The derivative of sin ^{-1}left(3 x-4 x^3righ | vedclass

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(B) Let $y = \sin ^{-1}\left(3 x-4 x^3ight)$.We know that $\sin(3	heta) = 3\sin	heta - 4\sin^3	heta$.Let $x = \sin	heta$,then $	heta = \sin^{-1

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$\frac{-3}{\sqrt{1-x^2}}$

Vedclass · Answer

(B) Let $y = \sin ^{-1}\left(3 x-4 x^3ight)$.We know that $\sin(3	heta) = 3\sin	heta - 4\sin^3	heta$.Let $x = \sin	heta$,then $	heta = \sin^{-1}x$.Since $\frac{1}{2} Then $3	heta$ lies in the range $(\frac{\pi}{2}, \frac{3\pi}{2})$.In this range,$\sin^{-1}(\sin(3	heta)) = \pi - 3	heta$.Thus,$y = \pi - 3\sin^{-1}x$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(\pi - 3\sin^{-1}x) = 0 - 3 	imes \frac{1}{\sqrt{1-x^2}} = \frac{-3}{\sqrt{1-x^2}}$.

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

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