# Leibniz's formula

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

• For better searchability (and to remove ignorable parts) please consider retyping that image using LaTeX code. Here's 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...) – Tobias Kienzler Sep 11 '12 at 11:48
• 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.) – Py42 Sep 11 '12 at 11:53
• Great spirit :-) Btw, if you want to make sure someone reads your reply, put @Username in it, that way they are notified. To mark a question as answered, 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 means something else, usually unsuitability). And, welcome to math.SE! – Tobias Kienzler Sep 11 '12 at 11:59
• 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. – S4M Sep 11 '12 at 11:59
• @S4M interesting case, I think for $p,q\not\in\mathbb N$ the mentioned cancellation won't occur – Tobias Kienzler Sep 11 '12 at 12:02

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 $<p-1$).