Derivative of $(Ax) \otimes y$ with respect to $x$ Suppose $A$ is an $ n \times n$ matrix and suppose that
$x,\, k$ are  $n \times 1$ vectors. Also suppose that
$k$ is a constant vector.
let
$$
y : = \left( Ax \right) \otimes k
$$
Note that by $\otimes$ in this context we mean the outer product of
two vectors given as $v\otimes u = vu^\top$.
I would like to find $\frac{\partial y}{\partial x}$. By this symbol
I mean to find the derivative of each entry of $y$ with respect
to each entry of $x$.
I know that,
$$
\mathrm{d}(x \otimes y) = (\mathrm{d}x)\otimes y +
x \otimes (\mathrm{d}y)
$$
Therefore (at least formally) we would have
\begin{align}
\frac{\partial y}{\partial x} = A \otimes k \label{A}
\end{align}
But then what I don't understand is (if the way I have found the
derivative is correct) what is meant by $A \otimes k$?. Does it mean
the Kronecker product of $A$ and $k$ in the usual sense?
Can someone please clarify?. Better yet, how does one find this
derivative?
EDIT
$Ax k^\top$ is an $n \times n$ matrix.
 A: Let $f: \mathbb{R}^n \rightarrow \mathbb{R}, \ f(x) = (Ax)k^T$. Note that $f$ is a linear map. Thus, $df=f$. Meaning the derivative of $f$ in the point $a\in \mathbb{R}^n$ is the linear map $df(a)[h]=f(h)$ for $h\in \mathbb{R}^n$.
A: Actually, $y$ is a matrix not a vector, whose components are
$$\eqalign{
 Y_{ij} &= A_{im}x_{m}k_{j} \cr
}$$
Its gradient with respect to $x$ is a 3rd order tensor
$$\eqalign{
G_{ijk} &= \frac{\partial Y_{ij}}{\partial x_k} = A_{im}\delta_{mk}k_{j} = A_{ik}k_{j} \cr\cr
}$$
Introducing the 4th order isotropic tensor
$$\eqalign{
{\mathcal B}_{ijkl} &= \delta_{il}\delta_{jk} \cr
}$$
allows this gradient to be written without subscripts
$$\eqalign{
G &= A\star k:{\mathcal B} \cr
dY &= G\cdot dx\cr
}$$
in which the symbol $(\star)$ denotes a tensor product, $(\cdot)$ denotes a single-dot product, and $(:)$ denotes a double-dot product.
A: Vectorization provides a straightforward solution method.
$$\eqalign{
{\rm vec}(Y) &= {\rm vec}(Axk^T) \cr
y &= (k\otimes A)\,x \cr
dy &= (k\otimes A)\,dx \cr
\frac{\partial y}{\partial x} &= (k \otimes A) \cr
}$$
In index notation this is just greg's answer: 
$\,\frac{\partial Y_{ij}}{\partial x_m} = A_{im}k_{j}$
