Proof of the Generalized Leibniz Rule | 신승민

Let $I$ be an interval, and let $u,v:I\to\mathbb{R}$ . For every positive integer $n$ , if $u$ and $v$ are $n$ times differentiable on $I$ , the generalized Leibniz rule states that

\frac{d^n}{dx^n}(uv) = \sum_{k=0}^{n} \binom{n}{k} u^{(k)}v^{(n-k)}

The proof proceeds by induction. The base case $n=1$ is the ordinary product rule:

\frac{d}{dx}(uv)=u'v+uv'

For the induction step, let $u$ and $v$ be $(n+1)$ times differentiable on $I$ , and assume that the formula holds for some positive integer $n$ . Each term on the right-hand side is then differentiable, so differentiating both sides and applying the product rule to each term gives

\frac{d^{n+1}}{dx^{n+1}}(uv) = \sum_{k=0}^{n} \binom{n}{k} u^{(k+1)}v^{(n-k)} + \sum_{k=0}^{n} \binom{n}{k} u^{(k)}v^{(n-k+1)}

Reindexing the first sum by letting $j=k+1$ , and then renaming $j$ as $k$ , gives

\frac{d^{n+1}}{dx^{n+1}}(uv) = \sum_{k=1}^{n+1} \binom{n}{k-1} u^{(k)}v^{(n+1-k)} + \sum_{k=0}^{n} \binom{n}{k} u^{(k)}v^{(n+1-k)}

For the interior terms $1\leq k\leq n$ , Pascal’s identity gives

\binom{n}{k-1}+\binom{n}{k} = \binom{n+1}{k}

The endpoint terms $k=0$ and $k=n+1$ also have the correct coefficients because

\binom{n}{0}=\binom{n+1}{0}=1

and

\binom{n}{n}=\binom{n+1}{n+1}=1

Therefore,

\frac{d^{n+1}}{dx^{n+1}}(uv) = \sum_{k=0}^{n+1} \binom{n+1}{k} u^{(k)}v^{(n+1-k)}

Thus, by mathematical induction, the generalized Leibniz rule holds for every positive integer $n$ .

Derivation of the generalized Leibniz rule, beginning with the product rule

Continuation of the proof using Pascal's identity to complete the induction

AI Use Disclosure: AI assisted with drafting and refining the proof explanation.