Nth Derivative of Reciprocal of Mth Power

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Theorem

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

The $n$th derivative of $\dfrac 1 {x^m}$ with respect to $x$ is:

$\dfrac {\d^n} {\d x^n} \dfrac 1 {x^m} = \dfrac {\paren {-1}^n m^{\overline n}} {x^{m + n}}$

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


Corollary

The $n$th derivative of $\dfrac 1 x$ with respect to $x$ is:

$\dfrac {\d^n} {\d x^n} \dfrac 1 x = \dfrac {\paren {-1}^n n!} {x^{n + 1} }$

where $n!$ denotes $n$ factorial.


Proof

Proof by induction:

For all $n \in \N_{>0}$, let $\map P n$ be the proposition:

$\dfrac {\d^n} {\d x^n} \dfrac 1 {x^m} = \dfrac {\paren {-1}^n m^{\overline n}} {x^{m + n}}$


Basis for the Induction

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

\(\ds \frac {\d} {\d x} \frac 1 {x^m}\) \(=\) \(\ds \frac {\d} {\d x} x^{-m}\)
\(\ds \) \(=\) \(\ds \paren {-m} x^{-m - 1}\) Power Rule for Derivatives
\(\ds \) \(=\) \(\ds \frac {-m} {x^{m+1} }\)

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 $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.


So this is our induction hypothesis:

$\dfrac {\d^k} {\d x^k} \dfrac 1 {x^m} = \dfrac {\paren {-1}^k m^{\overline k} } {x^{m + k} }$


Then we need to show:

$\dfrac {\d^{k + 1} } {\d x^{k + 1} } \dfrac 1 {x^m} = \dfrac {\paren {-1}^{k + 1} m^{\overline {k + 1} } } {x^{m + k + 1} }$


Induction Step

This is our induction step:

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

\(\ds \frac {\d^{k + 1} } {\d x^{k + 1} } \frac 1 {x^m}\) \(=\) \(\ds \map {\frac \d {\d x} } {\frac {\d^k} {\d x^k} \frac 1 {x^m} }\)
\(\ds \) \(=\) \(\ds \frac \d {\d x} \frac {\paren {-1}^k m^{\overline k} } {x^{m + k} }\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \paren {-1}^k m^{\overline k} \frac \d {\d x} \frac 1 {x^{m + k} }\) Derivative of Constant Multiple
\(\ds \) \(=\) \(\ds \paren {-1}^k m^{\overline k} \paren {\frac {-\paren {m + k} } {x^{m + k + 1} } }\) Basis for the Induction
\(\ds \) \(=\) \(\ds \frac {\paren {-1}^{k + 1} m^{\overline {k + 1} } } {x^{m + k + 1} }\) Definition of Rising Factorial

$\blacksquare$