Let $d \in \mathbb N$ and $f: \mathbb R^{d} \to \mathbb R$ a continuous partial differentiable function. Show that $\forall v \in \mathbb R^{d}-\{0\}$:
$\partial_{v}f=\langle v,\nabla f \rangle$, using the mean value theorem.
Thoughts: Until now we have only used the mean value theorem along one dimension, but I'm assuming that it goes along the lines of: $\exists \xi \in \mathbb R^{d}$, such that $\xi < v$:
$\partial_{v}f(x+h\xi)=\frac{f(x+hv)-f(x)}{h}$, (Note: I'm unsure which derivative to use, since the "first derivative" is not necessarily given in multi-dimensional spaces. It follows that:
$h\partial_{v}f(x+h\xi)=f(x+hv)-f(x)$.
Now looking at $\partial_{v}{f(x)}=\lim_{h\to0}\frac{f(x+hv)-f(x)}{h}=\lim_{h\to0}\frac{h\partial_{v}f(x+h\xi)}{h}=\lim_{h\to0}\partial_{v}f(x+h\xi)=\partial_{v}f(\lim_{h \to 0}x+h\xi)=\partial_{v}{f(x)}$.
I'm basically going in circles, so I do not see how the mean value theorem helps in this situation, as I am not closer to getting the scalar product.
Given $x,y\in \mathbb{R}^d$ define, $g:[0,1]\to \mathbb{R}$ as follows $$ g(t)=f\left((1-t)x+ty\right) $$ Using the MVT for 1D function $g(t)$, show that there exists $T\in [0,1]$ such that $$ \nabla f \left((1-T)x+Ty\right)\cdot (y-x)=f(y)-f(x) $$ Can you finish the job from here?