 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2008/paper-1/20/a/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

(f g)^{(n)} = \sum_{k = 0}^{n} (k n) f^{(k)} g^{(n - k)} .

n = 0

it is immediate. If it holds at

n

, differentiating each summand gives two sums. The coefficient of

f^{(k)} g^{(n + 1 - k)}

after regrouping is

(k n) + (k - 1 n) = (k n + 1)

by Pascal's identity, including the two endpoint terms. This proves the formula by mathematical induction.

Apply it to

(f g) h

and then expand the derivatives of

f g

. For

p + q + r = n

, the resulting coefficient is

(p + q n) (p p + q) = n! / (p! q! r!)

. Consequently

(f g h)^{(n)} = p + q + r = n \sum \frac{n !}{p ! q ! r !} f^{(p)} g^{(q)} h^{(r)},

where

p, q, r

range over nonnegative integers. This is the three-factor generalized Leibniz rule.

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