Why is the the $k$-th derivative a symmetric multilinear map?

I am having trouble understanding, why the $k$-th derivative of a map $F\colon\mathbb R^n \to\mathbb R^m$ is a symmetric multilinear map for each $x$ in $\mathbb R^n$. Can you please explain which vectors this map accepts as input where multilinearity comes from ? Also, why is symmetry mentioned ?

Thank you
readingframe

Solutions Collecting From Web of "Why is the the $k$-th derivative a symmetric multilinear map?"

Well, in the general case where $F:\mathbb R^n\to V$ for some (normed) vector space $V$, the derivative $F'(x_0)$ at some point $x_0\in\mathbb R^n$ is a linear map $\mathbb R^n\to^{\mathrm{Lin}} V$ such that
$$F(x_0+h)=F(x_0) + F'(x_0)(h) + o(h)$$
when $h\in\mathbb R^n$ goes toward $0$.

Thus, the function that takes every $x_0$ to the derivative at $x_0$ is of type $\mathbb R^n\to(\mathbb R^n\to^{\mathrm{Lin}} V)$. Because the space of $\mathbb R^n\to^{\mathrm{Lin}} V$ functions is itself a vector space, we can repeat the process to get higher-order derivatives. The $k$th derivative, if it exists, ends up having type
$$\mathbb R^n \to (\underbrace{\mathbb R^n\to^{\mathrm{Lin}}(\mathbb R^n\to^{\mathrm{Lin}}(\cdots\to^{\mathrm{Lin}}(\mathbb R^n\to^{\mathrm{Lin}}}_{k\text{ times}} V)\cdots)))$$
which (if you know tensor products) is isomorphic to
$$\mathbb R^n \to (\underbrace{\mathbb R^n\otimes\mathbb R^n\otimes\cdots\otimes\mathbb R^n}_{k\text{ times}}\to^{\mathrm{Lin}} V)$$
Thus, for each $x_0$, $F^{(k)}(x_0)$ is a multilinear map taking $k$ vectors in $\mathbb R^n$ to one vector in $v$.

That the map is symmetric means that
$$F^{(k)}(x_0)(h_1)(h_2)\cdots(h_k)=F^{(k)}(x_0)(h_{\sigma(1)})(h_{\sigma(2)})\cdots(h_{\sigma(k)})$$
for any permutation $\sigma$ of the $h_i$’s. (This is true under appropriate smoothness conditions for the original $F$).

Generalizes ordinary higher derivatives. When $n=1$, this all reduces to ordinary higher derivatives because $\mathbb R\to^{\mathrm{Lin}} V$ is naturally isomorphic to $V$ (each $f:\mathbb R\to^{\mathrm{Lin}} V$ is fully determined by $f(1)$).

Connection to partial derivatives. A multilinear map is determined by its values on the standard basis vectors, and
$$F^{(k)}(\vec x_0)(\mathbf e_i)(\mathbf e_j)\cdots(\mathbf e_l) =
\frac{\partial^k F}{\partial x_i \partial x_j \cdots \partial x_l} (\vec x_0) $$