TheoremStatement: AI-adaptedProof: AI-generated

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 chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$

Statement

Let $f : U \to V \subseteq R^{n}$ be totally differentiable at $a \in U$ and let $g : V \to R^{p}$ be totally differentiable at $f (a)$ . Then $g \circ f$ is totally differentiable at $a$ and

$D (g \circ f) (a) = D g (f (a)) \circ D f (a) .$

Facts & Assumptions

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

[L1]

The total-derivative definition supplies the normwise remainder estimate for every $ε > 0$ ; in positive source dimension this is equivalently the normalized remainder tending to zero as $h$ tends to zero (The total (Fréchet) derivative $D f (a)$ 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

1.1givenL1L2algebra

If the source dimension is zero or $p = 0$ , the derivative of the composite and the displayed composite of derivatives are the unique zero linear map, and the remainder estimate is identically zero. If the intermediate dimension $n = 0$ , then $f$ is constant with value $0$ , $D f (a) = 0$ , and $g \circ f$ is constant with zero derivative; again both sides agree. For the remaining proof all three dimensions are positive.

1.2L1L2

Write $f (a + h) = f (a) + D f (a) h + r_{f} (h)$ and $g (f (a) + k) = g (f (a)) + D g (f (a)) k + r_{g} (k)$ , with both normalized remainders tending to zero.

2.1step 1.2L2algebra

By [L2], $k = f (a + h) - f (a) = O (∥ h ∥_{2})$ ; boundedness of $D g (f (a))$ and the two remainder limits show both $D g (f (a)) r_{f} (h)$ and $r_{g} (k)$ are $o (∥ h ∥_{2})$ , including the case $k = 0$ .

3.1step 1.2step 2.1∎

Substitution into the two expansions leaves $g (f (a + h)) - g (f (a)) - D g (f (a)) D f (a) h = o (∥ h ∥_{2})$ , and the composite of linear maps is linear.

Depends on

Dependency tree · 2 levels

Sources

J. Lebl, Basic Analysis I, §8.3