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No assumption on the square matrix $H=(h_{ij})$ are needed. If $\Sigma:=\exp(-\rho H)$, then $\Sigma^{-1}=\exp(\rho H)$ with derivative wrto $\rho$ equal to $(\Sigma^{-1})'=H \exp(\rho H)$. By the Liouville formula for the solution of linear system, $\big(\det(\Sigma)\big)'=-(\operatorname{tr}H)\det(\Sigma) $, and $\big(\log \det(\Sigma)\big)'=-(\operatorname{tr}H).$