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The Total Derivative: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-02Open item page →

The polynomial map $(x,y)\mapsto(1+x+2y+x^2,\,2x+3y+xy)$ and its Jacobian

Example

Let $F(x,y)=(1+x+2y+x^2,\,2x+3y+xy)$ . Then $F$ is totally differentiable everywhere and

$DF(x,y)=JF(x,y)=\begin{pmatrix}1+2x&2\\2+y&3+x\end{pmatrix}.$

Facts & Assumptions

Given: The polynomial map $F:\mathbb R^2\to\mathbb R^2$ .

[L1]

The function $x\mapsto x^n$ is differentiable everywhere with derivative $nx^{n-1}$ for positive natural $n$ (For a natural $n \ge 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $\iota(n)\,x^{\,n-1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \ge 1$ the function $x \mapsto x^{-n}$ is differentiable at every $x \ne 0$ with derivative $-\iota(n)\,x^{-n-1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term).

[L2]

Continuous partial derivatives imply total differentiability with Jacobian derivative (If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative).

Verification

technique · direct

1.1

The four partial derivatives are $1+2x$ , $2$ , $2+y$ , and $3+x$ , by [L1] and derivative algebra.

L1L2

2.1

These polynomial partial derivatives are continuous everywhere, so [L2] gives total differentiability and identifies the derivative with the displayed Jacobian.

step 1.1L2algebra

3.1

Directly, the increment remainder is $(h_1^2,h_1h_2)$ , whose norm divided by $\|(h_1,h_2)\|_2$ tends to zero, agreeing with step 2.1.

step 1.1step 2.1∎

CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-02Open item page →

$xy/(x^2+y^2)$ has both partial derivatives at the origin but is discontinuous there

Statement refuted

If both partial derivatives of a real function exist at a point, then the function is continuous there.

Facts & Assumptions

Given: The function $f(0,0)=0$ and $f(x,y)=xy/(x^2+y^2)$ when $(x,y)\ne(0,0)$ .

[L1]

Partial derivatives are directional derivatives in the standard basis directions (Directional derivatives and partial derivatives of a map $U\subseteq\mathbb{R}^m\to\mathbb{R}^n$ ).

[L2]

A map is continuous at $a$ if its values tend to $f(a)$ as the input tends to $a$ (Continuity of a map between metric spaces, at a point and globally, in the $\varepsilon$ - $\delta$ form).

Counterexample

technique · direct

1.1

Both axis restrictions of $f$ are identically zero, so [L1] gives $\partial_xf(0,0)=\partial_yf(0,0)=0$ .

L1L2

2.1

On the punctured diagonal $(t,t)$ , $f(t,t)=1/2$ .

step 1.1L2algebra

3.1

As $(t,t)\to(0,0)$ but the values in step 2.1 do not tend to $f(0,0)=0$ , [L2] shows that $f$ is discontinuous at the origin.

step 1.1step 2.1∎

CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-02Open item page →

$x^2y/(x^4+y^2)$ tends to zero on every line through the origin but not along $y=x^2$

Statement refuted

If a function tends to its proposed value along every straight line through a point, then it is continuous at that point.

Facts & Assumptions

Given: $f(0,0)=0$ and $f(x,y)=x^2y/(x^4+y^2)$ away from the origin.

[L1]

A vector-valued map is continuous at $a$ when its limit at $a$ equals its value there (Vector-valued functions $f : A \to \mathbb{R}^m$ , their limits and continuity, with the dictionary to the metric notions).

Counterexample

technique · direct

1.1

On a line $(x,y)=(ta,tb)$ with $b\ne0$ , $f(ta,tb)=t a^2b/(t^2a^4+b^2)\to0$ ; for $b=0$ the restriction is identically zero.

2.1

Along the parabola $(x,y)=(t,t^2)$ with $t\ne0$ , $f(t,t^2)=1/2$ .

step 1.1algebra

3.1

The nonlinear path tends to the origin but its values do not tend to $0$ , so [L1] shows that $f$ is not continuous there despite all straight-line tests.

step 1.1step 2.1∎

CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-02Open item page →

$x^2y/(x^2+y^2)$ 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∎

CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-02Open item page →

The map $y(x^2+y^2)/x$ off the line $x=0$ , extended by zero on that line, has every directional derivative zero at the origin but is discontinuous there

Statement refuted

If every directional derivative at a point is zero, then the function is continuous there.

