I've found this proposition and I don't know how to prove it. The proposition is the following:
Let's have $f:U\subset E\rightarrow R^{n}$, $E$ normed, $U$ open in $E$. Let's pick $p,q \in U, [p,q] \in U$ and $v\in R^{n}$ arbitrary. Then $\exists \space \xi \in [p,q]$ such that $\langle v,f(p)-f(q)\rangle=\langle v, Df_{\xi}(p-q)\rangle$.
Or course we need to assume that $f$ is differentiable. Define $g(t)=\langle v, f\bigl(tp+(1-t)q\bigr) \rangle$. Then $g:[0,1]\to\mathbb R$ is differentiable, so by Lagrange theorem you have $$ \langle v,f(p)-f(q)\rangle = g(1)-g(0) = g'(s) = \langle v, Df_{sp+(1-s)q}(p-q)\rangle $$ for some $s\in(0,1)$. Of course, $\xi=sp+(1-s)q\in[p,q]$.