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

1.1baseL2

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

1.2ihassume-hyp

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

2.1step 1.2L1L3

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

3.1step 2.1L2L3algebra

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

4.1step 1.1step 3.1discharge-induction∎

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

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

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