Find the derivative: frac{d}{dx} left( frac{l | vedclass

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(B) To find the derivative of $\frac{\log x}{\sin x}$,we use the quotient rule: $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \

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$\frac{\frac{\sin x}{x} - \log x \cdot \cos x}{\sin^2 x}$

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(B) To find the derivative of $\frac{\log x}{\sin x}$,we use the quotient rule: $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$.Here,$u = \log x$ and $v = \sin x$.Then,$\frac{du}{dx} = \frac{1}{x}$ and $\frac{dv}{dx} = \cos x$.Applying the quotient rule:$\frac{d}{dx} \left( \frac{\log x}{\sin x} \right) = \frac{\sin x \cdot \frac{1}{x} - \log x \cdot \cos x}{(\sin x)^2}$.This simplifies to $\frac{\frac{\sin x}{x} - \log x \cdot \cos x}{\sin^2 x}$.

List-$I$	List-$II$
$A$. $\sec^{-1} x$	$I$. $\frac{1}{1-x^2}, x \in (-1, 1)$
$B$. $\tanh^{-1} x$	$II$. $\frac{-1}{\|x\| \sqrt{x^2+1}}, x \neq 0$
$C$. $\coth^{-1} x$	$III$. $\frac{1}{\|x\| \sqrt{x^2-1}}, \|x\| > 1$
$D$. $\operatorname{cosech}^{-1} x$	$IV$. $\frac{1}{1-x^2}, x \in R - [-1, 1]$
	$V$. $\frac{-1}{\|x\| \sqrt{1-x^2}}, \|x\| < 1, x \neq 0$

Find the derivative: $\frac{d}{dx} \left( \frac{\log x}{\sin x} \right)$

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