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

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(A) Given $y = 	an^{-1}\left( \frac{x}{1 + \sqrt{1 - x^2}} ight)$.Substitute $x = \sin 	heta$,where $	heta = \sin^{-1} x$.Then $y = 	an^{-1}\lef

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

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(A) Given $y = 	an^{-1}\left( \frac{x}{1 + \sqrt{1 - x^2}} ight)$.Substitute $x = \sin 	heta$,where $	heta = \sin^{-1} x$.Then $y = 	an^{-1}\left( \frac{\sin 	heta}{1 + \sqrt{1 - \sin^2 	heta}} ight) = 	an^{-1}\left( \frac{\sin 	heta}{1 + \cos 	heta} ight)$.Using trigonometric identities $\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$ and $1 + \cos 	heta = 2 \cos^2 \frac{	heta}{2}$,we get:$y = 	an^{-1}\left( \frac{2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}}{2 \cos^2 \frac{	heta}{2}} ight) = 	an^{-1}\left( 	an \frac{	heta}{2} ight) = \frac{	heta}{2}$.Substituting $	heta = \sin^{-1} x$,we have $y = \frac{1}{2} \sin^{-1} x$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\sqrt{1 - x^2}} = \frac{1}{2\sqrt{1 - x^2}}$.

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

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