TITLE: Derivative of multivariate normal distribution wrt mean and covariance
QUESTION [11 upvotes]: I want to differentiate this wrt $\mu$ and $\Sigma$ :
$${1\over \sqrt{(2\pi)^k |\Sigma |}} e^{-0.5 (x-\mu)^T \Sigma^{-1} (x-\mu)} $$  
I'm following the matrix cookbook here and also this answer . The solution given in the answer (2nd link), doesn't match with what I read in the cookbook.
For example, for this term, if I follow rule 81 from the linked cookbook, I get a different answer (differentiating wrt $\mu$) :
$(x-\mu)^T \Sigma^{-1} (x-\mu)$  
According to the cookbook, the answer should be : $-(\Sigma^{-1} + \Sigma^{-T}) (x-\mu)$ . Or, am I missing something here? Also, how do I differentiate $(x-\mu)^T \Sigma^{-1} (x-\mu)$
 with respect to $\Sigma$ ?

REPLY [5 votes]: I also had the same question as you. After trying equation 81 from the Matrix cookbook, I got this equation:
$$
\frac{\partial{f}}{\partial{\mu}} = -\frac{1}{2}(\Sigma ^{-1} + (\Sigma^{-1})^{T}) (x - \mu)*(-1)  
$$
Since $ \Sigma $ is the co-variance matrix, it is symmetrical. Inverse of a symmetrical matrix is also symmetric (Is the inverse of a symmetric matrix also symmetric?). Therefore, we have $ (\Sigma^{-1})^{T}  = \Sigma ^{-1} $.
Now, the above equation reduces to 
$$
\frac{\partial{f}}{\partial{\mu}} = \Sigma ^{-1}(x - \mu)  $$