Answer
$$\frac{{dy}}{{dx}} = 2\sin 2x$$
Work Step by Step
$$\eqalign{
& y = - \cos 2x + \cos \frac{\pi }{6} \cr
& {\text{differentiate with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ { - \cos 2x + \cos \frac{\pi }{6}} \right] \cr
& {\text{use sum rule for derivatives}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ { - \cos 2x} \right] + \frac{d}{{dx}}\left[ {\cos \frac{\pi }{6}} \right] \cr
& \frac{{dy}}{{dx}} = - \frac{d}{{dx}}\left[ {\cos 2x} \right] + \frac{d}{{dx}}\left[ {\cos \frac{\pi }{6}} \right] \cr
& {\text{using the chain rule for }}{D_x}\left( {\cos u} \right) = - \sin u \cdot {D_x}\left( u \right).{\text{ and }}\frac{d}{{dx}}\left[ k \right] = 0 \cr
& \frac{{dy}}{{dx}} = - \left( { - \sin 2x} \right)\frac{d}{{dx}}\left[ {2x} \right] + 0 \cr
& \frac{{dy}}{{dx}} = - \left( { - \sin 2x} \right)\left( 2 \right) + 0 \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dx}} = 2\sin 2x \cr} $$
