questions with matrix derivatives, dererminat and trace Recently i have some derivatives with matrix to work with. For Example:
$$\det \pmb{XA}$$
And i need to find the derivative  with respect to $\mathbf{X}$.
From Matrix cookbook i have 
$$\frac{\partial\ \det (\mathbf{Y})} {\partial x}= \det (\mathbf{Y}) Tr \left[ \mathbf{Y}^{-1} \frac{\partial\ \mathbf{Y}} {\partial x}\right]$$
So if:  $$\mathbf{Y} = \mathbf{XA}$$
$$\frac{\partial\ \mathbf{Y}} {\partial x}=\frac{\partial\ \mathbf{X}} {\partial x} \mathbf{A}$$
Then $$\frac{\partial\ \det (\mathbf{XA})} {\partial x}= \det (\mathbf{XA}) Tr \left[ (\mathbf{XA})^{-1} \frac{\partial\ \mathbf{X}} {\partial x} \mathbf{A} \right]$$
From cyclic property of trace: $$\frac{\partial\ \det (\mathbf{XA})} {\partial x}= \det (\mathbf{XA}) Tr \left[ \mathbf{A} (\mathbf{XA})^{-1} \frac{\partial\ \mathbf{X}} {\partial x} \right]$$
How can i proced with the $\frac{\partial\ \mathbf{X}} {\partial x} ?$
If i want relative to $\mathbf{X}$ can i do $\frac{\partial\ \mathbf{X}} {\partial \mathbf{X}}= \mathbf{I}$ ? So that :
$$\frac{\partial\ \det (\mathbf{XA})} {\partial \mathbf{X}}= \det (\mathbf{XA}) Tr \left[ \mathbf{A} (\mathbf{XA})^{-1}  \right] ?$$
I ask this because i see some other questions here and them just trow the Trace operator away like:
$$\frac{\partial\ \det (\mathbf{XA})} {\partial \mathbf{X}}= \det (\mathbf{XA}) \mathbf{A} (\mathbf{XA})^{-1}   $$
How is that possible?
Any help is apreciated also books recomendations of the subject!
 A: Let $M = XA$ and $f = \det(M)$ (assuming $M$ is invertible).
Utilizing Jacobi's formula, the differential of $f$ is
\begin{align}
df = d \det(M) = {\rm tr} \left( {\rm adj} \left(M \right) dM\right) = \det \left( M \right) {\rm tr} \left( M^{-1} dM\right)  .
\end{align}
And, $dM = dX A$. 
Moreover, we will utilize the following the identities as well


*

*Trace and Frobenius product relation $$A:B={\rm tr}(A^TB) := \langle A, B \rangle$$ or $$A^T:B={\rm tr}(AB)$$

*Cyclic property of Trace/Frobenius product $$\eqalign{
A:BC
 &= AC^T:B \cr
 &= B^TA:C \cr
 &= {\text etc.} \cr
}$$


We compute the differential of $f$ first and then the gradient.
So, 
\begin{align}
df &= d \det \left( M \right) \\  
   &= \det(M) {\rm tr} \left( M^{-1} dM \right)  \\
   &= \det(M) \left[ M^{-T} \ : \ dM \right]\hspace{8mm} \text{note: utilized trace and Frobenius relation} \\
   &= \det(M) \left[ M^{-T} \ : \ dX A \right] \\
  &= \det(M) \left[ M^{-T} A^T \ : \ dX \right]   \hspace{8mm} \text{note: utilized cyclic property of Frobenius product} \\
\end{align}
Hence, the gradient is
\begin{align}
\frac{\partial}{\partial X} f = \frac{\partial}{\partial X} \det \left(  XA \right) 
&=  \det(M) M^{-T} A^T  = \det\left( X A\right) \left( X A \right)^{-T} A^T.\\
\end{align}
Also, following the above procedure, one could show that
\begin{align}
\frac{\partial}{\partial X} f &= \frac{\partial}{\partial X} \det \left(  XA \right) \\
&= \det\left( X A\right) \left( X A \right)^{-T} A^T \\
&= {\rm adj}\left( XA \right)^T A^T .
\end{align}
