I am trying to understand one form of the higher order multivariate differential from the definition, and I would like to see if my derivation is correct. The definition goes as follows:
Suppose that $r \geq 1$ and all partial derivatives of $f$ of order $d \leq r-1$ are differentiable in a neighborhood of $x_{0} \in {D_{f}}^{0}$, where ${D_{f}}^{0}$ is the interior of $D_{f}$, then the $r$-th differential of $f$ at $x_{0}$, denoted by $\mathrm{d}_{x_{0}}^{\left(r\right)}f$ is defined by \begin{equation} \mathrm{d}_{x_{0}}^{\left(r\right)}f = \sum_{i_{1},i_{2},\dots,i_{r}=1}^{n} \frac{\partial^{r}f\left(x_{0}\right)}{\partial x_{i_{1}} \partial x_{i_{2}} \dots \partial x_{i_{r}}}\mathrm{d}x_{i_{1}}\mathrm{d}x_{i_{2}}\dots\mathrm{d}x_{i_{r}}, \end{equation} where $\mathrm{d}x_{s}$ is a functional defined by \begin{equation} \forall x \in \mathbb{R}^{n},\mathrm{d}x_{s}\left(x\right) = x_{s}. \end{equation}
It's understood that the value of \begin{equation} \frac{\partial^{r}f\left(x_{0}\right)}{\partial x_{i_{1}} \partial x_{i_{2}} \dots \partial x_{i_{r}}} \end{equation} is decided only by the number times $f$ is differentiated with respect to each variable, and not on the order of differentiation. As a result, we may consider a specific case where $f$ is differentiated $r_{1}$ times with respect to $x_{1}$, $r_{2}$ times with respect to $x_{2}$, and etc. Then in general $f$ is differentiated $r_{i}$ times with respect to $x_{i}$, with $1 \leq i \leq n$. Also, the following formula holds due to the order of differentiation: \begin{equation} \sum_{i=1}^{n}r_{i} = r. \tag{1} \label{eq: sum} \end{equation} We can construct a bijection from each \begin{equation} \frac{\partial^{r}f\left(x_{0}\right)}{\partial x_{i_{1}} \partial x_{i_{2}} \dots \partial x_{i_{r}}} \end{equation} to the $r$-tuple \begin{equation} \left(i_{1},i_{2},\dots,i_{r}\right), \end{equation} and the total number of cases where $f$ is differentiated $r_{i}$ times with respect to $x_{i}$ is equivalent to counting the number of arrangements of $\left(i_{1},i_{2},\dots,i_{r}\right)$ where there are $r_{i}$ occurrences of $i$ in the $r$-tuple. This number is a multinomial coefficient \begin{equation} \binom{r}{r_{1},r_{2},\dots,r_{n}} = \frac{r!}{r_{1}!r_{2}!\dots r_{n}!}. \end{equation} Note that we have only analyzed a specific case for $r_{1},r_{2},\dots,r_{n}$, and in general, all cases where $\eqref{eq: sum}$ holds should be included. This gives us the second expression for the $r$-th order differential: \begin{equation} \mathrm{d}_{x_{0}}^{\left(r\right)}f = \sum_{r_{1}+\dots+r_{n}=r}\frac{r!}{r_{1}!r_{2}!\dots r_{n}!}\frac{\partial^{r}f\left(x_{0}\right)}{\partial x_{1}^{r_{1}}\partial x_{2}^{r_{2}}\dots\partial x_{n}^{r_{n}}}\left(\mathrm{d}x_{1}\right)^{r_{1}}\left(\mathrm{d}x_{2}\right)^{r_{2}}\dots\left(\mathrm{d}x_{n}\right)^{r_{n}}. \end{equation}
P.S. Multivariate differential is much more abstract than the univariate case. I am wondering if there is any problem with this derivation, and how to memorize formulae in multivariate differential theory.