Q.1 (v) If y = log_x a then dy/dx = _____. A. -log a/x(log x)² B. xlog a C. -1/log_x a D. 1/xlog a

Answer: A. -log a/x(log x)² Answer: (a) -log a/x(log x)² By change of base, y = log a/log x = log a (log x)⁻¹ . Then dy/dx = log a · (-1)(log x)⁻²·1/x = -log a/x(log x)² .

If \(y = \log_x a\) then \(\frac{dy}{dx} =\) _____.

Q.1 (v) If \(y = \log_x a\) then \(\frac{dy}{dx} =\) _____.

A. \(\frac{-\log a}{x(\log x)^2}\)

B. \(x\log a\)

C. \(-\frac{1}{\log_x a}\)

D. \(\frac{1}{x\log a}\)

Doubtly Asked 28 Oct 2026 Board exam · June 2026 hsc 12th Save WhatsApp Report

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Answer: A. \(\frac{-\log a}{x(\log x)^2}\)

Answer: (a) \(\frac{-\log a}{x(\log x)^2}\)

By change of base, \(y = \frac{\log a}{\log x} = \log a\,(\log x)^{-1}\). Then \(\frac{dy}{dx} = \log a \cdot (-1)(\log x)^{-2}\cdot\frac{1}{x} = \frac{-\log a}{x(\log x)^2}\).

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

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If \(y = \log_x a\) then \(\frac{dy}{dx} =\) _____.

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