Nth Derivative of Reciprocal of Mth Power

Theorem

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

The $n$th derivative of $\dfrac 1 {x^m}$ w.r.t. $x$ is:

$\displaystyle \frac {d^n}{dx^n} \frac 1 {x^m} = \frac {\left({-1}\right)^n m^{\overline n}} {z^{m + n}}$

where $m^{\overline n}$ denotes the rising factorial.

Corollary

$\displaystyle \frac {d^n}{dx^n} \frac 1 x = \frac {\left({-1}\right)^n n!} {z^{n + 1}}$

where $n!$ denotes $n$ factorial.

Proof

Proof of Main Result

Proof by induction:

For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition:

$\displaystyle \frac {d^n}{dx^n} \frac 1 {x^m} = \frac {\left({-1}\right)^n m^{\overline n}} {z^{m + n}}$

Basis for the Induction

$P(1)$ is true, as this is the case:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac d {dx} \frac 1 {x^m}\)	\(=\)	\(\displaystyle \frac d {dx} x^{-m}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({-m}\right)x^{-m-1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Power Rule for Derivatives
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {-m} {x^{m+1} }\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

which matches the proposition as $m^{\overline 1} = m$ from the definition of rising factorial.

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:

$\displaystyle \frac {d^k}{dx^k} \frac 1 {x^m} = \frac {\left({-1}\right)^k m^{\overline k}} {z^{m + k}}$

Then we need to show:

$\displaystyle \frac {d^{k+1}}{dx^{k+1}} \frac 1 {x^m} = \frac {\left({-1}\right)^{k+1} m^{\overline {k+1}}} {z^{m + k + 1}}$

Induction Step

This is our induction step:

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

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac {d^{k+1} }{dx^{k+1} } \frac 1 {x^m}\)	\(=\)	\(\displaystyle \frac d {dx} \left({\frac {d^k}{dx^k} \frac 1 {x^m} }\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac d {dx} \frac {\left({-1}\right)^k m^{\overline k} } {z^{m + k} }\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by the induction hypothesis
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({-1}\right)^k m^{\overline k} \frac d {dx} \frac 1 {z^{m + k} }\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Derivative of Constant Multiple
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({-1}\right)^k m^{\overline k} \left({\frac {- \left({m+k}\right)} {z^{m + k + 1} } }\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by the basis for the induction
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {\left({-1}\right)^{k+1} m^{\overline {k+1} } } {z^{m + k + 1} }\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by the definition of rising factorial

$\blacksquare$

Proof of Corollary

Follows directly by putting $m = 1$.

$\blacksquare$