The question is to find the $n$th derivative of $f(x) = (e^{2x})/x$
So what I've done so far is work out derivatives of and
Which are: \begin{align*} u &= x^{-1},& v &= e^{2x},\\ u' &= -(x^{-2}),& v' &= 2e^{2x},\\ u'' &= 2x^{-3}, & v'' &= 4e^{2x},\\ u''' &= -6x^{-4}& v''' &= 8e^{2x},\\ &\vdots&&\vdots\\ u^{(n)} &= (-1)^{n}(n!)x^{-(n+1)}. &v^{(n)} &= 2^{n}e^{2x}.\end{align*}
And then plugging this into the Theorem you get:
$x^{-1}(2^{n}e^{2x})) + ((n!/(n-1)!)(-x^{-2})2e^{2x})+\cdots + ((-1)^{n}(n!)x^{-(n+1)}e^{2x}) $
My question is... have I done this right? Is this the sort of answer I'm looking for? Should i display more terms? There doesn't seem to be a pattern or anything of the like.