Answer
$$\frac{{dy}}{{dx}} = 4\cos 8x$$
Work Step by Step
$$\eqalign{
& y = \frac{1}{2}\sin 8x \cr
& {\text{differentiate with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{1}{2}\sin 8x} \right] \cr
& {\text{use multiple constant rule}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}\frac{d}{{dx}}\left[ {\sin 8x} \right] \cr
& {\text{using the chain rule for }}{D_x}\left( {\sin u} \right) = \cos u \cdot {D_x}\left( u \right).{\text{ then}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}\left( {\cos 8x} \right)\frac{d}{{dx}}\left[ {8x} \right] \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}\left( {\cos 8x} \right)\left( 8 \right) \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dx}} = 4\cos 8x \cr} $$
