If y = sin^{-1}left[cos sqrt{frac{1+x}{2}}rig | vedclass

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(C) Given $y = \sin^{-1}\left[\cos \sqrt{\frac{1+x}{2}}ight] + x^x$. Using $\cos 	heta = \sin\left(\frac{\pi}{2} - 	hetaight)$,we have: $y = \si

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$\frac{3}{4}$

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(C) Given $y = \sin^{-1}\left[\cos \sqrt{\frac{1+x}{2}}ight] + x^x$. Using $\cos 	heta = \sin\left(\frac{\pi}{2} - 	hetaight)$,we have: $y = \sin^{-1}\left[\sin\left(\frac{\pi}{2} - \sqrt{\frac{1+x}{2}}ight)ight] + x^x = \frac{\pi}{2} - \sqrt{\frac{1+x}{2}} + x^x$. Differentiating with respect to $x$: $\frac{dy}{dx} = 0 - \frac{1}{\sqrt{2}} \cdot \frac{d}{dx}((1+x)^{1/2}) + \frac{d}{dx}(x^x)$. For $x^x$,let $u = x^x$,then $\ln u = x \ln x$. $\frac{1}{u} \frac{du}{dx} = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1$,so $\frac{du}{dx} = x^x(1 + \ln x)$. Thus,$\frac{dy}{dx} = -\frac{1}{\sqrt{2}} \cdot \frac{1}{2\sqrt{1+x}} + x^x(1 + \ln x)$. At $x = 1$: $\left[\frac{dy}{dx}ight]_{x=1} = -\frac{1}{2\sqrt{2}\sqrt{2}} + 1^1(1 + \ln 1) = -\frac{1}{4} + 1(1 + 0) = -\frac{1}{4} + 1 = \frac{3}{4}$.

If $y = \sin^{-1}\left[\cos \sqrt{\frac{1+x}{2}}\right] + x^x$,then find $\frac{dy}{dx}$ at $x = 1$.

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