Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24
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.

sin⁡(xy) and its mixed partial derivatives

Example

For

f:R2⟶R,f(x,y)=sin⁡(xy),

the first partial derivatives and the two mixed partial derivatives exist everywhere and satisfy

∂xf(x,y)=ycos⁡(xy),∂yf(x,y)=xcos⁡(xy),

∂y∂xf(x,y)=∂x∂yf(x,y)=cos⁡(xy)−xysin⁡(xy).

These formulas hold without excluding either coordinate axis.

Facts & Assumptions

Given: The function f(x,y)=sin⁡(xy) on R2.

[L1]

The functions sin⁡ and cos⁡ are differentiable on R, with (sin⁡t)′=cos⁡t and (cos⁡t)′=−sin⁡t (The derivatives of sine and cosine are cosine and minus sine).

[L4]

A coordinate partial derivative is the derivative at zero of the corresponding coordinate-line restriction (Directional derivatives and partial derivatives of a map U⊆Rm→Rn).

Verification

technique · direct
1.1L1L2L4

Fixing y and differentiating the map x↦sin⁡(xy) gives ∂xf(x,y)=ycos⁡(xy).

1.2L1L2L4

Fixing x and differentiating the map y↦sin⁡(xy) gives ∂yf(x,y)=xcos⁡(xy).

2.1step 1.1L1L2L3

Differentiating the formula in step 1.1 with respect to y gives ∂y∂xf(x,y)=cos⁡(xy)−xysin⁡(xy).

2.2step 1.2L1L2L3

Differentiating the formula in step 1.2 with respect to x gives ∂x∂yf(x,y)=cos⁡(xy)−xysin⁡(xy).

3.1step 1.1step 1.2step 2.1step 2.2∎

The formulas in steps 1.1 through 2.2 are defined for every (x,y)∈R2, including x=0 or y=0, and the two mixed partials agree everywhere.

Remarks

The equality is obtained by direct calculation rather than by invoking Clairaut--Schwarz theorem for continuous second partial derivatives. It is therefore an explicit instance of that theorem, not an application used to determine the common formula.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

21 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources