The derivative of sin ^{-1}left(2 x sqrt{1-x^ | vedclass

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(A) Let $y = \sin ^{-1}\left(2 x \sqrt{1-x^2}ight)$ and $z = \sin ^{-1}\left(3 x-4 x^3ight)$.Substitute $x = \sin 	heta$,which implies $	heta =

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$\frac{2}{3}$

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(A) Let $y = \sin ^{-1}\left(2 x \sqrt{1-x^2}ight)$ and $z = \sin ^{-1}\left(3 x-4 x^3ight)$.Substitute $x = \sin 	heta$,which implies $	heta = \sin ^{-1} x$.Then,$y = \sin ^{-1}\left(2 \sin 	heta \sqrt{1-\sin ^2 	heta}ight) = \sin ^{-1}(2 \sin 	heta \cos 	heta) = \sin ^{-1}(\sin 2 	heta) = 2 	heta = 2 \sin ^{-1} x$.Similarly,$z = \sin ^{-1}\left(3 \sin 	heta - 4 \sin ^3 	hetaight) = \sin ^{-1}(\sin 3 	heta) = 3 	heta = 3 \sin ^{-1} x$.Now,differentiate $y$ and $z$ with respect to $x$:$\frac{dy}{dx} = \frac{2}{\sqrt{1-x^2}}$ and $\frac{dz}{dx} = \frac{3}{\sqrt{1-x^2}}$.Finally,the derivative of $y$ with respect to $z$ is given by $\frac{dy}{dz} = \frac{dy/dx}{dz/dx} = \frac{2/\sqrt{1-x^2}}{3/\sqrt{1-x^2}} = \frac{2}{3}$.

The derivative of $\sin ^{-1}\left(2 x \sqrt{1-x^2}\right)$ with respect to $\sin ^{-1}\left(3 x-4 x^3\right)$ is

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