Derivative of a determinant with respect to a matrix Can someone tell me the derivative of the following determinant ($\Psi\in\mathbb{R}^{p\times p}$, $Z\in\mathbb{R}^{p\times q}$, $\alpha\in\mathbb{R}^q$)


*

*$\frac{\partial}{\partial \Psi} \log|\Psi+(Z\alpha)(Z\alpha)^T|=?$

*$\frac{\partial}{\partial \alpha} \log|\Psi+(Z\alpha)(Z\alpha)^T|=?$


$\Psi$ is a symmetric positive definite matrix. Thank you very much!
 A: It's been quite a while since this question was posted so there may not be much use with this answer. However, I stumbled upon needing to perform a very similar derivative myself recently so figured I'd post it here to help in the future.
The first question is to evaluate the following:
$$%
{{\partial }\over{\partial \Psi}} \log \left| {\bf F}\left( \Psi \right)\right| %
$$
where ${\bf F}\left( \Psi \right)$ is some matrix function of the matrix variable $\Psi$. For your example ${\bf F }\left(\Psi\right) = \Psi + (Za)(Za)^{T}$. As mentioned in the comments the chain rule is the correct way to proceed. Namely, the fact that
$$%
{{\partial }\over{\partial \Psi}} \log \left| {\bf F}\left( \Psi \right)\right| = \left| {\bf F}\left( \Psi \right)\right|^{-1}   {{\partial \left| {\bf F}\left( \Psi \right)\right|}\over{\partial \Psi}} %
$$
From this stage, the determinant needs to be calculated. A few of the books I've seen performing this derivative seem to skip the general formulation of how to form this derivative explicitly for a matrix function argument. It is
$$%
{{\partial \left| {\bf F}\left( \Psi \right)\right|}\over{\partial \Psi}} = \left| {\bf F}\left( \Psi \right)\right| \mbox{tr} \left( {\bf F}\left(\Psi\right) ^{-1}  {{\partial {\bf F}\left( \Psi \right)}\over{\partial \Psi}}\right) %
$$
Therefore executing this identity for your problem (and noting the symmetry of the $\Psi$ variable) gives
\begin{align}%
{{\partial }\over{\partial \Psi}} \log \left| {\bf F}\left( \Psi \right)\right| &= {\bf F}\left(\Psi\right) ^{-1} + {\bf F}\left(\Psi\right) ^{-1^{T}} - \mbox{diag}\left({\bf F}\left(\Psi\right) ^{-1}\right) \\
&= 2{\bf F}\left(\Psi\right) ^{-1} - \mbox{diag}\left({\bf F}\left(\Psi\right) ^{-1}\right) %
\end{align}
The second example follows from this with the exact same machinery. Noting that the derivative of a trace with respect to a scalar will still give a scalar.
