Proof of Product Rule for Derivatives using Proof by Induction

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

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

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 $k$.