Q.21 Find dy/dx if y=tan⁻¹(√(3-x/3+x))

Answer: dy/dx=-1/2√(9-x²) Put x=3cos2θ , so θ=1/2cos⁻¹(x/3) with 0ltθltπ/2 . 3-x/3+x=3-3cos2θ/3+3cos2θ=2sin²θ/2cos²θ=tan²θ . y=tan⁻¹(tanθ)=θ=1/2cos⁻¹(x/3) . dy/dx=1/2·-1/√(1-x²/9)·1/3=-1/2√(9-x²) .

Chapter: Differentiation· Mathematics & Statistics Part 2

Find \(\dfrac{dy}{dx}\) if \(y=\tan^{-1}\left(\sqrt{\dfrac{3-x}{3+x}}\right)\)

Q.21 Find \(\dfrac{dy}{dx}\) if \(y=\tan^{-1}\left(\sqrt{\dfrac{3-x}{3+x}}\right)\)

Doubtly Asked 29 Oct 2026 Board exam · July 2024 hsc 12th Save WhatsApp Report

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Answer: \(\dfrac{dy}{dx}=-\dfrac{1}{2\sqrt{9-x^2}}\)

Put \(x=3\cos2\theta\), so \(\theta=\dfrac12\cos^{-1}\left(\dfrac x3\right)\) with \(0\lt\theta\lt\dfrac{\pi}{2}\).

\(\dfrac{3-x}{3+x}=\dfrac{3-3\cos2\theta}{3+3\cos2\theta}=\dfrac{2\sin^2\theta}{2\cos^2\theta}=\tan^2\theta\).

\(y=\tan^{-1}(\tan\theta)=\theta=\dfrac12\cos^{-1}\left(\dfrac x3\right)\).

\(\dfrac{dy}{dx}=\dfrac12\cdot\dfrac{-1}{\sqrt{1-\frac{x^2}{9}}}\cdot\dfrac13=-\dfrac{1}{2\sqrt{9-x^2}}\).

Doubtly AI · 29 Oct 2026 AI-generated · not yet verified Report

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Find \(\dfrac{dy}{dx}\) if \(y=\tan^{-1}\left(\sqrt{\dfrac{3-x}{3+x}}\right)\)

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