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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.

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Unless incomplete gamma functions are part of your repertoire, you can't get anything simpler than $2^n\frac{e^{2x}}{x}\sum\limits_{k=0}^n \left(-\frac12\right)^k \binom{n}{k}\frac{k!}{x^k}$... – J. M. Nov 17 '11 at 16:14

Try rewriting the equation as $xy = e^{2x}$ and then repeatedly differenting both sides. Incidentally, old calculus texts (which are usually freely available at google-books) are a good source for this topic. An especially good treatment is given in Articles 56-58 (pages 64-73) of the following book. In particular, see Exercises 2 and 3 in Article 57 (p. 70).

An Elementary Text-Book on the Differential and Integral Calculus by William Holding Echols (1902)

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Perhaps not the best link? – AD. Nov 17 '11 at 17:37
@AD. Clicking on the book cover photo takes you to a page where page numbers can be inserted (push "Enter" after typing in the desired page number), plus a pull-down menu there allows access to the various chapters (and gives their page ranges). Of course, the .pdf file can simply be downloaded. Finally, a URL for a specific page 'kmn' can be obtained by sticking '&pg=PAkmn' at the end of the "initial URL" that I gave. Thus, click on the URL I gave and then paste '&pg=PA70' (without quotes) at the end of it where it appears in your internet browser window, then push "Enter". – Dave L. Renfro Nov 17 '11 at 19:02
I'm not sure what could be suboptimal about the URL I gave, but it occurred to me just now that perhaps some books digitized by google and freely available in the U.S. might not be accessible in some other countries. With that in mind, here's another link to the Echols book: – Dave L. Renfro Nov 18 '11 at 15:12
Thanks Dave, yes that is probably the case. – AD. Nov 21 '11 at 20:22
@DaveL.Renfro Yes haha , 3 years later and it is still not available here . EDIT: It seems that it is not available even in US now – A Googler Mar 20 '15 at 20:01

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