The derivative of sin ^{-1}left(frac{sqrt{1+x | vedclass

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Question

(A) Let $y = \sin ^{-1}\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{2}ight)$ and $u = \cos ^{-1} x$. Substitute $x = \cos 	heta$,where $	heta = \cos ^{-1} x

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$\frac{1}{2}$

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(A) Let $y = \sin ^{-1}\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{2}ight)$ and $u = \cos ^{-1} x$. Substitute $x = \cos 	heta$,where $	heta = \cos ^{-1} x$. Then,$\frac{\sqrt{1+x}+\sqrt{1-x}}{2} = \frac{\sqrt{1+\cos 	heta}+\sqrt{1-\cos 	heta}}{2} = \frac{\sqrt{2 \cos ^{2} \frac{	heta}{2}}+\sqrt{2 \sin ^{2} \frac{	heta}{2}}}{2} = \frac{\sqrt{2} \cos \frac{	heta}{2} + \sqrt{2} \sin \frac{	heta}{2}}{2} = \frac{1}{\sqrt{2}} \cos \frac{	heta}{2} + \frac{1}{\sqrt{2}} \sin \frac{	heta}{2}$. Using $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,we get $\sin \frac{\pi}{4} \cos \frac{	heta}{2} + \cos \frac{\pi}{4} \sin \frac{	heta}{2} = \sin \left(\frac{\pi}{4} + \frac{	heta}{2}ight)$. Thus,$y = \sin ^{-1} \left(\sin \left(\frac{\pi}{4} + \frac{	heta}{2}ight)ight) = \frac{\pi}{4} + \frac{	heta}{2} = \frac{\pi}{4} + \frac{1}{2} u$. Now,differentiate $y$ with respect to $u$: $\frac{dy}{du} = \frac{d}{du} \left(\frac{\pi}{4} + \frac{1}{2} uight) = \frac{1}{2}$.

The derivative of $\sin ^{-1}\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right)$ with respect to $\cos ^{-1} x$ is

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