Matrix trace differentiation correctness 
I need some help in verifying if my derivation in matrix differentiation is correct.
$\alpha_{n\times n \times m}$  is a tensor, $F(w)_{n \times m} , Z_{n \times d}$ matrix independent of $\alpha$ , $\beta_{n \times m }$ matrix where each entry $\beta_{ij} = \sum_{k=1}^{n} \alpha_{ikj}$
$$ 
 \begin{equation}\label{eqn:alphaGradFw11Vec}
  \begin{split}
   \frac{\partial}{\partial \alpha}Tr\left[ F(w)\left(\beta Z^{\top} Z\right)^{\top} \right] &= \left(\frac{\partial}{\partial \alpha} \left(F(w)\left(\beta Z^{\top}Z\right)^{\top}\right)\right)^{\top} \\
   &= \left(F(w) Z^{\top} Z \left(\frac{\partial}{\partial \alpha}\beta^{\top}\right) \right)^{\top} \\
   &= \left(F(w) Z^{\top} Z \left(\frac{\partial}{\partial \alpha}\beta^{\top}\right) \right)^{\top} \\
   &= \left(\frac{\partial}{\partial \alpha}\beta^{\top}\right)^{\top} \left(F(w) Z^{\top} Z \right)^{\top}   \\
   &= \left(\frac{\partial}{\partial \alpha}\beta^{\top}\right)^{\top} \left(F(w) Z^{\top} Z \right)^{\top}   \\
   &= \left(\frac{\partial}{\partial \alpha}\beta\right) Z^{\top} Z\, F(w)^{\top} \\
  \end{split}
 \end{equation}
$$
For above I have used the result: $$\frac{\partial \left(Tr(g(\mathbf{X}))\right)}{\partial \mathbf{X}} = \left(g^{'}(\mathbf{X})\right)^{\top}$$
Please let me know if this derivation is correct and also the dimensionality of the result.
Thanks in advance
 A: $\def\o{{\large\tt1}}\def\p#1#2{\frac{\partial #1}{\partial #2}}$For
typing convenience, define $\o\in{\mathbb R}^{n}$ as the all-ones vector and the matrices
$$\eqalign{
M &= FZ^TZ \quad&\implies\quad M_{ij} = F_{ip}Z_{qp}Z_{qj} \\
\beta& \quad&\implies\quad \beta_{ij} = \o_k\,\alpha_{ikj} \\
}$$
NB: $\,$ For the equations to be dimensionally compatible, $Z$ must be a $(d\times m)$ matrix not $(n\times d)$ as stated in the question.
Write the trace function using index notation
$$\eqalign{
T &= {\rm Tr}(FZ^TZ\beta^T) \\
  &= M_{ij}\,\beta_{ij} \\
  &= M_{ij}\,\o_k\,\alpha_{ikj} \\
}$$
Then calculate its differential/gradient.
$$\eqalign{
dT &= M_{ij}\o_k\,d\alpha_{ikj} \\
\p{T}{\alpha_{ikj}}
 &= M_{ij}\o_k \;=\; F_{ip}Z_{qp}Z_{qj}\o_k \\
}$$
The gradient is obviously a third-order tensor, and therefore cannot be expressed using standard matrix notation, but it's rather straightforward using index notation.
By defining a tensor with components
$\,\gamma_{ikj}=M_{ij}\o_k,\,$ you can write an index-free equation
$$\p{T}{\alpha} = \gamma\\$$
It may be of interest to calculate the gradient of
$\beta$ with respect to $\alpha$
$$\eqalign{
\beta_{ij} &= \o_k\,\alpha_{ikj} \\
  &= \o_k\delta_{ip}\delta_{jq}\,\alpha_{pkq} \\
d\beta_{ij} &= \o_k\delta_{ip}\delta_{jq}\;d\alpha_{pkq} \\
\p{\beta_{ij}}{\alpha_{pkq}} &= \o_k\delta_{ip}\delta_{jq} 
\;\doteq\; \Gamma_{ijpkq} \\
}$$
where $\delta_{ip}$ is a Kronecker ${\rm symbol}.\;$ Note that this gradient $\left(\Gamma=\p{\beta}{\alpha}\right)\,$ is a $5^{th}$ order tensor!
But you can actually use $\Gamma$ to calculate the original derivative using double and triple dot products instead of index notation
$$\eqalign{
 T &= M:\beta \\
dT &= M:d\beta \\
   &= M:\big(\Gamma\therefore d\alpha\big) \\
\p{T}{\alpha} &= M:\Gamma \\
}$$
