# Nth Derivative of Mth Power

## Contents

## 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:

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

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 $\map P n$ be the proposition:

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

### Basis for the Induction

$\map P 1$ is true, as this just says:

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

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

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

This is our basis for the induction.

### Induction Hypothesis

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

So this is our induction hypothesis:

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

Then we need to show:

- $\dfrac {\d^{k + 1} } {\d x^{k + 1} } x^m = \begin{cases} m^{\underline {k + 1} } x^{m - \paren {k + 1} } & : k + 1 \le m \\ 0 & : k + 1 > m \end{cases}$

### Induction Step

This is our induction step:

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

\(\displaystyle \frac {\d^{k + 1} } {\d x^{k + 1} } x^m\) | \(=\) | \(\displaystyle \map {\frac \d {\d x} } {\frac {\d^k} {\d x^k} x^m}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map {\frac \d {\d x} } {m^{\underline k} x^{m - k} }\) | Induction Hypothesis | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle m^{\underline k} \map {\frac \d {\d x} } {x^{m - k} }\) | Derivative of Constant Multiple | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle m^{\underline k} \paren {m - k} \paren {x^{m - k - 1} }\) | Basis for the Induction | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle m^{\underline {k + 1} } x^{m - \paren {k + 1} }\) | 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:

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

Then by Derivative of Constant:

- $\dfrac {\d^{k + 1} } {\d x^{k + 1} } x^m = \dfrac \d {\d x} m! = 0$

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:

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

Then by Derivative of Constant:

- $\dfrac {\d^{k + 1} } {\d x^{k + 1} } = 0$

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

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

Therefore:

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

$\blacksquare$