Solving a limit with a determinant I've found this result:
$$ \lim_{h\to 0} \frac{\det\left(A^{-1}(t)A(t + h)\right)- 1}{h} = tr \left(A^{-1}(t)\frac{dA(t)}{dt}\right)$$
It occurred to me to do a Taylor expansion of $A(t + h)$, but I get
$$\lim_{h\to 0} \frac{\det\left(A^{-1}(t)\left(A(t) + \frac{dA}{dt}h + ...\right)\right)- 1}{h} = \lim_{h\to 0} \frac{\det \left(I + A^{-1}\frac{dA}{dt}h + ...\right) - 1}{h} $$
so I don't get to obtain the trace. Any help will be welcome. Thanks.
 A: Hint:
$$ \lim_{h\to 0} \frac{\det\left[A^{-1}(t)A(t + h)\right]- 1}{h} = \lim_{h\to 0} \frac{\det A^{-1}(t)\det A(t + h)- \det A^{-1}(t)\det A(t)}{h}$$
$$=\det A^{-1}(t)\lim_{h\to 0} \frac{\det A(t + h)- \det A(t)}{h}=\det A^{-1}(t)\dfrac{d}{dt}\det A(t).$$
As Clement C. pointed out using Jacobi's formula will result in:
$$\det A^{-1}(t)\dfrac{d}{dt}\det A(t)=\det A^{-1}(t)\operatorname{tr}\left[\operatorname{adj}A(t)\dfrac{dA(t)}{dt} \right].$$
The adjugate of $A(t)$ can be rewritten as $\operatorname{adj}A(t)=\det A(t)A^{-1}(t)$ and then we can extract $\det A(t)$ from the trace as it is a scalar. This leads to:
$$\det A^{-1}(t)\operatorname{tr}\left[\det A(t)A^{-1}(t)\dfrac{dA(t)}{dt} \right]=\operatorname{tr}\left[A^{-1}(t)\dfrac{dA(t)}{dt} \right].$$
A: An intuitive way to proceed is to use the identity we are about to find. 
For a $n$-dimensional square diagonalizable matrix:
$$
\rm{det}(A)=\prod_{i=1}^n\lambda_i
$$
with $\lambda_i $ eigenvalue for $A$
and if the eigenvalues are all positive (for the mixed case just factor signs away):
$$
\prod_{i=1}^n\lambda_i=\exp\left(\sum_{i=1}^n\log(\lambda_i)\right)=\exp\left(\rm{Tr}(\log(A))\right)
$$
so now for a diag matrix we have:
$$
\rm{det}(A)=\exp\left(\rm{Tr}(\log(A))\right)
$$
from this representation you can easily recover both Jacobi's identity and solve your problem (for this refer to MrYouMath answer).
Remark: This is not rigorous enough to be a proof, but it's easy to turn it into one.

A much simpler argument is to show that:
$$
\det(I+h\epsilon)=1+h \rm{Tr}(\epsilon)+O(h^2)
$$
you can show this using the identity above.
indeed:
$$
\rm{det}(I+h\epsilon)=\exp\left(\rm{Tr}(\log(I+h\epsilon))\right)\sim1+\rm{Tr}(\log(I+h\epsilon))\sim 1+h\rm{Tr}(\epsilon)
$$
The point of this answer is to provide an intuitive understanding and nothing more.
