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

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Question

(A) 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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) 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 $y = 3 \sin^{-1} x$.Now,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)) = $

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