Proof of Product Rule for Derivatives using Proof by Induction

Question

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.

Answer 1

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 $k$.

Answer 2

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

$$\rm\begin{eqnarray}\rm D(f_{n+1}\cdots f_1)\ &=&\rm\ (D\:f_{n+1})\ f_n\cdots f_1\ +\ f_{n+1}\:D(f_n\cdots f_1) \\[.7em] \Rightarrow\ \ \rm\ \dfrac{D(f_{n+1}\cdots f_1)} {f_{n+1}\cdots f_1}\ &=&\rm\ \dfrac{D\:f_{n+1}}{f_{n+1}}\ +\ \dfrac{D(f_{n}\cdots f_1)} {f_{n}\cdots f_1}\\[.6em] &=&\rm\ \dfrac{D\:f_{n+1}}{f_{n+1}}\ +\ \dfrac{D\:f_{n}}{f_{n}}\ +\ \dfrac{D(f_{n-1}\cdots f_1)} {f_{n-1}\cdots f_1} \\ &\vdots&\\ \Rightarrow\ \ \ \ \rm\dfrac{D(f_{n+1}\cdots f_1)} {f_{n+1}\cdots f_1}\ &=&\rm\ \dfrac{D\:f_{n+1}}{f_{n+1}}\ +\ \dfrac{D\:f_{n}}{f_{n}}\ +\ \cdots\ +\ \dfrac{D\:f_1}{f_1} \end{eqnarray}\qquad\qquad$$

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:

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

Obviously the same proof works for arbitrary length products yielding

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