= Solution
Factor $A+H=A(I+A^{-1}H)$. The <determinant> expansion near the identity is
$$
\det(I+K)=1+\operatorname{tr}K+O(\|K\|^2):
$$
the linear terms in the permutation formula come from choosing one diagonal increment and all remaining identity entries. Therefore
$$
\det(A+H)=\det A+\det A\operatorname{tr}(A^{-1}H)+O(\|H\|^2).
$$
The required <Fréchet derivative> is the <linear functional>
$$
\boxed{D(\det)_A(H)=\det(A)\operatorname{tr}(A^{-1}H).}
$$
The quadratic remainder proves full differentiability, rather than only the existence of each directional <derivative>. This is the <derivative of the determinant>.
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