Derivative of cross product Let $f:V_1\times\dots\times V_N \to W$ be multilinear. Then $f$ s differentiable and
$$df(a_1,\dots,a_n)(h_1+\dots+h_n)=f(h_1,a_2,\dots,a_n)+f(a_1,h_2,a_3,\dots,a_n)+f(a_1,\dots,a_{n-1},h_n)\tag{1}$$
Problem: Calculate the derivative of  the cross product $\mathbb R^3\times \mathbb R^3 \to \mathbb R^3$.
Solution: Let $f$ be the cross product. It's easy to show that $f$ is multilinear by a direct calculation thus we can use (1). We calculate the derivative at $(u,v)=((u_1,u_2,u_3),(v_1,v_2,v_3))$. We then have
$$df(u,v)(h_1+h_2)=h_1\times v + u\times h_2\tag{2}$$
Let $e_i, i=1,2,3$ be the unit vectors in $\mathbb R^3$. We calculate
$$e_1\times v=\begin{pmatrix}0\\ -v_3 \\ v_2\end{pmatrix}$$
$$e_2\times v=\begin{pmatrix}v_3\\ 0 \\ -v_1\end{pmatrix}$$
$$e_3\times v=\begin{pmatrix}-v_2 \\ v_1 \\ 0\end{pmatrix}$$
For $u$ we use the identity $u\times e_i=-e_i\times u$. So we get the following derivative matrix:
$$df(u,v)=\begin{pmatrix}0&v_3&-v_2&0&-u_3&u_2 \\ -v_3 & 0 & v_1 & u_3 & 0 & -u_1 \\ v_2 & -v_1 & 0 & -u_2 & u_1 & 0\end{pmatrix}$$
Question: 
1. Apprently, the $+$ means writing the result of $f$ as a column, so it's not the $+$ from e.g. $\mathbb R^3$. Should I see that from the definition above without the example? How should I know the dimension of the imge of $df$ without them specifying it somehow? 


*What $+$ is used in $f(a_i)(h_1+h_2)$? How does $h_1$ and $h_2$ and $h_1+h_2$ look like?

 A: For a multilinear mapping, it suffices to consider its Frechet derivative. Let $W$ be an $n$-D vector space, and each $V_i$ be an $m_i$-D vector space with $i=1,2,...,N$. Let $f:V_1\times V_2\times\cdots\times V_N\to W$ be multilinear. Then $\forall\left(v_1,v_2,...,v_N\right)\in V_1\times V_2\times\cdots\times V_N$, the Frechet derivative of $f$ at this location, denoted by $({\rm d}f)(v_1,v_2,...,v_N)$, is also a multilinear mapping, i.e.,
$$
({\rm d}f)(v_1,v_2,...,v_N):V_1\times V_2\times\cdots\times V_N\to W.
$$
According to Frechet, it follows that
\begin{align}
&({\rm d}f)(v_1,v_2,...,v_N)(h_1,h_2,...,h_N)\\
&=f(h_1,a_2,...,a_N)\\
&+f(a_1,h_2,...,a_N)\\
&+\cdots\\
&+f(a_1,a_2,...,h_N).
\end{align}
Recall that, if $g$ is linear, its entry-wise form reads
$$
g_i(v)=\sum_ja_{ij}v_j,
$$
and if $g$ is bilinear, its entry-wise form reads
$$
g_i(v_1,v_2)=\sum_{j_1,j_2}a_{ij_1j_2}v_{1j_1}v_{2j_2}.
$$
Inductively and formally, the above multilinear $f$ observes the following entry-wise form
$$
f_i(v_1,v_2,...,v_N)=\sum_{j_1=1}^{m_1}\sum_{j_2=1}^{m_2}\cdots\sum_{j_N=1}^{m_N}a_{ij_1j_2...j_N}v_{1j_1}v_{2j_2}...v_{Nj_N}
$$
for $i=1,2,...,m$, where each $v_{kj_k}$ denotes the $j_k$-th entry of $v_k\in V_k$, while $a_{ij_1j_2...j_N}$'s are the coefficients of $f$.
Thanks to this entry-wise form, we may then write down the entry-wise form of ${\rm d}f$ as well, which reads
\begin{align}
&({\rm d}f)_i(v_1,v_2,...,v_N)(h_1,h_2,...,h_N)\\
&=\sum_{j_1=1}^{m_1}\sum_{j_2=1}^{m_2}\cdots\sum_{j_N=1}^{m_N}a_{ij_1j_2...j_N}h_{1j_1}v_{2j_2}...v_{Nj_N}\\
&+\sum_{j_1=1}^{m_1}\sum_{j_2=1}^{m_2}\cdots\sum_{j_N=1}^{m_N}a_{ij_1j_2...j_N}v_{1j_1}h_{2j_2}...v_{Nj_N}\\
&+\cdots\\
&+\sum_{j_1=1}^{m_1}\sum_{j_2=1}^{m_2}\cdots\sum_{j_N=1}^{m_N}a_{ij_1j_2...j_N}v_{1j_1}v_{2j_2}...h_{Nj_N}.
\end{align}
In other words, as $a_{ij_1j_2...j_N}$'s are known, the entry-wise form of ${\rm d}f$ could be expressed straightforwardly as above.
Finally, the "$+$" in OP's original post, i.e., $(h_1+h_2+\cdots+h_N)$, is a convention in some context, which is exactly $(h_1,h_2,...,h_N)$ here. When there is free of ambiguity, both expressions can be used as per ones preference.
