# Leibniz's theorem to find nth derivatives

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 $2^n\frac{e^{2x}}{x}\sum\limits_{k=0}^n \left(-\frac12\right)^k \binom{n}{k}\frac{k!}{x^k}$... –  Guess who it is. 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).