Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02
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Sums and scalar multiples of totally differentiable maps are totally differentiable with the expected derivatives

Statement

If f,g:U→Rn are totally differentiable at a and c∈R, then f+g and cf are totally differentiable at a, with

D(f+g)(a)=Df(a)+Dg(a),D(cf)(a)=cDf(a).

Facts & Assumptions

Given: Total first-order expansions for f and g at 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]

A norm satisfies the triangle inequality and ∥cw∥=∣c∣ ∥w∥ (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

Proof

technique · direct
1.1

Add the two expansions and multiply the first by c to obtain remainders rf+rg for f+g and crf for cf, with the displayed candidate linear maps.

L1L2
2.1

By [L2], ∥rf(h)+rg(h)∥2/∥h∥2 is bounded by the sum of two quantities tending to zero, and ∥crf(h)∥2/∥h∥2=∣c∣ ∥rf(h)∥2/∥h∥2 tends to zero (also when c=0).

step 1.1L2algebra
3.1

Sums and scalar multiples of linear maps are linear, so step 2.1 verifies the definition with exactly the two stated derivatives.

step 1.1step 2.1∎

Depends on

Used by

Dependency tree · two levels

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Sources