If x = sqrt{e^{sin^{-1} t}} and y = sqrt{e^{c | vedclass

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(C) Given $x = \sqrt{e^{\sin^{-1} t}}$ and $y = \sqrt{e^{\cos^{-1} t}}$.Multiplying $x$ and $y$,we get:$xy = \sqrt{e^{\sin^{-1} t}} \cdot \sqrt{e^{\co

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$\frac{2y}{x^2}$

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(C) Given $x = \sqrt{e^{\sin^{-1} t}}$ and $y = \sqrt{e^{\cos^{-1} t}}$.Multiplying $x$ and $y$,we get:$xy = \sqrt{e^{\sin^{-1} t}} \cdot \sqrt{e^{\cos^{-1} t}} = \sqrt{e^{\sin^{-1} t + \cos^{-1} t}}$.Since $\sin^{-1} t + \cos^{-1} t = \frac{\pi}{2}$,we have $xy = \sqrt{e^{\pi/2}}$,which is a constant.Differentiating both sides with respect to $x$:$x \frac{dy}{dx} + y = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x} \quad ... (i)$.Differentiating again with respect to $x$:$\frac{d^2 y}{dx^2} = -\frac{d}{dx} \left( \frac{y}{x} \right) = -\left( \frac{x \frac{dy}{dx} - y}{x^2} \right)$.Substituting $\frac{dy}{dx} = -\frac{y}{x}$ from $(i)$:$\frac{d^2 y}{dx^2} = -\left( \frac{x(-\frac{y}{x}) - y}{x^2} \right) = -\left( \frac{-y - y}{x^2} \right) = -\left( \frac{-2y}{x^2} \right) = \frac{2y}{x^2}$.

If $x = \sqrt{e^{\sin^{-1} t}}$ and $y = \sqrt{e^{\cos^{-1} t}}$,then $\frac{d^2 y}{dx^2}$ is

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