If x = sin^{-1}(3t - 4t^3) and y = cos^{-1}(s | vedclass

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(D) Given $x = \sin^{-1}(3t - 4t^3)$ and $y = \cos^{-1}(\sqrt{1 - t^2})$.For $y = \cos^{-1}(\sqrt{1 - t^2})$,let $t = \sin 	heta$,then $\sqrt{1 - t^2

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$1/3$

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(D) Given $x = \sin^{-1}(3t - 4t^3)$ and $y = \cos^{-1}(\sqrt{1 - t^2})$.For $y = \cos^{-1}(\sqrt{1 - t^2})$,let $t = \sin 	heta$,then $\sqrt{1 - t^2} = \cos 	heta$.So,$y = \cos^{-1}(\cos 	heta) = 	heta = \sin^{-1} t$.For $x = \sin^{-1}(3t - 4t^3)$,let $t = \sin 	heta$,then $x = \sin^{-1}(3 \sin 	heta - 4 \sin^3 	heta) = \sin^{-1}(\sin 3 	heta) = 3 	heta = 3 \sin^{-1} t$.Now,differentiate $x$ and $y$ with respect to $t$:$\frac{dx}{dt} = 3 \cdot \frac{1}{\sqrt{1 - t^2}}$$\frac{dy}{dt} = 1 \cdot \frac{1}{\sqrt{1 - t^2}}$Therefore,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{1/\sqrt{1 - t^2}}{3/\sqrt{1 - t^2}} = \frac{1}{3}$.

If $x = \sin^{-1}(3t - 4t^3)$ and $y = \cos^{-1}(\sqrt{1 - t^2})$,then $\frac{dy}{dx}$ is equal to

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