d dx ( x x ) = - Vidyadip

MCQ

${d \over {dx}}\left( {{{\log x} \over {\sin x}}} \right) = $

A
${{{{\sin x} \over x} - \log x.\cos x} \over {\sin x}}$
✓
${{{{\sin x} \over x} - \log x.\cos x} \over {{{\sin }^2}x}}$
C
${{\sin x - \log x.\cos x} \over {{{\sin }^2}x}}$
D
${{{{\sin x} \over x} - \log x} \over {{{\sin }^2}x}}$

✓

Answer

Correct option: B.

${{{{\sin x} \over x} - \log x.\cos x} \over {{{\sin }^2}x}}$

b
(b) $\frac{d}{{dx}}\left( {\frac{{\log x}}{{\sin x}}} \right) = \frac{{\frac{{\sin x}}{x} - \log x.\cos x}}{{{{\sin }^2}x}}$.

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