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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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    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.