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

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Aicharya · Accepted Answer

Given, $\begin{aligned} f(x) & =\cos ^{-1}\left(\sin \sqrt{\frac{1+x}{2}}\right)+x^x \ & =\frac{\pi}{2}-\sin ^{-1}\left(\sin \sqrt{\frac{1+x}{2}}\right)+x^x \ & =\frac{\pi}{2}-\sqrt{\frac{1+x}{2}}+x^x \end{aligned}$ $\therefore \quad f^{\prime}(x)=\frac{-1}{\sqrt{2}} \cdot \frac{1}{2 \sqrt{1+x}}+x^x\left(\log _e(e x)\right)$ $\left[\because \frac{d\left(x^x\right)}{d x}=x^x \log _e(e x)\right]$ On putting $x=1$, we get $f^{\prime}(1)=\frac{-1}{4}+1=\frac{3}{4}$ Hence, option (c) is correct.

The derivative of \(f(x)=\cos ^{-1}\left[\sin \sqrt{\frac{1+x}{2}}\right]+x^x\) with respect to \(x\) at…

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