I was reading this post that explains why $$\frac{d}{dt} [\det A(t)]=\det A(t) \cdot \operatorname*{tr}[A^{-1}(t)\cdot \frac{d}{dt} A(t)]$$ However I don't understand exactly the following step :
$$\det A (t) \lim_{h \to 0} \frac{\det \bigl(A (t)^{-1} A (t + h)\bigr) - 1}{h}=\det A (t) \operatorname{tr} \Bigl(A (t)^{-1}\frac{d A}{dt} (t) \Bigr)$$
The post you mentioned takes for granted that the derivative of the determinant at the identity equals trace. That is, \begin{equation*} \det(\text{Id} + A) = 1 + \text{tr}(A) + o(\|A\|) \quad \text{as} \, \, \|A\| \to 0. \end{equation*}
From this, we see that \begin{align*} \det(A(t)^{-1}A(t +h)) &= \det(\text{Id} + (A(t)^{-1}A(t+h) - \text{Id})) \\ &= 1 + \text{tr}(A(t)^{-1}A(t +h) - \text{Id}) + o(\|A(t)^{-1}A(t+h) - \text{Id}\|) \end{align*} Now observe that $A(t+h) = A(t) + h \frac{d}{dt}A(t) + o(h)$ as $h \to 0$. Thus, \begin{align*} \text{tr}(A(t)^{-1}A(t+h) - \text{Id}) = h \text{tr}\left(A(t)^{-1} \frac{d}{dt}A(t)\right) + o(h) \end{align*} and \begin{equation*} o(\|A(t)^{-1}A(t+h) - \text{Id}\|) = o(h). \end{equation*} Putting it all together, we deduce that \begin{equation*} \text{det}(A(t)^{-1}A(t+h)) = 1 + h \text{tr} \left(A(t)^{-1} \frac{d}{dt}A(t) \right) + o(h), \end{equation*} or, equivalently, \begin{equation*} \lim_{h \to 0} \frac{\text{det}(A(t)^{-1}A(t+h)) - 1}{h} = \text{tr} \left(A(t)^{-1} \frac{d}{dt}A(t) \right). \end{equation*}