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Let $F:U\rightarrow\mathbb{R}^m$ be a function for some open $U\subseteq\mathbb{R}^n$. If all partial derivatives of $F$ exist and are continuous in a neighborhood of some point $x_0\in U$, then $F$ is differentiable at $x_0$.

Partial derivatives are the directional derivatives in the direction of the standard basis vectors of $\mathbb{R}^n$. I wondered if there is something special about partial derivatives as they seem to depend on the choice of this basis.
Is the statement still true if we choose a different basis? Could the above statement be generalized in the following way?
Let $F:U\rightarrow W$ be a funtion where $U\subseteq V$ open and $V,W$ finite dimensional (real) Banach spaces. Also, let $\{v_1,\dots,v_n\}$ be a basis for $V$. If the directional derivatives $\partial_{v_i}F$ exist for $i=1,\dots,n$ and are continuous in a neighbordhood of $x_0\in U$, then $F$ is differentiable in $x_0$.

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  • $\begingroup$ What kind of vector spaces are $V$ and $W$? You mention open subsets and directional derivatives of maps $V \to W$, but in general we don't have enough structure to define such notions. $\endgroup$ – Ulrik Apr 3 '15 at 23:52
  • $\begingroup$ I added this to my post. I meant the notions of derivatives, open subsets etc. with respect to any norm on these (real) vector spaces. $\endgroup$ – Pete Apr 4 '15 at 0:09
  • $\begingroup$ One may talk about derivatives in any Banach Space, using the natural generalization: en.wikipedia.org/wiki/Fr%C3%A9chet_derivative $\endgroup$ – Shalop Apr 4 '15 at 0:40
  • $\begingroup$ Have you tried mimicking the proof of the Euclidean case? $\endgroup$ – Ulrik Apr 4 '15 at 0:53
  • $\begingroup$ Yes, thats the derivative I meant. But I was wondering if there is a generalization of the statement at the top of my post since there are no partial derivatives in a Banach space. $\endgroup$ – Pete Apr 4 '15 at 0:55

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