= Solution
Give $G(H)$ its natural <trace-class> topology: differentiability means $F(t+h)=F(t)+h\dot F(t)+o_1(h)$, with remainder small in <trace norm>. The <Fredholm determinant> is the continuous extension of finite-rank <determinants>; it is multiplicative by finite-rank approximation and satisfies
$$
\det(I+K)=1+\operatorname{Tr}K+O(\|K\|_1^2)
$$
near zero. For example its exterior-power <series> bounds the remainder by $e^{\|K\|_1}-1-\|K\|_1$, proving that estimate directly.
Factor $F(t+h)=F(t)[I+hF(t)^{-1}\dot F(t)+o_1(h)]$. Multiplicativity and the preceding expansion give
$$
\boxed{\frac d{dt}\det F(t)=\det F(t)\operatorname{Tr}(F(t)^{-1}\dot F(t)).}
$$
The <determinant> is nonzero for invertible $F$, so division yields the required logarithmic <derivative>. Differentiability only in the bounded-operator <norm> would not justify this <trace> formula; the specified topology is essential.
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