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If y=x+1+x−1x+1−x−1y = \frac{\sqrt{x+1} + \sqrt{x-1}}{\sqrt{x+1} - \sqrt{x-1}}, then (x2−1)3/2d2ydx2=(x^2-1)^{3/2} \frac{d^2y}{dx^2} =

Solution

✅ Correct Option: 4

Rationalize: y=(x+1+x−1)2(x+1)−(x−1)=2x+2x2−12=x+x2−1y = \frac{(\sqrt{x+1}+\sqrt{x-1})^2}{(x+1)-(x-1)} = \frac{2x + 2\sqrt{x^2-1}}{2} = x + \sqrt{x^2-1}.

dydx=1+xx2−1\frac{dy}{dx} = 1 + \frac{x}{\sqrt{x^2-1}}

d2ydx2=x2−1−x⋅xx2−1x2−1=−1(x2−1)3/2\frac{d^2y}{dx^2} = \frac{\sqrt{x^2-1} - x \cdot \frac{x}{\sqrt{x^2-1}}}{x^2-1} = \frac{-1}{(x^2-1)^{3/2}}

So (x2−1)3/2d2ydx2=−1(x^2-1)^{3/2} \frac{d^2y}{dx^2} = -1.

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