\(y = x.\sin x + \sqrt {1 + {{\cos }^2}2x} \)
Giải chi tiết:
\(\begin{array}{l}y' = \sin x + x\cos x + \dfrac{{2\cos 2x.\left( { - 2\sin 2x} \right)}}{{2\sqrt {1 + {{\cos }^2}2x} }}\\\,\,\,\,\,\, = \sin x + x\cos x + \dfrac{{ - \sin 4x}}{{\sqrt {1 + {{\cos }^2}2x} }}\end{array}\)