Find nth derivative of 2x⋅ sin2x ⋅cos3x.

$\sin^2 x \cos^ x= \left ( \dfrac {1-\cos^2 x}{2} \right ) \left ( \dfrac {\cos 3 x + 3 \cos x }{4}\right ) $ $=\dfrac {1}{8} (\cos 3 x + 3 \cos x - \cos 3 x \cos 2 x - 3 \cos 2 x \cos x) $ $= \dfrac …