Exponential of a matrix and related derivative $\DeclareMathOperator{\tr}{tr}$I have $ X \in M(n,\mathbb R) $ to be fixed. I define $ g(t) = \det(e^{tX}) $
Then the author proceeds as follows:
$ g'(s) = \frac {d}{dt} g(s+t) $
= $ \frac {d}{dt} \det(e^{(s+t)X}) |_{t=0} $
= $ \frac{d}{dt}(\det(e^{sX})\det(e^{tX})|_{t=0} $
= $ g(s)\tr(X) $, as $ s $ is independent of $ t $
The author proves this in the course of proving 
$ \det(\exp(X))=\exp(\tr(X)) $
My question is why is the function $ g $ concocted for this proof? Also how is the very first step understood??I apologise if I am missing something quite simple and thanks in advance for any replies.
 A: $\DeclareMathOperator{\exp}{exp}\DeclareMathOperator{\det}{det}\DeclareMathOperator{\tr}{tr}$Here's my take on unpacking your source's argument. Observe that
$$
 \exp(\tr(tX)) = e^{\tr(X)t}, \quad t \in \mathbb{R},
$$
so that if the function $g(t) := \det(\exp(tX))$ satisfies the initial value problem
$$
 g^\prime(t) = \tr(X)g(t), \quad g(0) = 1,
$$
then by the basic theory of ODE,
$$
 \det(\exp(tX)) =: g(t) = g(0)e^{\tr(X) t} = \exp(\tr(tX)), \quad t \in \mathbb{R},
$$
yielding the desired result. So, your source's strategy is precisely to check that $g(t)$ does indeed satisfy the relevant initial value problem; since $g(0) = 1$ by a simple calculation, it suffices to check that $g^\prime(t) = \tr(X)g(t)$.
Now, fix $t \in \mathbb{R}$, and let $0 \neq h \in \mathbb{R}$. Then
$$
 g(t+h) = \det(\exp((t+h)X)) = \det(\exp(tX)\exp(hX)) = g(t)g(h),
$$
so that
$$
 \frac{g(t+h)-g(t)}{h} = \frac{g(h)-1}{h}g(t).
$$
Hence, to conclude that
$$
 g^\prime(t) := \lim_{h \to 0} \frac{g(t+h)-g(t)}{h} = \tr(X)g(t),
$$
it suffices to check that
$$
 g^\prime(0) := \lim_{h \to 0} \frac{g(h)-1}{h} = \tr(X),
$$
but this in turn can be easily checked on upper triangular matrices. Perhaps you could also use this answer by Avitus together with the chain rule, though using the Jacobi formula for the total derivative of the determinant might be tricky in its own right.
