Taking derivative with respect to a vector x $\textbf{x}$ is a n-by-1 vector, $F(\textbf{x})$ is function from $R^{n} \rightarrow R^{n\times n}$.
What is the derivative of $F(\textbf{x})\textbf{x}$ with respect to $\textbf{x}$?
$\dfrac{dF(\textbf{x})\textbf{x}}{d\textbf{x}}$ = ?
 A: Have a look at the reference here: https://en.m.wikipedia.org/wiki/Jacobian_matrix
A: Note $$
\begin{array}{l|rcl}
B : & \mathbb R^{n \times n} \times \mathbb R^n &\longrightarrow & \mathbb R^n\\
& (A,u) & \longmapsto & A.u\end{array}
$$
$B$ is a bilinear map. Hence its Fréchet derivative at point $(A,u)$ is defined by $$B^\prime(A,u).(h,k)= h.u+A.k.$$ In particular, $$B^\prime(F(\textbf{x}), \textbf{x}).(h,k)= h.\textbf{x} + F(\textbf{x}).k.$$ And the function for which you're looking for the derivative is $$f(\textbf{x})=F(\textbf{x}).\textbf{x}=B(F(\textbf{x}),\textbf{x}).$$ Applying the chain rule to this function composition, you find that $$f^\prime(\textbf{x}).y=[F^\prime(\textbf{x}).y].\textbf{x}+ F(\textbf{x}).y$$ which is a linear map from $\mathbb R^n$ to $\mathbb R^n$ i.e. an element of $\mathbb R^{n \times n}$.
A: You can write it in index notation as
$$ \left[\frac{d\left[F(\mathbf{x})\mathbf{x}\right]}{d\mathbf{x}}\right]_{ik} = \frac{\partial\left(F_{ij}x_j\right)}{x_k} = \frac{\partial F_{ij}}{\partial x_k} x_j + \delta_{jk} F_{ij} = \frac{\partial F_{ij}}{\partial x_k} x_j + F_{ik}$$
where I've assumed sums over matching indices and $\delta_{jk}$ is the Kronecker delta function. 
EDIT: To make clear what the first term looks like explicitly in matrix notation, take $\left[F(\mathbf{x})\right]_{ij}$ to be a matrix element of $F(\textbf{x})$ and $x_j$ to be an element of the vector $\mathbf{x}$. There is a sum over $j$ and this becomes the product of a matrix and a vector:
$$ \frac{\partial F_{ij}}{\partial x_k} x_j \longrightarrow \begin{pmatrix} \displaystyle{\frac{\partial \left[F(\mathbf{x})\right]_{00}}{\partial x_k}} & \displaystyle{\frac{\partial \left[F(\mathbf{x})\right]_{01}}{\partial x_k}} & \cdots \\ \displaystyle{\frac{\partial \left[F(\mathbf{x})\right]_{10}}{\partial x_k}} & \displaystyle{\frac{\partial \left[F(\mathbf{x})\right]_{11}}{\partial x_k}} & \cdots \\ \vdots & \vdots & \ddots\end{pmatrix}\begin{pmatrix}x_0 \\ x_1 \\ \vdots\end{pmatrix}$$
This is an $n$-dimensional vector and $x_k$, which we are taking derivatives with still has a free index, so this produces an $n\times n$ dimension result, as expected.
