$ ext {If } y=\sin ^{-1}\left[\cos \sqrt{\frac{1+x}{2}} ight]+x^x ext {, then } \frac{d y}{d x} ext { at } x=1 ext { is }$

Question

$	ext {If } y=\sin ^{-1}\left[\cos \sqrt{\frac{1+x}{2}}ight]+x^x 	ext {, then } \frac{d y}{d x} 	ext { at } x=1 	ext { is }$

Aicharya · Accepted Answer

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

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

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