If x = sqrt{10^{sin^{-1} t}} and y = sqrt{10^ | vedclass

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(C) Given $x = \sqrt{10^{\sin^{-1} t}}$ and $y = \sqrt{10^{\cos^{-1} t}}$.Multiply $x$ and $y$:$xy = \sqrt{10^{\sin^{-1} t} \cdot 10^{\cos^{-1} t}} =

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$-\frac{y}{x}$

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(C) Given $x = \sqrt{10^{\sin^{-1} t}}$ and $y = \sqrt{10^{\cos^{-1} t}}$.Multiply $x$ and $y$:$xy = \sqrt{10^{\sin^{-1} t} \cdot 10^{\cos^{-1} t}} = \sqrt{10^{\sin^{-1} t + \cos^{-1} t}}$.Since $\sin^{-1} t + \cos^{-1} t = \frac{\pi}{2}$,we have:$xy = \sqrt{10^{\pi/2}}$.Differentiating both sides with respect to $x$:$x \frac{dy}{dx} + y(1) = 0$.$x \frac{dy}{dx} = -y$.Therefore,$\frac{dy}{dx} = -\frac{y}{x}$.

If $x = \sqrt{10^{\sin^{-1} t}}$ and $y = \sqrt{10^{\cos^{-1} t}}$,then $\frac{dy}{dx} = $ . . . . . .

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