Biến tích thành tổng ta được :

\(\begin{array}{*{20}{l}} {y = c{\rm{osx}}{\rm{.\;cos2x\;}}{\rm{.cos3x}}}\\ {{\rm{\; = \;\;(cos3x}}{\rm{.\;cosx)}}{\rm{.\;cos2x\; = \;}}\frac{1}{2}{\rm{(cos4x\; + }}c{\rm{os2x)}}{\rm{.cos2x}}}\\ {{\rm{\; = \;\;}}\frac{1}{2}{\rm{cos4}}x.c{\rm{os2x}} + \frac{1}{2}.{\rm{co}}{{\rm{s}}^2}{\rm{2x\;\; = \;\;}}\frac{1}{2}.\frac{1}{2}\left( {{\rm{cos6x}} + c{\rm{os2x}}} \right) + \,\,\frac{1}{2}.\,\frac{{1 + c{\rm{os4x}}}}{2}}\\ { = \frac{1}{4}{\rm{cos6x\; + \;\;}}\frac{1}{4}{\rm{cos2x}} + \,\frac{1}{4} + \frac{1}{4}c{\rm{os4x}}}\\ { \Rightarrow y' = \frac{{ - 3}}{2}\sin 6x - \frac{1}{2}.\sin 2x - \sin 4x}\\ { \Rightarrow y = - 9\cos 6x - c{\rm{os2x\; - \;4cos\;4x\;}}} \end{array}\)

Chọn B