I have to give an example of a function $f \colon \mathbb{R}^2 \to \mathbb{R}$ that has partial derivatives $(\partial_1f)(0,0) = 1 = (\partial_2f)(0,0)$ but no total derivative in $(0,0)$.
I think that the function
$$
f \colon \mathbb{R}^2 \to \mathbb{R},
\quad
f(x,y) =
\begin{cases}
\frac{xy}{x^2 + y^2} + x + y & \text{if $(x,y) \neq (0,0)$} \\
0 & \text{if $(x,y) = (0,0)$}
\end{cases}
$$
is an example, I'm just not sure if my proof is correct.
We know that $f$ is total differentiable if $(x,y) \neq (0,0)$ because the partial derivatives exist and are continuous.
$f$ is also partial differentiable if $(x,y) = (0,0)$ because $f(x,0) = x$ and $f(0,y) = y$ and they are given by $(\partial_1 f)(x,0) = 1 = (\partial_2f)(0,y) = 1$.
To proof that $f$ is total differentiable in $(0,0)$ we now only have to proof that
$$
f(x,y)
= (1,1)
\begin{pmatrix}
x \\
y \\
\end{pmatrix}
+ o(\| (x,y) \|)
\qquad
\text{when $(x,y) \to (0,0)$}
$$
but this doesn't hold when $x=y=t$ and $t \to 0$ and so $f$ is not total differentiable when $(x,y) = (0,0)$.
Can someone confirm if this is correct?
Best Answer
Yes, you have it right.
Your function has partial derivatives at the origin without being continuous.