Q.21 If y = sin⁻¹ x , then show that : (1 - x²)d²y/dx² - x · dy/dx = 0 .

Answer: Differentiate √(1 - x²) dy/dx = 1 once more and multiply by √(1 - x²) . y = sin⁻¹x ⇒ dy/dx = 1/√(1 - x²) , i.e. √(1 - x²) dy/dx = 1 . Differentiating w.r.t. x : √(1 - x²) d²y/dx² + dy/dx · -2x/2√(1 - x²) = 0 . Multiplying by √(1 - x²) : (1 - x²)d²y/dx² - xdy/dx = 0 . Hence proved.

If \(y = \sin^{-1} x\), then show that : \((1 - x^2)\frac{d^2y}{dx^2} - x \cdot \frac{dy}{dx} = 0\).

Q.21 If \(y = \sin^{-1} x\), then show that : \((1 - x^2)\frac{d^2y}{dx^2} - x \cdot \frac{dy}{dx} = 0\).

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

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Answer: Differentiate \(\sqrt{1 - x^2}\,\frac{dy}{dx} = 1\) once more and multiply by \(\sqrt{1 - x^2}\).

\(y = \sin^{-1}x \Rightarrow \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}\), i.e. \(\sqrt{1 - x^2}\,\frac{dy}{dx} = 1\).

Differentiating w.r.t. \(x\): \(\sqrt{1 - x^2}\,\frac{d^2y}{dx^2} + \frac{dy}{dx} \cdot \frac{-2x}{2\sqrt{1 - x^2}} = 0\).

Multiplying by \(\sqrt{1 - x^2}\): \((1 - x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} = 0\). Hence proved.

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

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If \(y = \sin^{-1} x\), then show that : \((1 - x^2)\frac{d^2y}{dx^2} - x \cdot \frac{dy}{dx} = 0\).

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