= Solution
At a fixed $t$, write $A=f(t)$ and $B=f'(t)$. Differentiating the <matrix exponential> in the direction $B$ gives
$$
\boxed{\frac{d}{dt}e^{f(t)}=\int_0^1e^{(1-u)f(t)}f'(t)e^{uf(t)}\,du.}
$$
The order of factors matters because <matrix multiplication> need not be commutative.
To prove the <derivative of the matrix exponential> formula, expand its <power series>. The coefficient linear in $h$ in $(A+hB)^k$ is $\sum_{j=0}^{k-1}A^jBA^{k-1-j}$. On bounded sets of <matrices>, the exponential series and its directional derivative series converge uniformly: the latter is dominated in <operator norm> by $\|B\|\sum_{k\geq1}k\|A\|^{k-1}/k!$. Terms with two or more $hB$ factors have sum $O(h^2)$ locally. Hence the derivative is
$$
\sum_{k\geq1}\frac1{k!}\sum_{j=0}^{k-1}A^jBA^{k-1-j}.
$$
On the other hand, the integral in the box expands absolutely as
$$
\sum_{a,b\geq0}\frac{A^aBA^b}{a!b!}\int_0^1(1-u)^au^b\,du
=\sum_{a,b\geq0}\frac{A^aBA^b}{(a+b+1)!}.
$$
The integral identity follows by repeated <integration by parts>. Grouping by $a+b=k-1$ gives the same derivative. Finally differentiability means $f(t+h)=A+hB+o(h)$; the locally bounded derivative of the exponential sends the $o(h)$ error to $o(h)$, proving the chain rule even when $f'$ is not continuous. If $[f,f']=0$, the formula simplifies to $e^ff'$.
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