$\cos^2 x \sin x = \dfrac {\cos 2x -1}{2} \sin x =\dfrac {1}{2} \left [ \sin x \cos 2x- \sin x \right ] = \dfrac {1}{2} \left [ \dfrac {\sin 3x + \sin (-x)}{2} \right ] - \dfrac {\sin x}{2}$ $=\dfrac {\sin 3x}{4}- \sin x \left ( \dfrac {1}{4}- \dfrac {1}{2} \right ) = \dfrac {\sin 3x}{4}+ \dfrac {\sin x}{4}= \dfrac {1}{4} (\sin x + \sin 3x) $ $2^x \cos^2 x \sin x = \dfrac {1}{4} \left [ 2^x \sin x+2^x \sin 3x \right ] \ 2^x=e^{x\log 2} \ 2^x \cos^2 x \sin x =\dfrac {1}{4} \left [ e^{x \log 2} \sin x+e^{x \log 2} \sin 3x \right ]$ $Now, \ n^{th} \ derivation \ of \ e^{ax} \sin (bx+c)= r^{n}e^{ax} \sin (bx+c+na)$ $where \ r=\sqrt{a^2+b^2} \ and \ \alpha=\tan^{-1}\left ( \dfrac {b}{a} \right ), r^n = \left ( a^2 + b^2 \right )^{\frac {n}{2}} \ $ $\therefore \ n^{th} \ derivation \ of \ e^{x\log 2}\sin x= (1+ (\log 2)^2)^{\frac {n}{2}} e^{x \log 2}\sin \left ( x+ ntan^{-1} \left ( \dfrac {1}{\log 2} \right ) \right ) $ $and \ n^{th} \ derivation \ of \ e^{x \log 2} \sin 3x = \left ( 3^2 + (\log 2)^2 \right )^{\frac {n}{2}} e^{x \log 2}\sin \left ( x+ ntan^{-1} \left ( \dfrac {3}{\log 2} \right ) \right )$ $\therefore , \ the \ n^{th} \ derivation \ of \ 2^x \cos^2 x \sin x$ $= \dfrac {1}{4}\left ( 1+ (\log 2)^{2} \right )^{\frac {n}{2}} 2^x \sin \left ( x + ntan^{-1} \left ( \dfrac {1}{\log 2} \right ) \right )+ \dfrac {1}{4} \left ( 9 + (\log 2)^2 \right )^{\frac {n}{2}}2^x \sin \left ( x + ntan^{-1} \left ( \dfrac {3}{\log 2} \right ) \right )$