For a 0, t in left( 0, frac{pi}{2} right),l | vedclass

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(D) Given $x = \sqrt{a^{\sin^{-1} t}}$. Taking natural logarithm on both sides,$2 \ln x = (\sin^{-1} t) \ln a$. Differentiating with respect to $t$,$\

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

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(D) Given $x = \sqrt{a^{\sin^{-1} t}}$. Taking natural logarithm on both sides,$2 \ln x = (\sin^{-1} t) \ln a$. Differentiating with respect to $t$,$\frac{2}{x} \frac{dx}{dt} = \frac{\ln a}{\sqrt{1-t^2}}$. Thus,$\frac{dx}{dt} = \frac{x \ln a}{2\sqrt{1-t^2}}$. Similarly,$y = \sqrt{a^{\cos^{-1} t}} \Rightarrow 2 \ln y = (\cos^{-1} t) \ln a$. Differentiating with respect to $t$,$\frac{2}{y} \frac{dy}{dt} = -\frac{\ln a}{\sqrt{1-t^2}}$. Thus,$\frac{dy}{dt} = -\frac{y \ln a}{2\sqrt{1-t^2}}$. Using the chain rule,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-y \ln a / (2\sqrt{1-t^2})}{x \ln a / (2\sqrt{1-t^2})} = -\frac{y}{x}$. Therefore,$1 + \left( \frac{dy}{dx} \right)^2 = 1 + \left( -\frac{y}{x} \right)^2 = 1 + \frac{y^2}{x^2} = \frac{x^2 + y^2}{x^2}$.

For $a > 0, t \in \left( 0, \frac{\pi}{2} \right)$,let $x = \sqrt{a^{\sin^{-1} t}}$ and $y = \sqrt{a^{\cos^{-1} t}}$. Then,$1 + \left( \frac{dy}{dx} \right)^2$ equals

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