 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/ia/paper-1/11d/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

f (x + h) \to f (x)

. Add and subtract

f (x + h) g (x)

\frac{f ( x + h ) g ( x + h ) - f ( x ) g ( x )}{h} = f (x + h) \frac{g ( x + h ) - g ( x )}{h} + g (x) \frac{f ( x + h ) - f ( x )}{h} .

The two quotient limits exist by the given differentiability. Taking the limit and using continuity yields the product rule:

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

Thus the product is differentiable at the specified point; differentiability elsewhere is not needed.

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