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(disregard "part a" mention) According to the solution all terms in the Lebniz formula but one cancel out. Could someone please illustrate this?

Thanks in advance :)

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    For better searchability (and to remove ignorable parts) please consider retyping that image using LaTeX code. [Here's](http://math.stackexchange.com/editing-help#latex) how to do it. It would also be nice if you added some more information on what you tried so far and where you got stuck, so you can get an answer that actually helps you _understanding_ the solution instead of just imitating it ;-) (**TLDR version**: "Could someone please illustrate this" has this "please do my homework for me which I haven't even started to work at" taste...)2012-09-11
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    I have no homework (or even school), I'm just doing this for fun. Anyway I posted an answer, but I'm not sure how to mark a question as closed. (P.S. I included the solution solely because I wanted to understand it.)2012-09-11
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    Great spirit :-) Btw, if you want to make sure someone reads your reply, put @Username in it, that way they are [notified](http://math.stackexchange.com/editing-help#comment-formatting). To mark a question as [answered](http://meta.stackexchange.com/q/5234/146482), click on the checkmark next to the answer - however, to give other users a chance to post something helpful as well, you can only accept self-answers after your question turns two days old. (Also, [closing a question](http://meta.stackexchange.com/q/10582/146482) means something else, usually unsuitability). And, welcome to math.SE!2012-09-11
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    Can you confirm $p$ and $q$ are integer? I know $p+q$ is an integer but it doesnt mean $p$ and $q$ are integer as well.2012-09-11
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    @S4M interesting case, I think for $p,q\not\in\mathbb N$ the mentioned cancellation won't occur2012-09-11
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    @TobiasKienzler Thanks for the tip. About the second thing - it would be kind of weird to accept my own answer, I'll just leave it as is in this case.2012-09-11
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    @S4M Unfortunately that was not stated in the question.2012-09-11
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    @Py42 no, you _should_ accept your own answer in two days. That way there is a green box around the answer-counter in the overview so folks who have a similar question know they may find something useful here, and so those who might be able to answer it, too, can spend their (probably limited) time answering truly unanswered questions.2012-09-11
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    Here's another version of the Leibniz rule: $\displaystyle \frac{\partial^n}{\partial x_1\cdots\partial x_n} (uv) = \sum_S \frac{\partial^{|S|} u}{\prod_{i\in S}\partial x_i} \frac{\partial^{n-|S|} v}{\prod_{i\not\in S}\partial x_i}$. The one you cite is the special case in which the $n$ variables $x_1,\ldots,x_n$ are all the same variable.2012-09-12

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Oh I understand now:

Because $u^\left(p\right)=p!$ and $v^\left(q\right)=q!$,

whenever there is a term with a derivative higher than p (>p), you are actually differentiating p factorial (when p+1) or 0 (when $>p+1$). Whenever there is a term with a derivative lower than p, you are differentiating q factorial (when p-1) or 0 (when $).

So there is only one instance where you are not differentiating a constant.