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The value of $\frac{d}{d x}\left[\log \left(\sin \sqrt{\frac{x^2+1}{x^2+2}}\right)\right]$ when $x=\sqrt{2}$, is
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Verified Answer
The correct answer is:
$\frac{\sqrt{2} \cot \left(\frac{\sqrt{3}}{2}\right)}{8 \sqrt{3}}$
We have to find
$$
\begin{aligned}
& \frac{d}{d x}\left[\log \left(\sin \sqrt{\frac{x^2+1}{x^2+2}}\right)\right] \\
& \Rightarrow \frac{1}{\sin \sqrt{\frac{x^2+1}{x^2+2}}} \cdot \cos \sqrt{\frac{x^2+1}{x^2+2}} \cdot \frac{1}{2}\left(\frac{x^2+1}{x^2+2}\right)^{-\frac{1}{2}} \cdot \frac{2 x}{\left(x^2+2\right)^2} \\
&
\end{aligned}
$$
at $x=\sqrt{2}$, we get
$$
\frac{\sqrt{2} \cot \left(\frac{\sqrt{3}}{2}\right)}{8 \sqrt{3}}
$$
$$
\begin{aligned}
& \frac{d}{d x}\left[\log \left(\sin \sqrt{\frac{x^2+1}{x^2+2}}\right)\right] \\
& \Rightarrow \frac{1}{\sin \sqrt{\frac{x^2+1}{x^2+2}}} \cdot \cos \sqrt{\frac{x^2+1}{x^2+2}} \cdot \frac{1}{2}\left(\frac{x^2+1}{x^2+2}\right)^{-\frac{1}{2}} \cdot \frac{2 x}{\left(x^2+2\right)^2} \\
&
\end{aligned}
$$
at $x=\sqrt{2}$, we get
$$
\frac{\sqrt{2} \cot \left(\frac{\sqrt{3}}{2}\right)}{8 \sqrt{3}}
$$
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