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

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(D) Given $x = \sqrt{a^{\sin^{-1} t}}$ and $y = \sqrt{a^{\cos^{-1} t}}$.We know that $\sin^{-1} t + \cos^{-1} t = \frac{\pi}{2}$,so $\cos^{-1} t = \fr

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$-\frac{y}{x}$

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(D) Given $x = \sqrt{a^{\sin^{-1} t}}$ and $y = \sqrt{a^{\cos^{-1} t}}$.We know that $\sin^{-1} t + \cos^{-1} t = \frac{\pi}{2}$,so $\cos^{-1} t = \frac{\pi}{2} - \sin^{-1} t$.Thus,$y = \sqrt{a^{\frac{\pi}{2} - \sin^{-1} t}} = \sqrt{\frac{a^{\pi/2}}{a^{\sin^{-1} t}}} = \frac{\sqrt{a^{\pi/2}}}{\sqrt{a^{\sin^{-1} t}}} = \frac{k}{x}$,where $k = \sqrt{a^{\pi/2}}$ is a constant.Now,$xy = k$.Differentiating both sides with respect to $x$:$x \frac{dy}{dx} + y(1) = 0$.$x \frac{dy}{dx} = -y$.$\frac{dy}{dx} = -\frac{y}{x}$.

If $x = \sqrt{a^{\sin^{-1} t}}$ and $y = \sqrt{a^{\cos^{-1} t}}$,then $\frac{dy}{dx} = $

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