Proof of Lagrange's Mean Value Theorem? The Mean value theorem for Mapping says:

Let $f(x,y)$ be differentiable in $D$.  (D is open and connected).
For every $p=(x_1,y_1), q=(x_2,y_2)$ there exists a point $s \in [p,q]$ such that:
$$f(q)-f(p) = \nabla f(s)*(q-p)= f_x(s)(x_2-x_1)+f_y(s)(y_2-y_1)$$
Note: the interval $[p,q]$ is the following set of points $\{(1-t)p+tq |\ 0 \le t \le 1\}$

How may I prove that theorem? (it goes beyond my current level of studies)
Any help is appreciated :-)

Since I was almost finishig the proof I'll post the answer anyways.
I suppose $p, q\in D$. Consider the path $\gamma: [0, 1]\rightarrow \mathbb R^2$ given by
$$\gamma(t):=(1-t)p+tq.$$
Since $D$ is open and connected it is path-connected, hence the image of $\gamma$ falls into $D$. Notice that:
$$\gamma(0)=p\quad \textrm{and}\quad \gamma(1)=q.$$
Consider the composite:
$$f\circ \gamma: [0, 1]\rightarrow \mathbb R.$$
This map is continuous and differentiable. By the Mean value theorem in one variable there exists $c\in (0, 1)$ such that:
$$f(p)-f(q)=f(\gamma(1))-f(\gamma(0))=(f\circ \gamma)^\prime(c)(1-0)=(f\circ \gamma)^\prime(c).$$
But:
$$(f\circ \gamma)^\prime(c)=\nabla f(\gamma(c))\cdot \gamma^\prime(c).$$
Notice
$$\gamma^\prime(t)=-p+q=q-p$$ and therefore:
$$f(p)-f(q)=\nabla f(\gamma(c))\cdot q-p.$$
Now notice that $\gamma(c)\in [p, q]$ as required.
