If f(x) = sin^{-1}left(sqrt{frac{1-x}{2}}righ | vedclass

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(A) Given $f(x) = \sin^{-1}\left(\sqrt{\frac{1-x}{2}}\right)$.Method $1$: Using the chain rule:$f^{\prime}(x) = \frac{d}{dx}\left[\sin^{-1}\left(\sqrt

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

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(A) Given $f(x) = \sin^{-1}\left(\sqrt{\frac{1-x}{2}}ight)$.Method $1$: Using the chain rule:$f^{\prime}(x) = \frac{d}{dx}\left[\sin^{-1}\left(\sqrt{\frac{1-x}{2}}ight)ight] = \frac{1}{\sqrt{1-\left(\sqrt{\frac{1-x}{2}}ight)^{2}}} \cdot \frac{d}{dx}\left(\sqrt{\frac{1-x}{2}}ight)$$= \frac{1}{\sqrt{1-\frac{1-x}{2}}} \cdot \frac{1}{2\sqrt{\frac{1-x}{2}}} \cdot \left(-\frac{1}{2}ight)$$= \frac{1}{\sqrt{\frac{1+x}{2}}} \cdot \frac{1}{\sqrt{2}\sqrt{1-x}} \cdot \left(-\frac{1}{2}ight)$$= \frac{\sqrt{2}}{\sqrt{1+x}} \cdot \frac{1}{\sqrt{2}\sqrt{1-x}} \cdot \left(-\frac{1}{2}ight) = \frac{-1}{2\sqrt{1-x^{2}}}$.Method $2$: Using substitution:Let $x = \cos 	heta$,then $\sqrt{\frac{1-x}{2}} = \sqrt{\frac{1-\cos 	heta}{2}} = \sqrt{\frac{2\sin^{2}(	heta/2)}{2}} = \sin(	heta/2)$.So,$f(x) = \sin^{-1}(\sin(	heta/2)) = 	heta/2 = \frac{1}{2}\cos^{-1}(x)$.Differentiating with respect to $x$:$f^{\prime}(x) = \frac{1}{2} \cdot \left(-\frac{1}{\sqrt{1-x^{2}}}ight) = \frac{-1}{2\sqrt{1-x^{2}}}$.Thus,the correct option is $A$.

If $f(x) = \sin^{-1}\left(\sqrt{\frac{1-x}{2}}\right)$,then $f^{\prime}(x) = $

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