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If x2y2=aπ/2x^2y^2 = a^{\pi/2}, then d2ydx2=\dfrac{d^2y}{dx^2} =

Solution

✅ Correct Option: 3

Put k=aπ/2k = a^{\pi/2}, a constant. Then y2=k/x2y^2 = k/x^2, so y=k x−1y = \sqrt{k}\,x^{-1}.

Differentiating, dydx=−k x−2\dfrac{dy}{dx} = -\sqrt{k}\,x^{-2} and d2ydx2=2k x−3\dfrac{d^2y}{dx^2} = 2\sqrt{k}\,x^{-3}.

Since y3=k3/2/x3y^3 = k^{3/2}/x^3, we have 2y3k=2kx3\dfrac{2y^3}{k} = \dfrac{2\sqrt{k}}{x^3}, which is exactly the second derivative. Hence the answer is 2y3aπ/2\dfrac{2y^3}{a^{\pi/2}}.

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