I can see that by applying the product rule, all resulting expressions containing $(x-6)$ would become zero, hence leaving $(1)(x-1)(x-2)(x-3)(x-4)(x-5)(x-7)(x-8)(x-9)(x-10)$, which, at $x=6$, is $2880$. However, I'm not sure how to formally write out a concise method for finding this solution.

  • $\begingroup$ some summands of your derivative are zero if you plug in $$x=6$$ $\endgroup$ Oct 6, 2017 at 18:16
  • 1
    $\begingroup$ Your approach is correct. Your answer seems correct (it is wrong only if you have calculated it wrong). What is your question exactly? $\endgroup$ Oct 6, 2017 at 18:17
  • $\begingroup$ Is it incorrect to write out a solution in words only, without showing the mathematical analysis? If so, I'd like to be able to show mathematical analysis. $\endgroup$
    – Data
    Oct 6, 2017 at 18:21

2 Answers 2


$P(x)=\prod\limits_{k=1}^{10}(x-k)=(x-6)Q(x)$ where $Q(x)=\prod\limits_{k=1\\k\neq 6}^{10}(x-k)$


Thus $P'(6)=Q(6)+0=Q(6)=2880$

  • 2
    $\begingroup$ +1 This is easily the best answer. It may also help to note that $P'(6)=Q(6)+0=5!\cdot4!=2880$. I only mention this because I would hope that $5!\cdot4!$ would be an acceptable answer instead of actually computing $2880$ as an end result. $\endgroup$ Oct 6, 2017 at 18:36
  • $\begingroup$ @DanielW.Farlow yes, I did not detail this part since it did not seem to be an issue for the OP, but it is certainly the way to go to compute this value. $\endgroup$
    – zwim
    Oct 6, 2017 at 19:06
  • $\begingroup$ In fact, you don't even need $Q$ to be differentiable at $x=6$ to get the conclusion $P'(6) = Q(6)$ - with a direct calculation using the limit definition of derivative, you get it only assuming $Q$ is continuous at $x=6$. $\endgroup$ Oct 6, 2017 at 21:13
  • $\begingroup$ @DanielW.Farlow Why is $5!⋅4!$ preferable to $2880$? $\endgroup$
    – Data
    Oct 7, 2017 at 11:05
  • $\begingroup$ What a genius method!! $\endgroup$ Sep 10, 2020 at 19:26

Well, it depends on which properties you will take as given. Let's just say that from the usual product rule ($(fg)'=f'g+fg'$) you can obtain $$ (fgh)'=f'(gh)+f(gh)'=f'gh+fg'h+fgh',$$ and so on. In a general case, you could write $$\left(\prod_{k=1}^n f_k \right)'=\sum_{k=1}^n f_k' \cdot \prod_{j\not=k} f_j.$$

You may try from there (in your case, $f_k=x-k$ and n=10).

  • $\begingroup$ Could you explain the index “$j\neq k$”? $\endgroup$ Oct 9, 2017 at 23:41
  • 1
    $\begingroup$ Sure! I was actually being a little lazy. It should be $j$ from $1$ to $n$, and $j \neq k$. That is: you'll end up with an $n$-terms sum (because $k$ goes from $1$ to $n$), and in each of those terms you'll have a product of the derivative of $f_k$ and $n-1$ other factors, because you'll have $f_1 \cdot f_2 \cdot \ldots \cdot f_{k-1} \cdot f_{k+1} \cdot \ldots f_n$ ($f_1$ is omitted in the first term, $f_2$ in the second, and so on). Hope I didn't make it LESS clear. =S $\endgroup$ Oct 10, 2017 at 0:45
  • 1
    $\begingroup$ Let's just say $n=3$, then this will become: $$(f_1 \cdot f_2 \cdot f_3)' = f'_1 \cdot f_2 \cdot f_3 + f'_2 \cdot f_1 \cdot f_3 + f'_3 \cdot f_1 \cdot f_2. $$ $\endgroup$ Oct 10, 2017 at 0:46
  • $\begingroup$ Oh, now I understand! I had mixed up the argument of $\sum$. It's just shorthand for $$\sum_{k\in S} \left( f_k^\prime \prod_{j\in S\setminus\{k\}}f_j \right)$$ where $$S=\bigcup_{i=1}^{n}\{i\}$$ $\endgroup$ Oct 10, 2017 at 4:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.