Differentiating the following with respect to $x$: $\frac{\cos x}{\log x}, x > 0$

Let $y = \frac{\cos x}{\log x}$.
By using the quotient rule,$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$,we obtain:
$\frac{dy}{dx} = \frac{(\log x) \frac{d}{dx}(\cos x) - (\cos x) \frac{d}{dx}(\log x)}{(\log x)^2}$
Since $\frac{d}{dx}(\cos x) = -\sin x$ and $\frac{d}{dx}(\log x) = \frac{1}{x}$,we have:
$\frac{dy}{dx} = \frac{(\log x)(-\sin x) - (\cos x)(\frac{1}{x})}{(\log x)^2}$
$\frac{dy}{dx} = \frac{-\sin x \log x - \frac{\cos x}{x}}{(\log x)^2}$
Multiplying the numerator and denominator by $x$,we get:
$\frac{dy}{dx} = \frac{-(x \sin x \log x + \cos x)}{x(\log x)^2}, x > 0$