How to calculate Fisher Information (FI) matrix for Multivariate Normal Distribution (MN)

Question

Below is the gradient (score) of the MN log likelihood function L for n=1 observation. I originally attempted to calculate the Hessian matrix but ran into difficulty calculating 2nd order derivatives wrt μ and Σ, to obtain FI = -E(Hessian). MLEs for μ and Σ can be derived from these equations

$dL/du = Σ^{-1}(X−u) \quad,\quad dL/dΣ = −(1/2)Σ^{-1} + (1/2)Σ^{-1} (X−μ)(X−μ)′\ Σ^{-1}$

Alternatively, I calculated the expectation of the outer product of the gradient vector for the MN FI matrix as

                     | Σ^-1    0    |
              FI = n |              |
                     | 0   1/2 Σ^-2 |

under assumptions of E(X-μ)=0, E(X-μ)${^2}$ = E[(X-μ)(X-μ)']= Σ and [E(X-μ)(X-μ)']${^2}$ = E(X-u)${^4}$ = 3Σ${^2}$.

I think... this looks like the multivariate analog to the well known univariate normal distribution information matrix below, where Q${^2}$ is scalar variance parameter, E(x-μ)=0, E(x-μ)${^2}$ = E[(x-μ)(x-μ)]= Q${^2}$ and E(x-μ)${^4}$ = 3Q${^4}$ (kurtosis)

                     | Q^-2    0    |
              FI = n |              |
                     | 0   1/2 Q^-4 |

QUESTION: Can one infer a MN kurtosis of E(X-u)${^4}$ = 3Σ${^2}$ from the well known UN kurtosis of E(x-μ)${^4}$ = 3Q${^4}$ for this MN FI matrix derivation ???

The E(X-u)${^4}$ is needed for the [E(dL/dΣ)]${^2}$ element in outer product of gradient.

Any thoughts or comments please,
Thank you