$\sin (\log x)$

Let y $=$ sin (log x)
$\Rightarrow$ $\cfrac{{dy}}{{dx}} = \cos (\log x)\cfrac{1}{x}$

therefore, $\cfrac{{{d^2}y}}{{d{x^2}}} = \cos (\log x) \cdot \left( { - \cfrac{1}{{{x^2}}}} \right) + \cfrac{1}{x} \cdot \{ - \sin (\log x)\} \cdot \cfrac{1}{x}$

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

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