 Solution

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

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

Differentiability lets us write

f (a + h) = f (a) + f^{'} (a) h + r (h), \frac{r ( h )}{h} \to 0,

and, with

k \to 0

g (f (a) + k) = g (f (a)) + g^{'} (f (a)) k + s (k), \frac{s ( k )}{k} \to 0.

Substitute

k = f (a + h) - f (a) = f^{'} (a) h + r (h)

. Since

k = O (h)

s (k)

o (h)

. Thus

g (f (a + h)) - g (f (a)) = g^{'} (f (a)) f^{'} (a) h + o (h),

and the chain rule is

(g \circ f)^{'} (a) = g^{'} (f (a)) f^{'} (a) .

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