Continuity and Differentiability — Maths STD 12 Science — Question

$\therefore \frac{{dy}}{{dx}} = \frac{{\log x\frac{d}{{dx}}\left( {\cos x} \right) - \cos x\frac{d}{{dx}}\left( {\log x} \right)}}{{{{\left( {\log x} \right)}^2}}}$ [By quotient rule]

$= \frac{{\log x\left( { - \sin x} \right) - \cos x.\frac{1}{x}}}{{{{\left( {\log x} \right)}^2}}}$

$= \frac{{ - \left( {\sin x\log x + \frac{{\cos x}}{x}} \right)}}{{{{\left( {\log x} \right)}^2}}}$

$= \frac{{ - \left( {x\sin x\log x + \cos x} \right)}}{{x{{\left( {\log x} \right)}^2}}}$