\(
\text { If } y=\sin ^{-1}\left[\cos \sqrt{\frac{1+x}{2}}\right]+x^x \text {, then } \frac{d y}{d x} \text { at } x=1 \text { is }
\)

\(
\begin{aligned}
& \sin ^{-1}\left[\cos \sqrt{\frac{1+x}{2}}\right]+x^x \\
& =\sin ^{-1}\left[\sin \left(\frac{\pi}{2}-\sqrt{\frac{1+x}{2}}\right)\right]+x^x \\
& =\frac{\pi}{2}-\sqrt{\frac{1+x}{2}+x^x} \\
& \therefore \frac{d y}{d x}=0-\frac{1}{\sqrt{2}} \cdot \frac{d}{d x}(\sqrt{1+x})+\frac{d}{d x}\left(x^x\right)
\end{aligned}
\)
Let \(\mathrm{u}=\mathrm{x}^{\mathrm{x}} \Rightarrow \log \mathrm{u}=\mathrm{x} \log \mathrm{x}\)
\(
\begin{aligned}
& \therefore \frac{1}{u} \frac{d u}{d x}=\frac{x}{x}+\log x \Rightarrow \frac{d y}{d x}=x^x(1+\log x) \\
& \therefore \frac{d y}{d x}=\frac{-1}{\sqrt{2}}\left[\frac{1}{2 \sqrt{1+x}}\right]+x^x(1+\log x) \\
& \therefore\left[\frac{d y}{d x}\right]_{x=1}=\left(\frac{-1}{\sqrt{2}}\right)\left(\frac{1}{2 \sqrt{2}}\right)+1=\frac{-1}{4}+1=\frac{3}{4}
\end{aligned}
\)