infty} {a}_{k} k {\left(x - c\right)}^{k - 1}$ (there's no typo here; the $k = 0$ case on the right can be omitted)