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Assume $f$ and $g$ are differentiable at $x$. Prove that $(fg)^{(n)} = \sum_{k=0}^n \binom{n}{k}f^{(k)}(x)g^{(n-k)}(x)$

I am assuming here $fg = f(x) g(x)$. Then we can prove this via induction. If $n = 0$ we have $1 = 1$ which is true. Now assume it is true for some $m$ we need to show it is possible for $m+1$. We have $(fg)^m = \sum_{k=0}^m f^{(k)}(x)g^{m-k}(x)$. How do we show that?

NoChance
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user19405892
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  • Take the derivative of the expression $(fg)^{m}$ – Laars Helenius Mar 04 '16 at 21:16
  • If $f^{(n)}$ represents the $n$-th derivative of $f$, then it is not clear to me why for $n=0$ you get $1=1$. Do you realize what $f^{(0)}$ means? It means $f$ (underived). Therefore for $n=0$ you get $(fg)^{(0)} = (fg) = fg = f^{(0)} g^{(0)}$. Judging by your post, you seem not to understand the difference between $f^m$ and $f^{(m)}$, otherwise I cannot understand why plenty of parantheses are missing. – Alex M. Mar 04 '16 at 21:18
  • This is sometimes called Leibniz's Rule/One Variable - Proof by induction is here:https://proofwiki.org/wiki/Leibniz's_Rule/One_Variable - Another is here:https://www.lucaswillems.com/en/articles/6/nth-derivative-of-product – NoChance Mar 04 '16 at 21:19
  • Binomial coefficients are still missing in the title – Laurent Duval Mar 04 '16 at 21:19
  • Now I fixed it thanks. – user19405892 Mar 04 '16 at 21:20
  • Is there an easier proof of it? – user19405892 Mar 04 '16 at 21:21
  • Similar: https://math.stackexchange.com/questions/593282/prove-leibniz-s-formula-for-the-nth-derivitive-of-a-product-by-induction – Hans Lundmark Feb 07 '18 at 17:59

3 Answers3

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\begin{align*} (fg)^{m+1}(x) & = \left(\sum_{k=0}^m \binom{m}{k} f^{(k)}(x)g^{(m-k)}(x)\right)'\\ & = \left(\sum_{k=0}^m \binom{m}{k} f^{(k+1)}(x)g^{(m-k)}(x)\right)+\left(\sum_{k=0}^m \binom{m}{k} f^{(k)}(x)g^{(m-k+1)}(x)\right)\\ & = f^{(m+1)}(x)g(x) + \left(\sum_{k=0}^{m-1} \binom{m}{k} f^{(k+1)}(x)g^{(m-k)}(x)\right)+\left(\sum_{k=1}^{m} \binom{m}{k} f^{(k)}(x)g^{(m-k+1)}(x)\right) + f(x)g^{(m+1)}(x)\\ & = f^{(m+1)}(x)g(x) + \left(\sum_{k=0}^{m-1} \binom{m}{k} f^{(k+1)}(x)g^{(m-k)}(x)\right)+\left(\sum_{k=0}^{m-1} \binom{m}{k+1} f^{(k+1)}(x)g^{(m-k)}(x)\right) + f(x)g^{(m+1)}(x). \end{align*}

And to finish, use the property $$\binom{m}{k-1}+ \binom{m}{k} = \binom{m+1}{k}.$$

Alex M.
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C. Dubussy
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This is one instance where non-standard analysis helps understanding. In crude words, you work with infinitesimals, and standardize results (with the operator $\operatorname{st}$) to "real" objects.

Indeed, this goes back to basics, as Leibniz is one of the Godfathers of infinitesimals. Basically, the formula is similar to that of the binomial theorem, and a little investment, stuff can be cast to polynomial-like calculus. Here, this goes like this (only giving a sketch) for the first level:

$$\frac{d(fg)}{dx}=\operatorname{st}\left(\frac{(f + \mathrm df)(g + \mathrm dg) - fg}{\mathrm dx}\right) = \operatorname{st}\left(\frac{fg + f \cdot \mathrm dg + g \cdot \mathrm df + \mathrm dg \cdot \mathrm df -fg}{\mathrm dx}\right) ={f}\frac{dg}{dx} + {g}\frac{df}{dx}\,.$$

The idea is that standardization simplifies quantities. From that, you can apply induction and known results on polynomials.

Laurent Duval
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    here, I don't see how infinitesimals are simpler than $\lim_{h \to 0}$, in my opinion it is the opposite – reuns Mar 04 '16 at 22:52
  • Here, to me, the complexity is about the same for the $1$-level product rule. Only you do not have to travel with limits, and some notations may be lighter, because products of infinitesimals vanish with standardization. But is you dare to perfom the same calculus with the $n$-order derivative, the notations and standardization end up with classical binomial foormulae. – Laurent Duval Mar 04 '16 at 23:06
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  • Clue 1: $(fg)^{(m+1)}=((fg)^{(m)})^{(1)}$
  • Clue 2: use induction hypothesis
  • Clue 3: Collect terms. Use binomial coefficient properties
user155673
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