frac{d}{dx} sin^{-1}(3x - 4x^3) is equal to | vedclass

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(B) Let $y = \sin^{-1}(3x - 4x^3)$.Substitute $x = \sin 	heta$,which implies $	heta = \sin^{-1} x$.Then,$y = \sin^{-1}(3 \sin 	heta - 4 \sin^3 	he

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

Vedclass · Answer

(B) Let $y = \sin^{-1}(3x - 4x^3)$.Substitute $x = \sin 	heta$,which implies $	heta = \sin^{-1} x$.Then,$y = \sin^{-1}(3 \sin 	heta - 4 \sin^3 	heta)$.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 $	heta = \sin^{-1} x$,we have $y = 3 \sin^{-1} x$.Differentiating with respect to $x$:$\frac{dy}{dx} = 3 \cdot \frac{d}{dx}(\sin^{-1} x) = 3 \cdot \frac{1}{\sqrt{1-x^2}} = \frac{3}{\sqrt{1-x^2}}$.

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

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