## Derivative of Sum over Variable of derivative

I feel stupid for having to ask this, but does anybody have any idea how to handle $$\frac{d}{x}\sum_{n=k}^{g(x)}f(n,x)?$$ Example: $$\frac{d}{dx}\sum_{n=6}^{i^2+2i} \frac{1}{\ln{(i^2)}-\ln{\ln n}}.$$ If we were able to separate the summand into two functions, one with only $i$ as a variable, and one with only $n$ as a variable, this would be super simple. But it is not always the case. Any ideas?

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 I take it $i$ in your sum was meant to be $x$. Think about a simpler example, like ${d\over dx}(1+2+\dots+x)$, and you may see what's happening. – Gerry Myerson Mar 6 2011 at 23:03 @Gerry $1+2+\cdots+x$ is not a sum like the first one in this post. That is, unless $x$ is restricted to being an integer, and then I do not know the meaning of differentiation with respect to $x$. For $f(n,x)=n$ and $g(x)=x$, the resulting sum is $k+(k+1)+\cdots+\lfloor x\rfloor$. – Didier Piau Mar 7 2011 at 6:29 Didier, yes, I was hoping Alex would work out what you've put forward. What does one make of ${d\over dx}(1+2+\cdots+[x])$? – Gerry Myerson Mar 7 2011 at 6:34 @Gerry Then I should have kept quiet. Sorry. – Didier Piau Mar 7 2011 at 11:04 Didier, I didn't mean to criticize you, and you have nothing to apologize for. But where has Alex gone? – Gerry Myerson Mar 8 2011 at 12:10

This reminds me uncomfortably about a remark that Terry Tao made in this answer about the importance of teaching the derivative as a limit: at least one person tried to prove Fermat's Last Theorem by differentiating with respect to the exponent. It does not make sense.

The reason why it does not: the derivative is something that is properly defined for real (or complex) variables. If you're variable is an integer, derivatives will not make sense.

There may objects that are analogous to the derivative in some ways, but you cannot carelessly apply derivative rules.

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The easiest way to handle something like this particular example, is simply to expand and derive term by term. I see what you are saying. THank you

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I don't think that is what Gerry and Didier are saying. They are asking what, in your example, is the $x$ with respect to which you wish to derive. – Yemon Choi Mar 12 2011 at 1:36
@Alex, maybe you should think about the old joke which says that $x^2=x+x+\cdots+x$ with $x$ summands, so the derivative of $x^2$ is $1+1+\cdots+1$ with $x$ summands, which is $x$. – Gerry Myerson Mar 12 2011 at 4:06