# Nth Derivative of Mth Power

## Theorem

Let $m \in \Z$ be an integer such that $m \ge 0$.

The $n$th derivative of $x^m$ with respect to $x$ is:

$\dfrac {\mathrm d^n}{\mathrm d x^n} x^m = \begin{cases} m^{\underline n} x^{m - n} & : n \le m \\ 0 & : n > m \end{cases}$

where $m^{\underline n}$ denotes the falling factorial.

### Corollary

$\dfrac {\d^n} {\d x^n} x^n = n!$

where $n!$ denotes $n$ factorial.

## Proof

Proof by induction:

For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition:

$\dfrac {\mathrm d^n}{\mathrm d x^n} x^m = \begin{cases} m^{\underline n} x^{m - n} & : n \le m \\ 0 & : n > m \end{cases}$

### Basis for the Induction

$P(1)$ is true, as this just says:

$\dfrac {\mathrm d} {\mathrm dx} x^m = m x^{m - 1}$

This follows by Power Rule for Derivatives, which also includes the case:

$\dfrac {\mathrm d} {\mathrm d x} x^0 = 0$

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:

$\dfrac {\mathrm d^k}{\mathrm d x^k} x^m = \begin{cases} m^{\underline k} x^{m - k} & : k \le m \\ 0 & : k > m \end{cases}$

Then we need to show:

$\dfrac {\mathrm d^{k+1}}{\mathrm d x^{k+1}} x^m = \begin{cases} m^{\underline {k+1}} x^{m - \left({k+1}\right)} & : {k+1} \le m \\ 0 & : {k+1} > m \end{cases}$

### Induction Step

This is our induction step:

First, let $k < m$. Then we have:

 $\displaystyle \frac {\mathrm d^{k+1} }{\mathrm d x^{k+1} } x^m$ $=$ $\displaystyle \frac {\mathrm d}{\mathrm d x} \left({\frac {\mathrm d^k}{\mathrm d x^k} x^m}\right)$ $\displaystyle$ $=$ $\displaystyle \frac {\mathrm d}{\mathrm d x} \left({m^{\underline k} x^{m - k} }\right)$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle m^{\underline k} \frac {\mathrm d} {\mathrm d x} \left({ x^{m - k} }\right)$ Derivative of Constant Multiple $\displaystyle$ $=$ $\displaystyle m^{\underline k} \left({m-k}\right) \left({ x^{m - k - 1} }\right)$ Basis for the Induction $\displaystyle$ $=$ $\displaystyle m^{\underline {k+1} } x^{m - \left({k+1}\right)}$ Definition of Falling Factorial

At this stage, as $k < m$, we have that $k+1 \le m$. So far so good.

Now suppose that $k = m$.

Then by the induction hypothesis:

$\dfrac {\mathrm d^k}{\mathrm d x^k} x^m = m^{\underline k} x^0 = m!$

Then $\dfrac {\mathrm d^{k+1}}{\mathrm d x^{k+1}} x^m = \dfrac {\mathrm d}{\mathrm d x} m! = 0$ by Derivative of Constant.

At this stage, as $k = m$, we have that $k+1 > m$. Again, so far so good.

Finally, suppose that $k > m$.

Then by the induction hypothesis:

$\dfrac {\mathrm d^k}{\mathrm d x^k} x^m = 0$.

Then $\dfrac {\mathrm d^{k+1}}{\mathrm d x^{k+1}} = 0$ by Derivative of Constant.

At this stage, as $k > m$, we have that $k+1 > m$. This is all that is needed.

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:

$\dfrac {\mathrm d^n}{\mathrm d x^n} x^m = \begin{cases} m^{\underline n} x^{m - n} & : n \le m \\ 0 & : n > m \end{cases}$

$\blacksquare$