Facts & Assumptions

Given: $f(x,y)=y(x^2+y^2)/x$ for $x\ne0$ , and $f(0,y)=0$ .

[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]

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).

Counterexample

technique · direct

1.1

Along $(ta,tb)$ , the restriction is $t^2b(a^2+b^2)/a$ when $a\ne0$ and is $0$ when $a=0$ ; [L1] therefore gives zero directional derivative in every direction.

L1L2

2.1

Along the curve $(x,y)=(t^3,t)$ with $t\ne0$ , $f(t^3,t)=1+t^4\to1$ .

step 1.1L2algebra

3.1

Thus $f$ is discontinuous at the origin, and by [L2] it cannot be totally differentiable there either.

step 1.1step 2.1∎

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-02Open item page →

A locally constant step map on the disconnected open set $\mathbb{R}\setminus\{0\}$ has zero total derivative but is not globally Lipschitz

Statement refuted

A uniform total-derivative bound on every open domain implies a global Lipschitz bound on that domain.

Facts & Assumptions

Given: $U=\mathbb R\setminus\{0\}$ and $f:U\to\mathbb R$ defined by $f(x)=0$ for $x<0$ and $f(x)=1$ for $x>0$ .

[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 map is Lipschitz with constant $L$ when $|f(x)-f(y)|\le L|x-y|$ for every pair of points in its domain (Lipschitz map, $\alpha$ -Hölder map for rational $0 < \alpha \le 1$ , and contraction).

Counterexample

technique · direct

1.1

Each $x\in U$ has a small interval contained in its own component of $U$ , on which $f$ is constant; hence $Df(x)=0$ by [L1].

L1L2

2.1

For any $L\ge0$ , take $t=1/(2(L+1))$ . The points $-t,t\in U$ satisfy $|f(t)-f(-t)|=1>2Lt=L|t-(-t)|$ , so [L2] fails for that $L$ .

step 1.1L2algebra

3.1

The segment from $-t$ to $t$ contains $0\notin U$ , so $U$ is not convex; this is exactly the omitted hypothesis of the mean-value inequality.

step 1.1step 2.1∎

Sources

Standard references

Recommended treatments; not extraction sources.

The Total Derivative: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

The polynomial map (x,y)↦(1+x+2y+x2, 2x+3y+xy)(x,y)\mapsto(1+x+2y+x^2,\,2x+3y+xy)(x,y)↦(1+x+2y+x2,2x+3y+xy) and its Jacobian

Example

Facts & Assumptions

Verification

xy/(x2+y2)xy/(x^2+y^2)xy/(x2+y2) has both partial derivatives at the origin but is discontinuous there

Statement refuted

Facts & Assumptions

Counterexample

x2y/(x4+y2)x^2y/(x^4+y^2)x2y/(x4+y2) tends to zero on every line through the origin but not along y=x2y=x^2y=x2

Statement refuted

Facts & Assumptions

Counterexample

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

Statement refuted

Facts & Assumptions

Counterexample

The map y(x2+y2)/xy(x^2+y^2)/xy(x2+y2)/x off the line x=0x=0x=0, extended by zero on that line, has every directional derivative zero at the origin but is discontinuous there

Statement refuted

Facts & Assumptions

Counterexample

A locally constant step map on the disconnected open set R∖{0}\mathbb{R}\setminus\{0\}R∖{0} has zero total derivative but is not globally Lipschitz

Statement refuted

Facts & Assumptions

Counterexample

Sources

Standard references

The polynomial map $(x,y)\mapsto(1+x+2y+x^2,\,2x+3y+xy)$ and its Jacobian

$xy/(x^2+y^2)$ has both partial derivatives at the origin but is discontinuous there

$x^2y/(x^4+y^2)$ tends to zero on every line through the origin but not along $y=x^2$

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

The map $y(x^2+y^2)/x$ off the line $x=0$ , extended by zero on that line, has every directional derivative zero at the origin but is discontinuous there

A locally constant step map on the disconnected open set $\mathbb{R}\setminus\{0\}$ has zero total derivative but is not globally Lipschitz