frac{d}{dx} sin^{-1}(2axsqrt{1 - a^2x^2}) = | vedclass

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(C) Let $y = \sin^{-1}(2ax\sqrt{1 - a^2x^2})$.Substitute $ax = \sin 	heta$,which implies $	heta = \sin^{-1}(ax)$.Then $y = \sin^{-1}(2 \sin 	heta \

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $y = \sin^{-1}(2ax\sqrt{1 - a^2x^2})$.Substitute $ax = \sin 	heta$,which implies $	heta = \sin^{-1}(ax)$.Then $y = \sin^{-1}(2 \sin 	heta \sqrt{1 - \sin^2 	heta}) = \sin^{-1}(2 \sin 	heta \cos 	heta) = \sin^{-1}(\sin 2	heta) = 2	heta$.Substituting back,$y = 2 \sin^{-1}(ax)$.Now,differentiate with respect to $x$:$\frac{dy}{dx} = 2 \cdot \frac{1}{\sqrt{1 - (ax)^2}} \cdot \frac{d}{dx}(ax) = \frac{2}{\sqrt{1 - a^2x^2}} \cdot a = \frac{2a}{\sqrt{1 - a^2x^2}}$.

$\frac{d}{dx} \sin^{-1}(2ax\sqrt{1 - a^2x^2}) = $

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