CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02

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$x^2y/(x^2+y^2)$ has every directional derivative at the origin but is not totally differentiable there

Statement refuted

If every directional derivative of a real function exists at a point, then the function is totally differentiable there.

Facts & Assumptions

Given: $f(0,0)=0$ and $f(x,y)=x^2y/(x^2+y^2)$ away from the origin.

[L1]

The directional derivative is the derivative of $t\mapsto f(a+tv)$ at zero (Directional derivatives and partial derivatives of a map $U\subseteq\mathbb{R}^m\to\mathbb{R}^n$ ).

[L2]

A total derivative computes every directional derivative, so $v\mapsto D_vf(a)$ would be the linear map $Df(a)$ applied to $v$ (A total derivative computes every directional derivative, and its matrix is the Jacobian).

Counterexample

technique · direct

1.1

For $v=(a,b)\ne0$ , $f(ta,tb)=t\,a^2b/(a^2+b^2)$ , so [L1] gives $D_vf(0,0)=a^2b/(a^2+b^2)$ ; for $v=0$ it is $0$ .

L1L2

2.1

This direction map has value $0$ at $(1,0)$ and $(0,1)$ but value $1/2$ at their sum $(1,1)$ , so it is not additive.

step 1.1L2algebra

3.1

By [L2], total differentiability would make the directional-derivative map linear, contradicting step 2.1. Thus $f$ is not totally differentiable at the origin.

step 1.1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 66 results over 20 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

J. Lebl, Basic Analysis I, §8.3.3 (standard reference, not scraped)

x2y/(x2+y2)x^2y/(x^2+y^2)x2y/(x2+y2) has every directional derivative at the origin but is not totally differentiable there