Derivative of matrix logarithm Let $A(x)$ be a differentiable matrix-valued function with $\det A(x)\ne 0\,\forall x$. I understand that
$$\frac{d}{dx}\log A(x)$$
does not have a simple expression in terms of $A$ and $dA/dx$ unless these two things commute, in which case the expression is
$$A(x)^{-1}\frac{dA}{dx}.$$
Let's say I don't want to assume that $A$ and $dA/dx$ commute but I only care about the trace. Is it true for any differentiable non-singular function $A(x)$ that
$$\text{tr}\left(A(x)^{-1}\frac{dA}{dx}\right)=\frac{d}{dx}\text{tr}\,\log A(x)=\frac{d}{dx}\log\det A(x)$$
 A: Yes, it's true.
Let $F(t)$ and $f(t)=\tfrac{dF}{dt}\,\,$ define a function and its first derivative wrt a scalar argument. 
Now apply the function to a matrix argument and take the trace 
$$\eqalign{
 \phi &= {\rm \,tr}(F(A)) \cr
}$$
The differential of this function is given by
$$\eqalign{
 d\phi &= d{\rm \,tr}(F(A)) = f(A^T):dA\cr
}$$
where colon represents the trace/Frobenius product, i.e. 
$\,\,A:B={\rm tr}(A^TB).$
The specific case $F(t)=\log(t)$ yields
$$\eqalign{
d{\rm \,tr}(\log(A))
 &= (A^T)^{-1}:dA \cr
 &= (A^T)^{-1}:\tfrac{dA}{dx}\,dx \cr
\cr
\frac{d{\rm \,tr}(\log(A))}{dx}
 &= (A^T)^{-1}:\tfrac{dA}{dx} \cr
 &= {\rm tr}(A^{-1}\tfrac{dA}{dx}) \cr\cr
}$$
The formula 
$$\log(\det(e^L)) = {\rm tr}(L)$$
is due to Jacobi.
If $\{\lambda_k\}$ are the eigenvalues of $L$, then the eigenvalues of $e^L$ are $\{e^{\lambda_k}\}$ and Jacobi's formula simply states that 
$$\eqalign{
 \log\Big(\prod_k \exp(\lambda_k)\Big)
 =  \log\Big(\exp\big(\sum_k\lambda_k\big)\Big)
 = \sum_k \lambda_k  
}$$
Setting $L=\log(A)\,\,$ recovers the final equality in your question.
