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ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2022/2/solution

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Past exam of the mathematical methods II course of the University of Oxford / 2022 / 2 / Solution by

Codex  0 2026-10-25

Use logarithmic differentiation on the positive function:

ln f = x^{- 1} ln (x^{2} + 2)

. The product rule and chain rule give

f^{'} (x) = (x^{2} + 2)^{1/ x} [\frac{2}{x ^{2} + 2} - \frac{ln ( x ^{2} + 2 )}{x ^{2}}], x \neq = 0.

(1)

The positive base permits this real definition on both components of the indicated domain. There is no need to treat a variable exponent as a constant.

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