Q.1 (v) If y = sec(tan⁻¹ x) , then dy/dx at x = 1 is _____. A. 1/2 B. 1 C. 1/√(2) D. √(2)

Answer: C. 1/√(2) Answer: (c) 1/√(2) Let tan⁻¹x = θ , so tanθ = x and y = secθ = √(1 + x²) . Then dy/dx = x/√(1 + x²) , and at x = 1 this is 1/√(2) .

If \(y = \sec(\tan^{-1} x)\), then \(\dfrac{dy}{dx}\) at \(x = 1\) is _____.

Q.1 (v) If \(y = \sec(\tan^{-1} x)\), then \(\dfrac{dy}{dx}\) at \(x = 1\) is _____.

A. \(\dfrac{1}{2}\)

B. \(1\)

C. \(\dfrac{1}{\sqrt{2}}\)

D. \(\sqrt{2}\)

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

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Answer: C. \(\dfrac{1}{\sqrt{2}}\)

Answer: (c) \(\dfrac{1}{\sqrt{2}}\)

Let \(\tan^{-1}x = \theta\), so \(\tan\theta = x\) and \(y = \sec\theta = \sqrt{1 + x^2}\). Then \(\dfrac{dy}{dx} = \dfrac{x}{\sqrt{1 + x^2}}\), and at \(x = 1\) this is \(\dfrac{1}{\sqrt{2}}\).

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

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If \(y = \sec(\tan^{-1} x)\), then \(\dfrac{dy}{dx}\) at \(x = 1\) is _____.

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