# log multivariate normal differentiation (MLE)

I've come across a lot of explanations of how to differentiate the multivariate normal, but they all appear to skip the step that I'm stuck on. Here's what I've got so far.

By logging and removing terms unrelated to $\mu$, I got to this stage which I know to be correct: $$- \sum_{n = 1} \frac{1}{2}(x_i - \mu)^T \Sigma^{-1}(x_i - \mu)$$ I multiplied it out to get: $$= - \sum_{n = 1} \frac{1}{2}(x_i^T\Sigma^{-1}x_i - \mu^T\Sigma^{-1}x_i - x_i^T\Sigma^{-1}\mu + \mu^T\Sigma^{-1}\mu)$$ Then I tried to differentiate with respect to $\mu$ but ran into problems: I used the identity $\frac{\delta}{\delta z}z^TAz = (A + A^T)z$. $$= - \frac{1}{2}\sum_{n = 1} -x_i^T{\Sigma^{-1}}^T - x_i^T\Sigma^{-1} + (\Sigma^{-1} + {\Sigma^{-1}}^T)\mu$$ $$= - \frac{1}{2}\sum_{n = 1} (-{\Sigma^{-1}} - {\Sigma^{-1}}^T)x_i + (\Sigma^{-1} + {\Sigma^{-1}}^T)\mu$$

I think I shouldn't be getting the ${\Sigma^{-1}}^T$s, but I'm not sure how to change what I'm doing to get the correct answer - any help would be appreciated!

(1) $A^{-1^T}= A^{T^{-1}}$
(2) The covariance matrix is symmetric: $\Sigma^T = \Sigma$