 Solution

ID: past-exam-of-the-probability-statistics-and-options-ii-course-of-the-university-of-oxford/2011/f/f2/b/iii/solution

D g (t) h = h (cos t - sin t) .

Its image lies on the unit circle, so

f (g (t)) = 1/2

and

D (f \circ g) (t) = 0

. Meanwhile

J_{f} (g (t)) = (- sin t /2 - cos t /2),

J_{g} (t)

gives

- sin t cos t /2 + cos t sin t /2 = 0

. This verifies the chain rule explicitly.

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