 Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2011/f/f2/a/iii/solution

The chain rule states that

f \circ g

is differentiable at

x

and

D (f \circ g) (x) = D f (g (x)) \circ D g (x) .

J_{f \circ g} (x) = J_{f} (g (x)) J_{g} (x) .

The factors have dimensions

k \times m

and

m \times n

, respectively, giving the required

k \times n

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