Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02
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If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative

Statement

Let U⊆Rm be open and let f:U→Rn. Suppose every partial derivative ∂jf exists on a neighbourhood of a∈U and is continuous at a. Then f is totally differentiable at a, and Df(a) is the linear map with matrix Jf(a).

Facts & Assumptions

Given: The stated neighbourhood existence and continuity hypotheses for all vector partial derivatives.

[L1]

Coordinate-by-coordinate increments stay inside a Euclidean ball and telescope the total increment (Small coordinate-by-coordinate increments stay inside a Euclidean ball and telescope the total increment).

[L2]

The vector mean-value inequality says ∥f(b)−f(a)∥2≤M(b−a) on a real interval when the derivative norm is bounded by M (The mean value inequality: if f:[a,b]→Rm is continuous and differentiable on (a,b) with ∥f′∥2≤M, then ∥f(b)−f(a)∥2≤M(b−a)).

Proof

technique · direct
1.1

Choose a ball around a on which the partial derivatives exist. Given ε>0, continuity at a gives a smaller ball on which every ∥∂jf(z)−∂jf(a)∥2<ε/m.

L1L2
2.1

For h in that smaller ball, [L1] writes the increment as coordinate segments. On each segment apply [L2] to the one-variable map obtained after subtracting the fixed linear term Jf(a); its derivative norm is at most ε/m.

step 1.1L2algebra
3.1

Summing the segment bounds gives ∥f(a+h)−f(a)−Jf(a)h∥2≤(ε/m)∑j∣hj∣≤ε∥h∥2/m≤ε∥h∥2. Since ε is arbitrary, the normalized remainder tends to zero and Jf(a) is Df(a).

step 1.1step 2.1L3∎

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