frac{d}{d x} sin ^{-1}left(3 x-4 x^3right) is | vedclass

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(B) Let $y = \sin^{-1}(3x - 4x^3)$.Substitute $x = \sin 	heta$, which implies $	heta = \sin^{-1} x$.Using the trigonometric identity $\sin 3	heta =

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

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(B) Let $y = \sin^{-1}(3x - 4x^3)$.Substitute $x = \sin 	heta$, which implies $	heta = \sin^{-1} x$.Using the trigonometric identity $\sin 3	heta = 3\sin 	heta - 4\sin^3 	heta$, we get:$y = \sin^{-1}(\sin 3	heta) = 3	heta$.Substituting back the value of $	heta$:$y = 3\sin^{-1} x$.Differentiating with respect to $x$:$\frac{dy}{dx} = 3 \cdot \frac{d}{dx}(\sin^{-1} x) = 3 \cdot \frac{1}{\sqrt{1-x^2}} = \frac{3}{\sqrt{1-x^2}}$.

$\frac{d}{d x} \sin ^{-1}\left(3 x-4 x^3\right)$ is equal to

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