Derivative of $\mathrm{tr}(X^\top A) A$ Let A be a constant matrix. Define
$$
Y:= \mathrm{tr}(X^\top A) A
$$
I want to find
\begin{align}
\frac{\partial Y}{\partial X} \label{A}
\end{align}
I know that 
$$
\frac{\partial\,\mathrm{tr}(X^\top A)}{\partial X} = A \label{B}
$$
Of course, I should be able to extend this to the case I want. But I am a bit confused as to how I should notate the result. What I did to find \eqref{A} was to vectorize both sides and then use the fact in \eqref{B}. 
$$
\mathrm{vec(Y)} = \mathrm{tr}(X^\top A)\mathrm{vec}(A)
$$
I also knwo that I can write 
$$
\mathrm{tr}(X^\top A) = \mathrm{vec}(A)^\top \mathrm{vec}(X)
$$ 
So the desired the derivative should be
$$
\mathrm{vec}(A) \otimes \mathrm{vec}(A)
$$
Is this correct?. 
 A: Your result seems correct. You may verify it by using the entry-wise form.
Let the entry-wise form of the matrices $A$, $X$ and $Y$ be
\begin{align}
A&=\left(a_{jk}\right)_{jk},\\
X&=\left(x_{jk}\right)_{jk},\\
Y&=\left(y_{jk}\right)_{jk}.
\end{align}
Then $Y=\text{tr}\left(X^{\top}A\right)A$ reads
$$
y_{jk}=\left(\sum_{i,r}x_{ir}a_{ir}\right)a_{jk}.
$$
With this notation,
\begin{align}
\frac{\partial y_{jk}}{\partial x_{pq}}&=\frac{\partial}{\partial x_{pq}}\left[\left(\sum_{i,r}x_{ir}a_{ir}\right)a_{jk}\right]\\
&=\sum_{i,r}\frac{\partial x_{ir}}{\partial x_{pq}}a_{ir}a_{jk}\\
&=\sum_{i,r}\delta_{ip}\delta_{rq}a_{ir}a_{jk}\\
&=a_{pq}a_{jk}.
\end{align}
Thus if one regards $A$ as a tensor, one may write
$$
\frac{\partial Y}{\partial X}=A\otimes A.
$$
Alternatively, one may use the differential form
$$
{\rm d}y_{jk}=\left(\sum_{p,q}a_{pq}{\rm d}x_{pq}\right)a_{jk},
$$
which, in its matrix form, reads
$$
{\rm d}Y=\text{tr}\left(A^{\top}{\rm d}X\right)A.
$$
A: The problem can be approached in either tensor or vector form
$$\eqalign{
Y &= (A\star A):X
&\implies\,&\quad 
\frac{\partial Y}{\partial X} &= (A\star A)
\\
y &= (aa^T)\,x
&\implies\,&\quad
\frac{\partial y}{\partial x} &= (aa^T)
\\
}$$
where 
$$a={\rm vec}(A),\quad x={\rm vec}(X),\quad y={\rm vec}(Y)$$
and $(:\!|\star)$ are the trace and dyadic products, i.e.
$$\eqalign{
&\lambda = B:C &\implies \lambda = \sum_{i}\sum_{j}B_{ij}C_{ij} \\
&{\mathcal L} = B\star C \quad&\implies {\mathcal L}_{ijpq}=B_{ij}C_{pq} \\
}$$
