Derivative of a power

Take a look at the following table and try to find the general rule:

$f (x)$	$f^{'} (x)$
$x^{2}$	$2 x$
$x^{3}$	$3 x^{2}$
$x^{5}$	$5 x^{4}$
$x^{\frac{1}{2}}$	$\frac{1}{2} x - \frac{1}{2}$
$2 x^{2}$	$4 x$
$2 x^{3}$	$6 x^{2}$
$5 x^{6}$	$30 x^{5}$
$x^{n}$	?
$A x^{n}$	?

Solution: $\begin{array}{ll} f (x) = x^{n} & f^{'} (x) = n x^{n - 1} \\ f (x) = A x^{n} & f^{'} (x) = A n x^{n - 1} \end{array}$

Now verify the derivatives in the table by trying to identify what is the $A$ and what is the $n$ in each of the cases.

We have thus obtained a general formula. We need to emphasize that this formula is only applicable when $n$ is a rational number. We will see some examples that will show that we need to bear in mind this fact. Note also the following:

Example

If we have a function with a square or cubic root or any type of root we can rewrite it with a power, and we can then apply the rule.
When $n = 0$ the derivative is zero, since any number raised to $0$ is $1$ , which is a constant, and therefore the derivative is zero.

Summing up, then, the general formula has been deduced to derive three types of fundamental functions: constant function, linear function and any power. Check it in the following table:

$f (x) = A$	$f^{'} (x) = 0$
$f (x) = A x + b$	$f^{'} (x) = A$
$f (x) = A x^{n}$	$f^{'} (x) = A \cdot n \cdot x^{n - 1}$

and look at the following examples:

Example

a) $\begin{array}{ll} f (x) = 30 x + 5 & f^{'} (x) = 30 \end{array}$

b) $\begin{array}{ll} f (x) = 4 (x + 1) & f^{'} (x) = 4 \end{array}$

c) $\begin{array}{ll} f (x) = 3 (5 x + 2) & f^{'} (x) = 15 \end{array}$

d) $\begin{array}{ll} f (x) = 6 (x^{4} + 5) & f^{'} (x) = 6 \cdot 4 x^{3} = 24 x^{3} \end{array}$

e) $f (x) = \sqrt{x} = x^{\frac{1}{2}}$ $f^{'} (x) = \frac{1}{2} x^{\frac{1}{2} - 1} = \frac{1}{2} x^{- \frac{1}{2}} = \frac{1}{2 \sqrt{x}}$

f) $f (x) = \sqrt[3]{\sqrt{x^{2}}}$ $f^{'} (x) = \frac{2}{3} x^{\frac{2}{3} - 1} = \frac{2}{3} x^{- \frac{1}{3}} = \frac{2}{3 \sqrt[3]{x}}$