Câu hỏi 6

\(\begin{array}{l}y' = \left( {\cos 3x} \right)'.\sin 2x + \cos 3x\left( {\sin 2x} \right)' =  - \sin 3x.\left( {3x} \right)'.\sin 2x + \cos 3x.\cos 2x\left( {2x} \right)'\\=  - 3\sin 3x\sin 2x + 2\cos 3x\cos 2x\\ \Rightarrow y'\left( {\frac{\pi }{3}} \right) =  - 3\sin \pi .\sin \frac{{2\pi }}{3} + 2\cos \pi .\cos \frac{{2\pi }}{3} =  - 2.\left( { - \frac{1}{2}} \right) = 1\end{array}\)

Chọn D.