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Product Rule的General Case

今天上機率時,老師有提到相乘要用什麼方式可以讓他們變成相加,老師提到了取 log 之後,我才突然想起 log 這東西真的很好用。
在微積分討論微分時,會提到一個公式稱作Product Rule(乘法法則):

f(x)g(x)=f(x)g(x)+f(x)g(x)f(x)g(x)=f^\prime(x)g(x)+f(x)g^\prime(x)

但是,假設當 k=1,2,,nk=1,\,2,\,\dots,\,n 時,fkf_k 是一個在區間內可微函數。如果要問你找 k=1nfk(x)\displaystyle\prod_{k=1}^nf_k(x) 的微分,用公式去暴力一個一個解開,一定很耗時間吧。
這時,我們就需要用到取 log 了。

Theorem (General Product Rule)

Let fif_i be a differentiable functions on an open interval for i=1,2,,ni=1,\,2,\,\dots,\,n. Then the derivative of k=1nfk(x)\displaystyle\prod_{k=1}^nf_k(x) is (k=1nfk(x))=(k=1nfk(x))(k=1nfk(x)fk(x)).\displaystyle\left(\prod_{k=1}^nf_k(x)\right)^\prime=\left(\prod_{k=1}^nf_k(x)\right)\cdot\left(\sum_{k=1}^n\frac{f^\prime_k(x)}{f_k(x)}\right).

接下來,我們會證明這個公式,也算是讓你推導說如何求出兩個以上函數相乘的微分。
Proof. It is easy to solve (fg)=fg+fg(fg)^\prime=f^\prime g+fg^\prime by product rule. However, it is difficult to solve when more than two function. We know that

ln(k=1nfk(x))=k=1nlnfk(x).\ln{\left(\prod_{k=1}^nf_k(x)\right)}=\sum_{k=1}^n\ln{f_k(x)}.

Differentiate both side, we obtain

[ln(k=1nfk(x))]=(k=1nlnfk(x))    1k=1nfk(x)(k=1nfk(x))=k=1nfk(x)fk(x)    (k=1nfk(x))=(k=1nfk(x))(k=1nfk(x)fk(x)).\begin{align*} \left[\ln{\left(\prod_{k=1}^nf_k(x)\right)}\right]^\prime=\left(\sum_{k=1}^n\ln{f_k(x)}\right)^\prime &\iff\frac{1}{\displaystyle\prod_{k=1}^nf_k(x)}\left(\prod_{k=1}^nf_k(x)\right)^\prime=\sum_{k=1}^n\frac{f_k^\prime(x)}{f_k(x)}\\ &\iff\left(\prod_{k=1}^nf_k(x)\right)^\prime=\left(\prod_{k=1}^nf_k(x)\right)\cdot\left(\sum_{k=1}^n\frac{f^\prime_k(x)}{f_k(x)}\right). \end{align*}

Hence

(k=1nfk(x))=(k=1nfk(x))(k=1nfk(x)fk(x)).\left(\prod_{k=1}^nf_k(x)\right)^\prime=\left(\prod_{k=1}^nf_k(x)\right)\cdot\left(\sum_{k=1}^n\frac{f^\prime_k(x)}{f_k(x)}\right).

當想到使用取對數時,這個廣義的 Product Rule 就會變得很好解,也算是突然想到可以試著找有沒有廣義的 Product Rule 公式然後去證明。

Product Rule的General Case
https://nickterty.vercel.app/posts/product_rule_generalize/
Author
NickTerty
Post on
2026-06-03
License
CC BY-NC-SA 4.0