 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-2/37b/b/solution

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

Differentiate

f (λ x_{1}, \dots, λ x_{n}) = λ^{k} f (x_{1}, \dots, x_{n})

with respect to

λ

. The chain rule gives

\sum_{i} x_{i} \frac{\partial f}{\partial x _{i}} (λ x_{1}, \dots, λ x_{n}) = k λ^{k - 1} f (x_{1}, \dots, x_{n}) .

Setting

λ = 1

i \sum x_{i} \frac{\partial f}{\partial x _{i}} = k f .

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