$y =sin x\ sin 2x\ sin 3x \cdots (Given)$ $\ y = sin 2x\ sin 3x\ sin x \times \dfrac {2} 2$ $=\dfrac {1} 2 sin 3x. (2sin 2x\sin x )$ $=\dfrac {1} 2 sin 3x. (cos x - cos 3x ) \times \dfrac {2} 2$ $=\dfrac {1} 4 (2sin 3x\ cos x - 2sin 3x\cos 3x )$ $=\dfrac {1} 4 [sin 4x\ + sin2x\ - sin 6x ]$ $Now,\ taking\ n^{th} order\ derivative,\ we\ get$ $y_n=\dfrac {1} 4 \bigg[4^n sin \bigg (4x + \dfrac {n \Pi} 2 \bigg )+ 2^n sin \bigg (2x + \dfrac {n \Pi} 2 \bigg) - 6^n sin \bigg (6x + \dfrac {n \Pi} 2 \bigg) \bigg]$