TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-01

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The general Leibniz rule for the $n$ -th derivative of a product

Statement

If $n \in N$ and $f, g$ are $n$ -times differentiable on an interval $I$ , then $(f g)^{(n)} = \sum_{j = 0}^{n} ι ⁣ (\binom{n}{j}) f^{(j)} g^{(n - j)} .$

Facts & Assumptions

Given: $n \in N$ and functions $f, g$ with all derivatives through order $n$ .

[L1]

Products satisfy $(u v)^{'} = u^{'} v + u v^{'}$ (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

[L2]

Pascal's rule says $(\binom{k + 1}{j}) = (\binom{k}{j}) + (\binom{k}{j - 1})$ , with the boundary values from The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ (Pascal's rule $(\binom{n + 1}{k + 1}) = (\binom{n}{k}) + (\binom{n}{k + 1})$ , and the hockey-stick identity $\sum_{i \leq n} (\binom{i}{k}) = (\binom{n + 1}{k + 1})$ ).

[L3]

Finite sums split and reindex as stated in Finite sums and finite products, by recursion and Laws of finite sums and finite products, and the canonical embedding $ι$ preserves natural addition (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing).

Proof

technique · induction

1.1

For $n = 0$ , the displayed sum is $ι (\binom{0}{0}) f^{(0)} g^{(0)} = f g = (f g)^{(0)}$ .

baseL2

1.2

Assume the formula holds at an index $k$ , and assume $f, g$ have derivatives through order $k + 1$ .

ihassume-hyp

2.1

Differentiating the finite sum gives $(f g)^{(k + 1)} = \sum_{j = 0}^{k} ι (\binom{k}{j}) (f^{(j + 1)} g^{(k - j)} + f^{(j)} g^{(k - j + 1)})$ .

step 1.2L1L3

3.1

Shift $j + 1$ in the first sum, retain $j$ in the second, and combine the two interior coefficients by Pascal's rule; the two boundary coefficients are $1$ . The result is $(f g)^{(k + 1)} = \sum_{j = 0}^{k + 1} ι (\binom{k + 1}{j}) f^{(j)} g^{(k + 1 - j)}$ .

step 2.1L2L3algebra

4.1

Thus the formula holds for every $n \in N$ .

step 1.1step 3.1discharge-induction∎

Depends on

Used by

Dependency tree · two levels

53 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. Lebl, Basic Analysis I, Taylor's theorem and related calculus (standard reference, not scraped)
MIT OpenCourseWare 18.100B Real Analysis, Spring 2025 lecture notes (standard reference, not scraped)
UTSA Mathematics, Differentiation rules (standard reference, not scraped)
University of Chicago MATH 395 notes (standard reference, not scraped)

The general Leibniz rule for the nn-th derivative of a product