Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02
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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 chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a)

Statement

Let f:U→V⊆Rn be totally differentiable at a∈U and let g:V→Rp be totally differentiable at f(a). Then g∘f is totally differentiable at a and

D(g∘f)(a)=Dg(f(a))∘Df(a).

Facts & Assumptions

Given: The total first-order expansions of f at a and g at f(a).

[L1]

In the total-derivative definition, the normalized remainder tends to zero as h tends to zero (The total (Fréchet) derivative Df(a) as the linear first-order approximation with o(∥h∥2) remainder).

[L2]

Total differentiability gives a local O(∥h∥2) increment bound and therefore continuity (Total differentiability gives a local O(∥h∥2) increment bound and therefore continuity).

Proof

technique · direct
1.1

Write f(a+h)=f(a)+Df(a)h+rf(h) and g(f(a)+k)=g(f(a))+Dg(f(a))k+rg(k), with both normalized remainders tending to zero.

L1L2
2.1

By [L2], k=f(a+h)−f(a)=O(∥h∥2); boundedness of Dg(f(a)) and the two remainder limits show both Dg(f(a))rf(h) and rg(k) are o(∥h∥2), including the case k=0.

step 1.1L2algebra
3.1

Substitution into the two expansions leaves g(f(a+h))−g(f(a))−Dg(f(a))Df(a)h=o(∥h∥2), and the composite of linear maps is linear.

step 1.1step 2.1∎

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Dependency tree · two levels

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