Multivariate Taylor's formula (simplest case k=1) This was a theorem presented in my Applied (Functional) Analysis course.
Let $\Omega$ be an open subset of $\mathbb{R}^d$, $f: \Omega \to \mathbb{R}$ belongs to $C^k(\Omega)$. Let $x,y \in \Omega$ such that the segment $[x,y] \subset \Omega$. Then, we have:
$$
f(y) = f(x) + \int_0^1 \sum_{j=1}^d \partial_{x_j} f(x+t(y-x))(y_j-x_j) dt.
$$
The proof starts: Consider the function $t\in [0,1] \to \phi(t) = f(x+t(y-x))$. Then, by the chain rule
$$
\frac{d\phi}{dt} = \sum_{j=1}^d \partial_{x_j} f(x+t(y-x))(y_j-x_j).
$$
We integrate both sides with respect to $t$ and by the fundamental theorem of calculus, the proof is complete. (Note that $f(y)-f(x) = \phi(1) - \phi(0) = \int_0^1 \phi'(t) dt$)
Question 1: Surely, calculating total differentials, we should get (under chain rule)
$$
\frac{d\phi(\alpha_1, \dots, \alpha_n)}{dt} = \sum_{j=1}^d \frac{\partial\phi}{\partial\alpha_i} \frac{\partial \alpha_i}{\partial t},
$$
so shouldn't the above formula read
$$
\frac{d\phi}{dt} = \sum_{j=1}^d \frac{\partial f(x+t(y-x))}{\partial (x_j+t(y_j-x_j))} \frac{\partial (x_j+t(y_j-x_j))}{\partial t}= \sum_{j=1}^d \partial_{x_j+t(y_j-x_j)} f(x+t(y-x))(y_j-x_j)?
$$
There is no reason for $\partial_{x_j+t(y_j-x_j)} = \partial_{x_j}$, so I'm confused about the presented proof.
Question 2: In the expression
$$
f(y) = f(x) + \int_0^1 \sum_{j=1}^d \partial_{x_j} f(x+t(y-x))(y_j-x_j) dt,
$$
Is the $x_j$ in $\partial_{x_j}$ the j'th coordinate of $x \in \Omega$? I'm starting to suspect that some notations are off here.
Partial answers (answering only one of the questions) are welcome!
 A: You are mixing up the directions of differentiation with the point of evaluation. Adding the point of evaluation to the abstract chain rule, one obtains
$$
\frac{d\phi\big(\alpha_1(t), \ldots, \alpha_n(t)\big)}
{dt} 
= 
\sum_{i=1}^d 
\frac{\partial\phi}{\partial x^i} 
\big(\alpha_1(t), \ldots, \alpha_n(t)\big)
\frac{\partial \alpha_i}{\partial t}(t).
$$
The functions $\alpha_1, \ldots, \alpha_n \! : \mathbb{R} \longrightarrow \mathbb{R}$ get plugged into $\phi$ on the left hand side. On the right hand side, you first partially differentiate $\phi$ in the $i$-th direction and then plug in the $\alpha_i$. Similarly, the function $\alpha_i$ first gets differentiated into the (only) $t$ direction and then you plug in $t$. Note that this now really works if you replace the evaluation point $t$ with yet another function of $t$, yielding another chain rule. For your specific application, we have
$$
\partial_{x_i} f \big(x+t(y-x)\big)
=
\frac{\partial f}{\partial x^i}
\big(x+t(y-x)\big),
$$
so one should really write $\big( \partial_{x_i} f \big) (x+t(y-x))$ instead to avoid this kind of confusion.
