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Answer: \(\dfrac{d^ny}{dx^n} = \dfrac{(-1)^{n-1}(n - 1)!}{x^n}\)

Let \(y = \log x\). Then

\(y_1 = \dfrac1x = x^{-1}\), \(y_2 = (-1)x^{-2} = \dfrac{(-1)^1\,1!}{x^2}\), \(y_3 = (-1)(-2)x^{-3} = \dfrac{(-1)^2\,2!}{x^3}\), \(y_4 = \dfrac{(-1)^3\,3!}{x^4}\), …

Following the pattern, \(y_n = \dfrac{(-1)^{n-1}(n - 1)!}{x^n}\).

Doubtly AI · 28 Oct 2026 AI-verified · Oct 2026 Report
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