Past exam of the options I course of the University of Oxford / 2004 / F / F1 / c / Solution

Here $\exp (I)=eI$. Applying part (a) at that matrix gives $D{s}_{eI}(K)=e(K^t+K)$, while part (b) gives $D{\exp }_I(H)=eH$. ThereforeBoth the outer evaluation point and the inner derivative contribute a factor of $e$.