Question concerning an example of higher derivatives and binomes

Question

I got this exercise form OCW 18.03SC - problem 1G-5b:

What is the solution of the following derivative?:

$$\dfrac{d^{p+q}}{dx^{p+q}}x^p(1+x)^q$$

I used Leibniz' formula and the only non-zero term is $$\binom{p+q}{q}p!q!=\frac{(p+q)!}{q!(p+q-q)!}p!q!=\frac{(p+q)!}{p!q!}p!q!=(p+q)!$$ Is this right?

In the solutions there is the following: $\binom{n}{p}p!q!$
I assume $n$ to be $p+q$, but is this a proper solution because there is this $\binom{n}{p}$ instead of $\binom{n}{q}$?

I have to think that's a typo; otherwise, what is $n$ supposed to be? Though, using Leibniz here seems like overkill: expanding that function gives all but one term with degree less than $p+q$. — Semiclassical, Aug 13 '14 at 19:37
Your approach is correct. You are using this formula and the product rule. — Mhenni Benghorbal, Aug 13 '14 at 20:46

score 1 · Accepted Answer · answered Aug 13 '14 at 20:07

$$\dfrac{d^{p+q}}{dx^{p+q}}x^p(1+x)^q=\dfrac{d^{p+q}}{dx^{p+q}}x^p\sum_{k=0}^q {q\choose k}x^k=\dfrac{d^{p+q}}{dx^{p+q}}\sum_{k=0}^q {q\choose k}x^{k+p}=\dfrac{d^{p+q}}{dx^{p+q}}\left(\sum_{k=0}^{q-1} {q\choose k}x^{k+p}+x^{p+q}\right)=(p+q)!$$

Since $\frac{d^n}{dx^n}x^k=0$ if $k<n$.

Question concerning an example of higher derivatives and binomes

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