y$ =sin⁡x \; sin⁡2x\; sin⁡3x \\ \; \\ \therefore y=\dfrac{1}{2} (2 \;sin⁡x\; sin⁡2x ) \; sin⁡3x=\dfrac{1}{2} [(cos⁡x-cos⁡3x ) \; sin⁡3x ] \\ \;\;\ldots(\because cos⁡C-cos⁡D=2 sin⁡(\dfrac{C+D}{2}) sin⁡(\dfrac{C-D}{2}) \\ \; \\ \; \\ \therefore y= \dfrac{1}{2} \; [cos⁡x \; sin⁡3x-cos⁡3x\; sin⁡3x ] \;= \; \dfrac{1}{4}[2 cos⁡x \; sin⁡3x-2 cos⁡3x \; sin⁡3x] \\ \; \\ \; \\ \therefore y=\dfrac{1}{4} [sin⁡4x+sin⁡2x-sin⁡6x ] \\ \; \\ \ \ \ \ \ \ldots (\because sin⁡C+sin⁡D=2 sin(\dfrac{C+D}{2})⁡ cos⁡(\dfrac{C-D}{2}) \; \;\& \;sin⁡6x=2 sin⁡3x cos⁡3x ) \\ \; \\ \; \\ Differentiating \; y \; w. r. t. \; x \; ‘n’ \; times, \\ \; \\ \; \\ \therefore y_n=\dfrac{1}{4} [4^n sin⁡(4x+n \dfrac{\pi}{2})-2^n sin⁡(2x+n \dfrac{\pi}{2})-6^n sin⁡(6x+n\dfrac{\pi}{2}) \\ \; \\ \; \\ \ \ \ \ \ \ \ldots ( \because y=sin⁡(bx+c), \;then \;y_n=b^n \; sin⁡(bx+c+n \dfrac{\pi}{2}) ) $