A) \[\frac{\frac{\sin x}{x}-\log x.\cos x}{\sin x}\]
B) \[\frac{\frac{\sin x}{x}-\log x.\cos x}{{{\sin }^{2}}x}\]
C) \[\frac{\sin x-\log x.\cos x}{{{\sin }^{2}}x}\]
D) \[\frac{\frac{\sin x}{x}-\log x}{{{\sin }^{2}}x}\]
Correct Answer: B
Solution :
\[\frac{d}{dx}\left( \frac{\log x}{\sin x} \right)=\frac{\frac{\sin x}{x}-\log x.\cos x}{{{\sin }^{2}}x}\].
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