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) $ $= \dfrac {1}{8} \left ( \cos 3 x +3 \cos x - \dfrac {\cos (3x+2x)+ \cos (3x-2x)}{2} -3 \dfrac {\cos (2x+x)+ \cos (2x-x)}{2} \right )$ $=\dfrac {1}{16} (2 \cos 3x+6 \cos x - \cos 5 x - \cos x -3 \cos 3 x - 3 \cos x) $ $ = \dfrac {1}{16} (2 \cos x - \cos 3x - \cos 5 x)$ $2^x \sin^2 x \cos^3 x = \dfrac {2^x}{16} (2 \cos x - \cos 3 x - \cos 5 x), \ and \ 2^x = e^{x \ln 2} $ $=\dfrac {1}{8} \left [ e^{x\ln 2} \cos x \right ] - \dfrac {1}{16} \left [ e^{x\ln 2} \cos 3 x \right ]- \dfrac {1}{16} \left [ e^{x \ln 2} \cos 5x \right ]$ $Now, \dfrac {d^n}{dx^n} \left ( e^{ax}\cos (bx+c) \right )= r^n e^{ax}\cos (bx+c+n\theta)$ $where , \ r= \sqrt{a^2+b^2}, \theta = \tan^{-1} \left ( \dfrac {b}{a} \right )$ $\therefore, \ \dfrac {d^n}{dx^n} (e^{x\ln 2}\cos (x)) = r_1^n e^{x \ln 2 } \cos (x+n\theta_1) = f_1 (x) \ \cdots \ \cdots (A)$ $where \ r_1 = \sqrt{1+(\ln 2)^2}, \theta_1 = \tan^{-1} \left ( \dfrac {1}{\ln 2} \right ) $ $\dfrac {d^n}{dx^n} (e^{x \ln 2} \cos (3x)) = r_2^n e^{x\ln 2}\cos (x+n\theta_2)= f_2 (x)\ \cdots \ \cdots (B)$ $where, \ r_2 = \sqrt{9+(\ln 2)^2}, \theta_2 = \tan^{-1} \left ( \dfrac {3}{\ln 2} \right ) $ $\dfrac {d^n}{dx^n} (e^{x\ln 2} \cos (5x)) = r_3^n e^{x \ln 2}\cos (x+n\theta_3)= f_3 (x)\ \cdots \ \cdots (C)$ $where, \ r_3=\sqrt{25+(\ln 2)^2} , \theta_3 = \tan^{-1} \left ( \dfrac {5}{\ln 2} \right )$ $\therefore \dfrac {d^n}{dx^n} [2^x \sin x \cos^2 x] = \dfrac {1}{8} f_1 (x) - \dfrac {1}{16}f_2 (x) - \dfrac {1}{16}f_3 (x)$ Where $f_1, f_2, f_3$ are as defined in (A), (B), (C)