How to calculate the $\frac{\partial det(\mathbf X)}{\partial \mathbf X}$ and $\frac{\partial tr(\mathbf X^n)}{\partial \mathbf X}$ How to calculate the $\frac{\partial det(\mathbf X)}{\partial \mathbf X}$ and $\frac{\partial tr(\mathbf X^n)}{\partial \mathbf X}$ by using  Frobenius product?i tried to begin the calculation,but i stuck here at once
$ det(\mathbf X)=I_{mn}:X$ and $ tr(\mathbf X^n)=I_{mn}:X^n$,but i don't know how to calculate it.
 A: Let's start by finding the gradient of a small concrete example.
$$\eqalign{
 \phi &= {\rm tr}(X^3) = I:X^3 \cr
d\phi
 &= I:(dX\,X\,X+X\,dX\,X+X\,X\,dX) \cr
 &= X^TX^T:dX + X^TX^T:dX + X^TX^T:dX \cr
 &= (3X^2)^T:dX \cr
\frac{\partial\phi}{\partial X} &= (3X^{2})^T \cr
}$$
This can be immediately extended to higher powers and scalar coefficients.
$$\eqalign{
 \phi &= {\rm tr}(\alpha X^k) \cr
d\phi &= (n\alpha X^{k-1})^T:dX \cr
\frac{\partial\phi}{\partial X} &= (k\alpha X^{k-1})^T \cr
}$$
Then extended again to any function expressed as a power series.
$$\eqalign{
 F &= \sum_k\alpha_kX^k \implies
 F' &= \sum_kk\alpha_kX^{k-1} \cr
 \phi &= {\rm tr}(F) \cr
d\phi &= (F')^T:dX \cr
\frac{\partial\phi}{\partial X} &= (F')^T \cr\cr
}$$
To handle the determinant, consider the following
$$\eqalign{
\exp({\rm tr}(X)) &= 
\exp\Big(\sum_k\lambda_k\Big) = 
\prod_k\exp(\lambda_k) = 
\det(\exp(X)) \cr
{\rm tr}(X) &= \log(\det(\exp(X))) \cr
}$$
Now assume $X=\log(Y)$ in which case $Y=\exp(X)$ and
$$
{\rm tr}(\log(Y)) = \log(\det(Y))
$$
This allows us to use the above trace-trick to find the gradient of
$$\eqalign{
 \phi &= \log(\det(X)) = {\rm tr}(\log(X)) \cr
d\phi &= (X^{-1})^T:dX \cr
\frac{\partial\phi}{\partial X} &= (X^{-1})^T \cr
}$$
To find the gradient of the determinant, note that the preceding was a logarithmic derivative, so
$$\eqalign{
\frac{\partial\phi}{\partial X}
 &= \frac{\tfrac{\partial\,\det(X)}{\partial X}}{\det(X)} \cr
\frac{\partial\,\det(X)}{\partial X} &= \det(X)\,(X^{-1})^T \cr
}$$
