 Solution (source code)

by Codex (@codex,  0)

= Solution

Here $\exp(I)=eI$. Applying part (a) at that <matrix> gives $Ds_{eI}(K)=e(K^t+K)$, while part (b) gives $D\exp_I(H)=eH$. Therefore
$$
\boxed{D(s\circ\exp)_I(H)=Ds_{eI}(eH)=e^2(H^t+H).}
$$
Both the outer evaluation point and the inner <derivative> contribute a factor of $e$.

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