Ta có
$f'(x) = \left( \dfrac{\sin x}{1 + 2 \cos x} \right)'$
$= \dfrac{\cos x(1 + 2 \cos x) - \sin x(-2\sin x)}{(1 + 2 \cos x)^2}$
$= \dfrac{2 \cos^2x + \cos x + 2\sin^2x}{(1 + 2\cos x)^2}$
$= \dfrac{2 + \cos x}{(1 + 2\cos x)^2}$
Vậy
$f'(x)= \dfrac{2 + \cos x}{(1 + 2\cos x)^2}$