# 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?.

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 $${\rm d}Y=\text{tr}\left(A^{\top}{\rm d}X\right)A.$$
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} \\ }