# How do you find the nth derivative of the function f(x)=x^n?

Jan 28, 2017

n!

#### Explanation:

$f \left(x\right) = {x}^{n}$
$f ' \left(x\right) = n {x}^{n - 1}$'
$f ' ' \left(x\right) = n \left(n - 1\right) {x}^{n - 2}$
...
${f}^{\left(n\right)} \left(x\right) = n \left(n - 1\right) \ldots 3.2 .1 {x}^{0}$
=n!

Or by induction on $n$ if you want a formal proof.

Jan 28, 2017

Each derivative gives us a pattern.

$f ' \left(x\right) = n {x}^{n - 1}$

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

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

and so on until $n - k = 0$ where $k$ is the order of the derivative. When we finish, we get:

${f}^{\left(k\right)} \left(x\right) = n \left(n - 1\right) \left(n - 2\right) \cdots \left(n - k + 1\right) {x}^{n - k}$

When we go all the way to $n = k$, then:

$\textcolor{b l u e}{{f}^{\left(n\right)} \left(x\right) = n \left(n - 1\right) \left(n - 2\right) \cdots \left(1\right) {\cancel{{x}^{0}}}^{1}}$

which is a constant equaling color(blue)(n!), as n! = n(n-1)(n-2)cdots(2)(1), and ${x}^{0} = 1$ which doesn't affect n!.