$\cfrac{{\cos x}}{{\log x}},x > 0$

Let $y = \cfrac{{\cos x}}{{\log x}}$

therefore, $\cfrac{{dy}}{{dx}} = \cfrac{d}{{dx}}\left( {\cfrac{{\cos x}}{{\log x}}} \right) = \cfrac{{\log x\cfrac{d}{{dx}}(\cos x) - \cos x\cfrac{d}{{dx}}(\log x)}}{{{{(\log x)}^2}}}$

$= \cfrac{{\log x( - \sin x) - \cos x\left( {\cfrac{1}{x}} \right)}}{{{{(\log x)}^2}}} = - \left[ {\cfrac{{\sin x\log x + \cfrac{1}{x}\cos x}}{{{{(\log x)}^2}}}} \right]$

$= \cfrac{{ - (x\sin x \cdot \log x + \cos x)}}{{x{{(\log x)}^2}}}$