Estimation in multivariate normal I need to show $\underline{\hat{\mu}}=\underline{\bar{y}}$:
$$L(y_1, y_2, ...) = \prod _{i=1}^n f(y_i,\underline{\mu},\underline{\Sigma})$$
$$ = \prod _{i=1}^n \frac{1}{\biggl(\sqrt{2 \pi}\biggl)^p |\Sigma|^{1/2}}e^{-\frac{1}{2}(y_i -\mu)'\Sigma ^{-1}(y_i -\mu)} $$
Taking log:
$$= -np \log{\biggl(\sqrt{2\pi}\biggl)}-\frac{n}{2}\log{(\Sigma)}-\frac{1}{2}\sum_{i=1}^n(y_i -\mu)'\Sigma ^{-1}(y_i -\mu)$$
Maximising:
$$ 0 = \frac{-1}{2} \times 2 \sum^{n}_{i=1} (y_i -\mu)$$
Then, I do not know how to proceed. 
I presume 
$$ n\mu =\sum^{n}_{i=1} y_i $$
 A: Recall that for a  vector $x \in \mathbb{R} ^ p$, and real symmetric matrix $A$,
$$
\frac{\partial}{\partial x} x ^ T A x = 2 A x,
$$
hence, 
$$
\frac{\partial}{\partial \mu} \ln L ( \cdot ; \mu, \Sigma )
 =
\sum \frac{1}{2}\frac{\partial}{\partial \mu}(y_i-\mu)^T\Sigma^{-1} (y_i-\mu) =  - \Sigma^{-1} \sum_{i=1}^n(y_i-\mu) = 0,
$$
multiplying by $\Sigma$ and re-arranging the equation you get 
$$
\hat{\mu} = \frac{1}{n}\sum_{i=1}^n y_i.
$$
Don't forget that $y_i$ are vectors, hence $\hat{\mu}$ is $p \times 1$ vector of sample means. 
A: You have done everything correct. You've reached 
$$0 = \frac{-1}{2} \times 2 \sum^{n}_{i=1} (y_i -\mu)$$
or more formally (since your maximum is attained at the MLE)
$$0 = \frac{-1}{2} \times 2 \sum^{n}_{i=1} (y_i - \hat{\mu})$$
where $\hat{\mu}$ is your MLE. Then 
$$0 =  \sum^{n}_{i=1} (y_i - \hat{\mu})$$
i.e.
$$0 =  \sum^{n}_{i=1} y_i -\sum^{n}_{i=1}  \hat{\mu}$$
But $\hat{\mu}$ is a constant being summed up $n$ times, so
$$0 =  \sum^{n}_{i=1} y_i - n \hat{\mu}$$
Finally,
$$\hat{\mu} = \frac{1}{n} \sum^{n}_{i=1} y_i = \bar{y}$$
is the MLE estimate, i.e. the sample mean of the data is the MLE estimate of $\mu$.
