I'm trying to show that $\log p(x) = -\frac{1}{2}(x-\mu)^{T} \Sigma^{-1}(x-\mu)$ is concave. How would I go about this in $\mathbb{R}^n$? I've tried taking derivatives but I'm getting stuck once I get to $$\nabla\log p(x) = - \det[ (x-\mu)^{T}\Sigma^{-1}(x-\mu) ] * \Sigma^{-1}(x-\mu) * [(x-\mu)^{T}\Sigma^{-1}(x-\mu)]^{-1}.$$ I still would need to take another gradient with respect to $x\in \mathbb{R}^n$ so this takes me nowhere. Can someone help?
Btw, $\mu$ is the mean and $\Sigma$ is the covariance matrix.