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The n-th derivative of the product of three functions can be expressed using the multinomial pattern.

The formula is given in the answer to this question: The $n$-th derivative of the product of three functions

How can you prove that the formula holds for each successive derivation by using induction?

Edit:

here is my attempt at proving the formula - up to the point where i got stuck

gt6989b
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user1505
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    What have you tried? What is the base case? Does it fit? Can you show your attempt at the inductive step so we know where you are getting stuck? – gt6989b Aug 03 '20 at 13:18
  • @gt6989b I have attached an image of my attempt in my question under "Edit:" – user1505 Aug 03 '20 at 13:48
  • great, now try to regroup the terms to show this is the same as $$\sum_{i+j+k=m+1} u^i v^j z^k$$ Note in particular that in all of your terms the coefficients now add to $m+1$... It may help to rewrite the sum as $\sum_{i=0}^? \sum_{j=0}^? ... u^iv^jz^{m-i-j}$ – gt6989b Aug 03 '20 at 23:37

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If we introduce the operator notation $$D_f(fgh):=f'gh$$ and similar for $g,h$, we have the formal equivalence

$$(fgh)'=(D_f+D_g+D_h)(fgh)=f'gh+fg'h+fgh'.$$

We can iterate,

$$(fgh)^{(n)}=(D_f+D_g+D_h)^n(fgh)$$ and use the multinomial theorem.

  • Thank you, but do you mind clarifying on how to proceed with the proof after iterating using the operator notation. Is there anyway to do it with only the multinomial theorem and induction? – user1505 Aug 03 '20 at 14:03
  • @user1505: it suffices you show that $(D_f+D_g+D_h)^n$ has the same effect as the expansion of $(D_f+D_g+D_h)^n$. This can be done by induction (I already showed the base case.) –  Aug 03 '20 at 14:08