TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-01

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced — the exact statement appears in a cited source; only wording and notation differ.
AI-adapted — a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated — a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

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

Statement

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

Facts & Assumptions

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

[L1]

Products satisfy $(uv)'=u'v+uv'$ (Sums, scalar multiples, products and quotients: $(f+g)'(c) = f'(c) + g'(c)$ , $(\alpha f)'(c) = \alpha f'(c)$ , $(fg)'(c) = f'(c)g(c) + f(c)g'(c)$ , and $(f/g)'(c) = \bigl(f'(c)g(c) - f(c)g'(c)\bigr)/g(c)^{2}$ when $g(c) \ne 0$ ).

[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 $\binom{n}{k} := \lvert [n]^{k}\rvert$ (Pascal's rule $\binom{n+1}{k+1} = \binom{n}{k} + \binom{n}{k+1}$ , and the hockey-stick identity $\sum_{i \le n}\binom{i}{k} = \binom{n+1}{k+1}$ ).

[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 $\iota$ preserves natural addition (The canonical natural $\iota(n) = n \cdot 1_F$ of a field, Canonical naturals are positive and strictly increasing).

Proof

technique · induction

1.1

For $n=0$ , the displayed sum is $\iota\binom00 f^{(0)}g^{(0)}=fg=(fg)^{(0)}$ .

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 $(fg)^{(k+1)}=\sum_{j=0}^{k}\iota\binom kj\bigl(f^{(j+1)}g^{(k-j)}+f^{(j)}g^{(k-j+1)}\bigr)$ .

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 $(fg)^{(k+1)}=\sum_{j=0}^{k+1}\iota\binom{k+1}{j}f^{(j)}g^{(k+1-j)}$ .

step 2.1L2L3algebra

4.1

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

step 1.1step 3.1discharge-induction∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 106 results over 27 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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 nnn-th derivative of a product