If y = tan^{-1}left( frac{x}{sqrt{1 - x^2}} r | vedclass

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(C) Given $y = 	an^{-1}\left( \frac{x}{\sqrt{1 - x^2}} ight)$.Let $x = \sin 	heta$,then $	heta = \sin^{-1} x$.Substituting $x = \sin 	heta$ in t

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

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(C) Given $y = 	an^{-1}\left( \frac{x}{\sqrt{1 - x^2}} ight)$.Let $x = \sin 	heta$,then $	heta = \sin^{-1} x$.Substituting $x = \sin 	heta$ in the expression for $y$:$y = 	an^{-1}\left( \frac{\sin 	heta}{\sqrt{1 - \sin^2 	heta}} ight)$$y = 	an^{-1}\left( \frac{\sin 	heta}{\cos 	heta} ight)$$y = 	an^{-1}(	an 	heta) = 	heta$Since $	heta = \sin^{-1} x$,we have $y = \sin^{-1} x$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1 - x^2}}$.

If $y = \tan^{-1}\left( \frac{x}{\sqrt{1 - x^2}} \right)$,then $\frac{dy}{dx} = $

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