 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/ia/paper-4/2c/solution

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

The base case

n = 1

is the product rule. Suppose the Leibniz rule holds at order

n

. Taking one more derivative of each term and collecting the coefficient of

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

gives

(f g)^{(n + 1)} = \sum_{r = 0}^{n + 1} [(r n) + (r - 1 n)] f^{(n + 1 - r)} g^{(r)} .

0 \leq r \leq n

are zero. For

1 \leq r \leq n

(r n) + (r - 1 n) = \frac{n !}{r ! ( n - r )!} + \frac{n !}{( r - 1 )! ( n - r + 1 )!} = \frac{( n + 1 )!}{r ! ( n + 1 - r )!} = (r n + 1) .

r = 0, n + 1

, the identity is

1 + 0 = 1

. Substitution gives the required order-

n + 1

formula, completing the induction.

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