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I am trying to understand the proof of the General Result for the Product Rule for Derivatives by reading this.

Relevant parts are as follows:

Basis for the induction $$ D_x \left({f_1 \left({x}\right) f_2 \left({x}\right)}\right) = D_x \left({f_1 \left({x}\right)}\right) f_2 \left({x}\right) + f_1 \left({x}\right) D_x \left({f_2 \left({x}\right)}\right) $$

Induction Hypothesis $$ D_x \left({\prod_{i=1}^k f_i \left({x}\right)}\right) = \sum_{i=1}^k \left({D_x \left({f_i \left({x}\right)}\right) \prod_{j \ne i} f_i \left({x}\right)}\right) $$

Induction Step $$ \begin{align} \tag{1} \kern-30pt D_x \left({\textstyle\prod\limits_{i=1}^{k+1} f_i \left({x}\right)}\right) &= D_x \left({\left({\textstyle\prod\limits_{i=1}^k f_i \left({x}\right)}\right) f_{k+1} \left({x}\right)}\right) \\ &= \tag{2} D_x \left({f_{k+1} \left({x}\right)}\right) \left({\textstyle\prod\limits_{i=1}^k f_i \left({x}\right)}\right) + D_x \left({\textstyle\prod\limits_{i=1}^k f_i \left({x}\right)}\right) f_{k+1} \left({x}\right) \\ &=\tag{3} D_x \left({f_{k+1} \left({x}\right)}\right) \left({\textstyle\prod\limits_{i=1}^k f_i \left({x}\right)}\right) + \left({\sum_{i=1}^k \left({D_x \left({f_i \left({x}\right)}\right) \textstyle\prod\limits_{j \ne i} f_i \left({x}\right)}\right)}\right) f_{k+1} \left({x}\right) \\ &= \tag{4} \sum_{i=1}^{k+1} \left({D_x \left({f_i \left({x}\right)}\right)\textstyle \prod\limits_{j \ne i} f_i \left({x}\right)}\right) \end{align} $$

Question

I am stuck at (3). How do I go from (3) to (4)? Specifically, what sort of algebraic manipulations need to be done and what are the motivations for doing those algebraic manipulations in order to arrive at (4)? To put it in another way, I would like to know that is the thought process that one goes through when simplifying (3) to (4).

Note: I hope someone can correct my LaTeX typesetting. I was under the impression that the align environment would automatically number the formulas I write. Thanks in advance.

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How's that (I had to make the products small to make things fit)? –  David Mitra Dec 11 '11 at 18:54
    
@DavidMitra: Thank you. –  Sara Dec 15 '11 at 3:19
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2 Answers 2

up vote 4 down vote accepted

Note that $$ D_x (f_{k+1}(x))\left( \prod_{i=1}^k f_i(x))\right) = \sum_{i=k+1}^{k+1} D_x(f_i(x)) \left( \prod_{j \neq i}^{k+1} f_i(x) \right) $$

and

$$ \left( \sum_{i=1}^k \left( D_x(f_i(x)) \prod_{j \neq i}^k f_i(x)\right) \right) f_{k+1}(x) = \sum_{i=1}^{k} D_x(f_i(x)) \left( \prod_{j \neq i}^{k+1} f_i(x) \right)$$, bringing the $f_{k+1}(x)$ term under the product. Adding these up, the result follows.

If still unclear, you can always try writing it out for small values of $k$.

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HINT $\ $ It is much clearer upon scaling, where it becomes additivity of logarithmic derivatives.

$$\rm\begin{eqnarray} D(f_{n+1}\cdots f_1)\ &=&\rm\ (D\:f_{n+1})\ f_n\cdots f_1\ +\ f_{n+1}\:D(f_n\cdots f_1) \\ \\ \Rightarrow\quad\rm\ \dfrac{D(f_{n+1}\cdots f_1)} {f_{n+1}\cdots f_1}\ &=&\rm\ \dfrac{D\:f_{n+1}}{f_{n+1}}\ +\ \dfrac{D(f_{n}\cdots f_1)} {f_{n}\cdots f_1} \\ \\ &=&\rm\ \dfrac{D\:f_{n+1}}{f_{n+1}}\ +\ \dfrac{D\:f_{n}}{f_{n}}\ +\ \dfrac{D(f_{n-1}\cdots f_1)} {f_{n-1}\cdots f_1} \\ \\ \\ &\cdots& \\ \\ \Rightarrow\qquad\rm\dfrac{D(f_{n+1}\cdots f_1)} {f_{n+1}\cdots f_1}\ &=&\rm\ \dfrac{D\:f_{n+1}}{f_{n+1}}\ +\ \dfrac{D\:f_{n}}{f_{n}}\ +\ \cdots\ +\ \dfrac{D\:f_1}{f_1} \end{eqnarray}$$

Multipying both sides above by $\rm\ f_{n+1}\cdots f_1\ $ yields the sought result.

Key is this. With $\rm\ L\: f\: :=\: D\:f/f\ $ we have $\rm\ L(f\:g)\ =\ L(f) + L(g)\:.\: $ The above proof is simply the inductive extension to a product of $\rm\:n+1\:$ terms, i.e. $\rm\ L(f_{n+1}\cdots f_1)\:=\: L(f_{n+1})+\:\cdots\:+L(f_1)\:.$ Multiplying this through by $\rm\:f_{n+1}\cdots f_1\:$ yields the sought derivative product rule (but, alas, obfuscates said key homomorphic property of the logarithmic derivative).

Perhaps you might also find helpful this hint from one of my prior posts.

logarithmic differentiation makes the n-ary generalization obvious:

$$\rm (abc)'\:=\ abc \; log(abc)'\: =\ abc \;(log\; a + log\; b + log\; c)'\: =\ abc \; \bigg(\frac{a'}{a} + \frac{b'}{b} + \frac{c'}{c}\bigg) $$

Obviously the same proof works for arbitrary length products yielding

$$\rm (abc\: \cdots\: f)'\: =\ \: abc\:\cdots f\:\ \bigg(\frac{a'}{a} + \frac{b'}{b} + \frac{c'}{c} +\:\cdots\:+ \frac{f\:'}{f}\bigg) $$

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:The answer looks interesting even if I am yet able to comprehend it, but thank you for answering anyway. –  Sara Dec 15 '11 at 3:20
    
@Sara I've expanded the answer. Please let me know if anything is still not clear. –  Bill Dubuque Dec 15 '11 at 4:18
    
My main problem is that my syllabus have not covered logarithmic differentiation yet. Sorry for being dense but there are also certain parts of the wikipedia link that I don't understand. I will get back to you once I have more time to study it in more detail. –  Sara Dec 19 '11 at 5:29
    
@Sara All you need to know is the product rule for derivatives - that's all I use in the first half of the post. While it is convenient to view it in terms of logs, as I explain in the second half, you can safely ignore that for now. The logarithmic derivative of $f$ is simply $f\:'/f$. That's equal to $(\log f)'$ but you don't need to employ that to see the additivity of logarithmic derivatives. –  Bill Dubuque Dec 19 '11 at 6:15
    
Thank you for not giving up on me. I can see the additivity of the logathrithmic derivatives now. Now I just need to understand why the derivative of $f$ is $f^\prime /f$. Anyway, thank you very much for your time. –  Sara Dec 25 '11 at 11:27
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