# Leibniz product rule

$(x\cdot x)' = x' \cdot x' = 1 \cdot 1 = 1$

$F(x)' = \lim_{t \to 0} \frac{F(x+t) - F(x)}{t}$

$limt → 0 \frac{F(x+t) - F(x)}{t} = limt → 0 \frac{ f(x+t)g(x+t) f(x)g(x) }{t}$

\begin{aligned} &\lim_{t \to 0} \frac{ f(x+t)g(x+t) - f(x)g(x) }{t} \\=& \lim_{t \to 0} \frac{ f(x)g(x+t) - f(x)g(x+t) + f(x+t)g(x+t) - f(x)g(x) }{t} \\=& \lim_{t \to 0} \frac{ (f(x)g(x+t) - f(x)g(x)) + (f(x+t)g(x+t) - f(x)g(x+t)) }{t} \\=& \lim_{t \to 0} \frac{ f(x)[g(x+t) - g(x)] + [f(x+t) - f(x)]g(x+t) }{t} \\=& \lim_{t \to 0} \frac{f(x)[g(x+t) - g(x)]}{t} + \lim_{t \to 0} \frac{[f(x+t) - f(x)]g(x+t)}{t} \\=& f(x) \cdot \lim_{t \to 0} \frac{[g(x+t) - g(x)]}{t} + \lim_{t \to 0} \frac{[f(x+t) - f(x)]g(x+t)}{t} \\=& f(x) \cdot \lim_{t \to 0} \frac{[g(x+t) - g(x)]}{t} + \lim_{t \to 0} \frac{[f(x+t) - f(x)]}{t} \cdot \lim_{t \to 0}g(x+t) \\=& f(x)g(x)' + f(x)'g(x) \end{aligned}

NOTE: $$\lim_{t \to 0}g(x+t) = g(x)$$ follows from the fact that differentiable functions are continuous.

## 1. Generalizations

$fgh' + (fg)'h \\= fgh' + (fg' + f'g)h \\= fgh' + fg'h + f'gh$

$(f1f2 ⋯ fk fk+1)' \\= (g fk+1)' \\= g' fk+1 g fk+1'$

\begin{aligned} \prod_{i=1}^{k} f_i =& f_1f_2 \cdots f_k \\=& (f_1'f_2 \cdots f_k) \;\;\; 第 1 項\\ +& (f_1f_2' \cdots f_k) \;\;\; 第 2 項\\ +& \cdots \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 第 i 項\\ +& (f_1f_2 \cdots f_k') \;\;\; 第 k 項\\ \end{aligned}

## 2. 運用 Generalizations

$Id(x)'=1 \\= n \cdot f^{n-1} \cdot f'$

$f' = \frac{1}{n \cdot f^{n-1}} \\= \frac{1}{n} \cdot f^{1-n} \\= \frac{1}{n} \cdot (x^\frac{1}{n})^{1-n} \\= \frac{1}{n} \cdot x^{\frac{1}{n}-1}$

Date: 2022-06-30 Thu 00:00