 Solution

ID: past-exam-of-the-options-i-course-of-the-university-of-oxford/2004/f/f1/c/solution

Here

exp (I) = e I

. Applying part (a) at that matrix gives

D s_{e I} (K) = e (K^{t} + K)

, while part (b) gives

D exp_{I} (H) = eH

. Therefore

D (s \circ exp)_{I} (H) = D s_{e I} (eH) = e^{2} (H^{t} + H) .

Both the outer evaluation point and the inner derivative contribute a factor of

e

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