The value of frac{d}{d x}left[log left(sin sq | vedclass

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(C) Let $y = \log \left(\sin \sqrt{\frac{x^2+1}{x^2+2}}\right)$.Using the chain rule, $\frac{dy}{dx} = \frac{1}{\sin \sqrt{\frac{x^2+1}{x^2+2}}} \cdot

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$\frac{\sqrt{2} \cot \left(\frac{\sqrt{3}}{2}\right)}{8 \sqrt{3}}$

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(C) Let $y = \log \left(\sin \sqrt{\frac{x^2+1}{x^2+2}}\right)$.Using the chain rule, $\frac{dy}{dx} = \frac{1}{\sin \sqrt{\frac{x^2+1}{x^2+2}}} \cdot \cos \sqrt{\frac{x^2+1}{x^2+2}} \cdot \frac{1}{2\sqrt{\frac{x^2+1}{x^2+2}}} \cdot \frac{d}{dx} \left( \frac{x^2+1}{x^2+2} \right)$.Calculating the derivative of the inner quotient: $\frac{d}{dx} \left( \frac{x^2+1}{x^2+2} \right) = \frac{2x(x^2+2) - 2x(x^2+1)}{(x^2+2)^2} = \frac{2x^3+4x-2x^3-2x}{(x^2+2)^2} = \frac{2x}{(x^2+2)^2}$.Substituting $x = \sqrt{2}$, we have $\frac{x^2+1}{x^2+2} = \frac{2+1}{2+2} = \frac{3}{4}$.Thus, $\frac{dy}{dx} = \cot \left( \sqrt{\frac{3}{4}} \right) \cdot \frac{1}{2 \sqrt{3/4}} \cdot \frac{2\sqrt{2}}{(2+2)^2} = \cot \left( \frac{\sqrt{3}}{2} \right) \cdot \frac{1}{2 \cdot \frac{\sqrt{3}}{2}} \cdot \frac{2\sqrt{2}}{16} = \cot \left( \frac{\sqrt{3}}{2} \right) \cdot \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{2}}{8} = \frac{\sqrt{2} \cot \left( \frac{\sqrt{3}}{2} \right)}{8 \sqrt{3}}$.

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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