Derivative of a power

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

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

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

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:

• 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:

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

and look at the following examples:

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

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

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

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

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

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