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I am trying to find the second derivative of a product series $\prod_{i=1}^{n} f_i(x) $

I found an answer from Arturo Magidin in First derivative product series, which is pretty convincing and easy to understand.

I tried to follow the same steps to get the second derivative and this is what I got:

$\sum_{i=1}^{n} f_i^{''}(x) \prod_{j=1, j\neq i}^{n} f_j(x)$ + $\sum_{i=1}^{n} f_i^{'}(x) \sum_{j=1, j\neq i}^{n} f_j^{'}(x) \prod_{k=1, k\neq j}^{n} f_k(x)$

My question is, is my derivation correct? If not, where am I making a mistake? And is there any way to simplify this more?

Thanks in advance,

Arturo Magidin
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Rawan
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    Welcome to MSE. Arturo is a male name. – José Carlos Santos Jul 07 '19 at 07:43
  • Look at yoyo's comment on the linked question, which gives a compact formula for the $n$th derivative. – saulspatz Jul 07 '19 at 07:57
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    Tangential: If you're unsure as to the person's title, it's probably best not to use "Mr/Ms" and simply refer to the person by name. – parsiad Jul 07 '19 at 07:59
  • Heh; “Arturo” is most definitely a male name; it’s just Spanish or “Arthur”. – Arturo Magidin Jul 08 '19 at 11:04
  • Yes, you can simplify it. If you think about each of the summands after the first derivative, taking the derivative of that will give you a sum in which one term has a second derivative, and all other terms have two first derivatives. So your first summand takes care of all the second derivatives that occur. The second summand can be simplified because you will get two summands for every pair of first derivatives, so you can rewrite the second summand as $2\left(\sum_{1\leq i\lt j\leq n}f_i’(x)f_j’(x)\prod_{k=1,k\neq i,k\neq j}^n f_k(x)\right)$. – Arturo Magidin Jul 08 '19 at 11:21
  • However, your formula is almost but not quite right: your final $k$ must also be different from $i$, not just from $j$. – Arturo Magidin Jul 08 '19 at 11:22

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