Problem with Understanding Proof of the Multivariable Mean Value Theorem I have some questions concerning a proof of the Mean Value Theorem from Bredon's book that I find rather cryptical.

Theorem:  Let $f : \mathbb{R}^n \to \mathbb{R}$ be $C^1$. Let $x := (x_1, \dots , x_n)$ and let $x' := (x'_{1}, \dots , x'_n)$. Then there exists a $x^* := (x^*_{1}, \dots , x^*_n)$ on the line segment between $x$ and $x'$ such that $$ f(x) - f(x') = \sum^{n}_{i = 1} \frac{\partial f}{\partial x_i} (x^*) (x - x').$$ 
Proof: Apply the Mean Value Theorem found in any calculus book to the function $\mathbb{R} \to \mathbb{R}$ defined by $t \mapsto f(tx + (1-t)x')$ and use the Chain Rule:
  $$ \frac{d f(tx + (1-t)x')}{dt} \Bigg|_{t = t^*} = %
\sum^{n}_{i = 1} \frac{\partial f}{\partial x_i} (x^*) \frac{d (tx_i + (1-t)x'_i)}{dt} \Bigg|_{t = t^*} = %
\sum^{n}_{i = 1} \frac{\partial f}{\partial x_i} (x^*) (x_i - x'_i )$$
  with $x^* := t^* x + (1 - t^* )x' $.


I have the following problems.


*

*Why $t \mapsto f(tx + (1-t)x')$ is defined on $\mathbb{R} \to \mathbb{R}$ ?
I think it should rather be defined on $[0,1] \to \mathbb{R}$ (i.e., it should be convex and not affine), since we want to capture that $x^* \in f(t^* x + (1 - t^* )x')$ for some $t^* \in [0,1]$.

*If $t \mapsto f(tx + (1-t)x')$ is defined on $[0,1]$ then I can almost get what happens in the equation. That is, things should work as follows: start from $t \mapsto f(tx + (1-t)x')$ and apply calculus Mean Value Theorem getting
\begin{align}
\frac{f(x) - f(x')}{?} & = \frac{df}{dt} (t^*) \\
& = \sum^{n}_i \frac{\partial f}{\partial x_i} (x^*).
\end{align}
Notice that at the denominator on the LHS I put a question mark because I don't see how it should work. We should get $(x - x')$, but I don't really see how. Or better, I see it, if the denominator works as
$$ (1x + (1-1)x' - 0x + (1-0)x' ),$$ which is indeed equal to $(x - x')$, but to me this is not clear at all. Indeed it should simply be equal to $1$, because we are acting on the domain of $t \mapsto f(tx + (1-t)x')$, that should be just $[0, 1]$, not on its codomain.
[The RHS should be OK, since we have that there is a $t^*$ and then we use it to define $x^*$.]

How should this actually work?
Any feedback would be greatly appreciated since I am self-taught and I think I never really got how these analysis proofs with differentiation actually work.
Thanks a a lot for your time.
 A: In a sense, the MVT here is NOT applied to $f(x)$, at least not directly. You have several functions here:


*

*$f(x):\mathbb{R}^n\to\mathbb{R}$, the given one;

*$h(t):\mathbb{R}\to\mathbb{R}^n$ via $h(t)=tx+(1-t)x'$ for the fixed points $x$ and $x'$, describing the line segment connecting them;

*and their composition $g(t):\mathbb{R}\to\mathbb{R}$ via $g(t)=f(h(t))=f(tx+(1-t)x')$.


It's the last one to which we can apply the MVT on the interval $[0,1]$. According to the MVT (I'm skipping verification of its conditions), there exists a point $t^{*}\in(0,1)$ such that
$$g'(t^{*})=\frac{g(1)-g(0)}{1-0}=g(1)-g(0).$$
The right-hand side is
$$g(1)-g(0)=f(h(1))-f(h(0))=f(x)-f(x'),$$
i.e. the desired difference at the endpoints. For the left-hand side we apply the Multivariate Chain Rule to $g=f(h(t))$, where $h(t)=(h_i(t))_{i=1}^n=(tx_i+(1-t)x'_i)_{i=1}^n$:
$$\frac{dg}{dt}=\sum_{i=1}^n \frac{\partial f}{\partial h_i}\cdot\frac{dh_i}{dt}=\sum_{i=1}^n \frac{\partial f}{\partial x_i}\cdot(x_i-x'_i).$$
So if we denote $x^{*}=h(t^{*})=(t^{*}x_i+(1-t^{*})x'_i)_{i=1}^n$ to be the point on this line segment with the "coordinate" $t^{*}$, we have the desired result that
$$\sum_{i=1}^n \frac{\partial f}{\partial x_i}(x^{*})\cdot(x_i-x'_i)=f(x)-f(x').$$
A: I think the sequence for the proof must be,
$$\frac{d f(tx + (1-t)x')}{dt} \Bigg|_{t = t^*} = %
\sum^{n}_{i = 1} \frac{\partial f}{\partial x_i} (x^*) \frac{d (tx_i + (1-t)x'_i)}{dt} \Bigg|_{t = t^*} = %
\sum^{n}_{i = 1} \frac{\partial f}{\partial x_i} (x^*) (x_i - x'_i )
$$
