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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 $$\text{tr}\left(A(x)^{-1}\frac{dA}{dx}\right)=\frac{d}{dx}\text{tr}\,\log A(x)=\frac{d}{dx}\log\det A(x)$$

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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.

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  • $\begingroup$ Thanks! Why is it true that $d\,\text{tr}(F(A))=f(A^T):dA$? $\endgroup$ – aaa Jun 30 '18 at 22:51
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    $\begingroup$ Work it out for a simple polynomial like $F(A)=A^3$, i.e. $$\eqalign{d{\rm\,tr}(AAA)&={\rm\,tr}(dA\,AA) + {\rm\,tr}(A\,dA\,A) + {\rm\,tr}(AA\,dA)\cr&={\rm\,tr}(3AA\,dA)={\rm\,tr}(3A^2\,dA)=3(A^T)^2:dA}$$Couple this with the fact that (due to Cayley-Hamilton) any analytic function of $A\in{\mathbb C}^{n\times n}$ can be written as an $n^{th}$ degree polynomial, and you'll reach the desired conclusion. $\endgroup$ – greg Jul 1 '18 at 2:22

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