I have a matrix $\Sigma$ with element $(i,j)$ 


$$\Sigma\_{i,j}= \exp(-h\_{i,j}\rho).$$


The matrix is positive definite and symmetric (it is a covariance matrix).
Now I need to evaluate
$$\frac{\partial \log(\det(\Sigma))}{\partial \rho} \text{ and } \frac{\partial \Sigma^{-1}}{\partial \rho}.$$


Someone can help me? 



Derivative of log determinant and inverse
[*modified according to the clarification given in comments*].


In general, for an invertible square matrix $\Sigma=\Sigma(\rho)$, differentiably depending on the real variable $\rho$, we have: $(\Sigma^{-1})'=-\Sigma^{-1} \Sigma' \Sigma^{-1}$, and $\big(\det(\Sigma)\big)'=\operatorname{tr} (\Sigma^{-1} \Sigma')\det(\Sigma)$, so that 
$\big(\log \det(\Sigma)\big)'=\operatorname{tr} (\Sigma^{-1} \Sigma')$. 


Here, $\Sigma'$ is the matrix with entries $-h\_{ij}e^{-\rho h\_{ij}}$, the Hadamard product of the matrices $-(h\_{ij})\_{ij}$ and $\Sigma$; and since both $\Sigma^{-1}$ and $\Sigma'$ are symmetric, $\operatorname{tr} (\Sigma^{-1} \Sigma')$ is just their Frobenius scalar product. I do not see other simplifications, unfortunately.


