Given : $y=e^{2x} \times \cos^2x \times \sin x$ and I have to find it's $n^{th}$ derivative
I have managed to break $y$ down to:
$y= \frac{1}{4} \times (e^{2x}.\sin 3x + e^{2x}.\sin x)$
But I don't know how to apply Leibnitz Rule in cases like $(e^{ax} \times \sin bx)$.
Please guide
Hint: Use the representation \begin{align*} \cos x=\frac{e^{ix}+e^{-ix}}{2}\qquad\qquad \sin x=\frac{e^{ix}-e^{-ix}}{2i} \end{align*} expand \begin{align*} y(x)=e^{2x}\cos^2(x)\sin(x) \end{align*} and differentiate.
Alternative:
Recall the general Leibniz rule and consider \begin{align*} \frac{d^n}{dx^n}\left(e^{ax}\sin(bx)\right)=\sum_{k=0}^n\binom{n}{k}\frac{d^k}{dx^k}\left(e^{ax}\right)\frac{d^{n-k}}{dx^{n-k}}\left(\sin(bx)\right) \end{align*}
$$4y=2e^{2x}(1+\cos2x)\sin x=2e^{2x}\sin x+e^{2x}[\sin3x-\sin x]=e^{2x}\sin3x+e^{2x}\sin x$$
$e^{ax}\sin bx=$imaginary part of $$e^{x(a+ib)}$$
Now see $100$-th derivative of the function $f(x)=e^{x}\cos(x)$
$y= \dfrac{1}{4} (e^{2x}\sin 3x + e^{2x}\sin x)=\dfrac{1}{4}{\bf Im}\left(e^{(2+3i)x} + e^{(2+i)x}\right)$ so $$y^{(n)}= \dfrac{1}{4}{\bf Im}\left((2+3i)^ne^{(2+3i)x} + (2+i)^ne^{(2+i)x}\right)$$
