In my textbook I have an example of a function that doesn't have a total differential, but it has partial derivatives :
Let f be a function from $\mathbb R^2$ to $\mathbb R$
$$f(x,y)=\begin{cases} 1 &\text{ if $x=0$ or $y=0$} \\ 0&\text{otherwise } \end{cases}$$
Partial derivatives at $(0,0)$ are:
$$
\frac{\partial f}{\partial x}(0,0) =\, 0 = \frac{\partial f}{\partial y}(0,0)
$$
on the other hand, $f$ isn't continuous at $(0,0)$, so its total differential doesn't exist.
And below this there is a theorem saying if a function has continuous partial derivatives at a point $a$, then f has a total differential at $a$.
So partial derivatives of $f$ (example above) aren't continuous? It seems to me that partial derivatives are continuous because it is always zero. Where is the problem? Thank you.
Best Answer
The partial derivatives don't even exist in any neighborhood of $(0,0)$. For example the second partial derivative at $(1,0)$ is $\lim_{h \to 0} \frac {0-1} h$ which does not exist.
You can conclude that the function is differentiable at $(0,0)$ if you know that both partial derivatives exist is some open disk around the origin and that they are continuous functions.