**Lemma 1** Suppose $A(t)$ and $M(t)$ are square matrices satisfying $$\f{d}{dt}A(t) = M(t)A(t).$$ Then $$\f{d}{dt}(\text{ln(det(}A(t))) = \text{Trace}(M(t)).$$ //Proof// We write $A=(a_{ij})$ and $M=(m_{ij})$, and let $A_{ij}$ be the [[http://en.wikipedia.org/wiki/Minor_(linear_algebra)|co-factor]] of $a_{ij}$. Then standard linear algebra tells that $$(3)\hspace{30pt}\sum_{j}a_{ij}A_{kj} = \delta_{ik}\text{det}(A).$$ By direct computation, we have $$\f{d (\text{det}(A)) }{dt} = \sum_{i,j}\f{d a_{ij}}{dt} A_{ij}.$$ We then calculate \begin{equation} \begin{split} \f{d}{dt}(\text{ln(det(}A(t)))&=\text{det}(A)^{-1}\sum_{i,j} \f{d a_{ij}}{dt} A_{ij}~\\ &=\text{det}(A)^{-1}\sum_{i,j,l}m_{il}a_{lj}A_{ij}~\\ &=\text{det}(A)^{-1}\sum_{i,l}m_{il} \delta_{il} \text{det}(A)~\\ &=\text{Trace}(M), \end{split} \end{equation} where we used $(3)$ in the last step. $\square$