 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-1/10e/b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Write

f (a + h) g (a + h) - f (a) g (a) = [f (a + h) - f (a)] g (a + h) + f (a) [g (a + h) - g (a)] .

Divide by

h

and let

h \to 0

. Differentiability implies continuity, so

g (a + h) \to g (a)

. Therefore the product rule gives

(f g)^{'} (a) = f^{'} (a) g (a) + f (a) g^{'} (a) .

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