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If $y = x\sin x,$then
A${1 \over y}{{dy} \over {dx}} = {1 \over x} + \cot x$
B${{dy} \over {dx}} = {1 \over x} + \cot x$
C${1 \over y}{{dy} \over {dx}} = {1 \over x} - \cot x$
DNone of these
Solution
(a) $y = x\sin x$==>$\frac{{dy}}{{dx}} = x\cos x + \sin x$ ==>$\frac{1}{y}\frac{{dy}}{{dx}} = \cot x + \frac{1}{x}$.
Standard 11
Physics
