# 除法定则

$\int_M \mathrm{d}\omega = \oint_{\partial M} \omega$

$f(x) = \frac{g(x)}{h(x)}$

$\frac{d}{dx}f(x) = f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{{h(x)}^2}$

## 例子

$(4x - 2)/(x^2 + 1)$的导数为：
 $\frac{d}{dx} \frac{(4x - 2)}{x^2 + 1}$ $=\frac{(x^2 + 1)(4) - (4x - 2)(2x)}{(x^2 + 1)^2}$ $=\frac{(4x^2 + 4) - (8x^2 - 4x)}{(x^2 + 1)^2}$ $=\frac{-4x^2 + 4x + 4}{(x^2 + 1)^2}$
$f(x) = \frac{2x^2}{x^3}$的导数为：
 $f'(x)\,$ $=\frac {\left(4x \cdot x^3 \right) - \left(2x^2 \cdot 3x^2 \right)} {\left(x^3\right)^2}$ $=\frac{4x^4 - 6x^4}{x^6}$ $=\frac{-2x^4}{x^6}$ $=-\frac{2}{x^2}$

## 证明

### 从牛顿差商推出

$f(x) = g(x)/h(x)$$h(x)\neq 0$，且$g$$h$均可导。
$f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} = \lim_{\Delta x \to 0} \frac{\frac{g(x + \Delta x)}{h(x + \Delta x)} - \frac{g(x)}{h(x)}}{\Delta x}$
$= \lim_{\Delta x \to 0} \frac{1}{\Delta x} \left( \frac{g(x+\Delta x)h(x)-g(x)h(x+\Delta x)}{h(x)h(x+\Delta x)} \right)$
$= \lim_{\Delta x \to 0} \frac{1}{\Delta x} \left( \frac{(g(x+\Delta x)h(x)-g(x)h(x))-(g(x)h(x+\Delta x)-g(x)h(x))}{h(x)h(x+\Delta x)} \right)$
$= \lim_{\Delta x \to 0} \frac{1}{\Delta x} \left( \frac{h(x)(g(x+\Delta x)-g(x))-g(x)(h(x+\Delta x)-h(x))}{h(x)h(x+\Delta x)} \right)$
$= \lim_{\Delta x \to 0} \frac{\frac{g(x+\Delta x)-g(x)}{\Delta x}h(x)-g(x)\frac{h(x+\Delta x)-h(x)}{\Delta x}}{h(x)h(x+\Delta x)}$
$= \frac{\lim_{\Delta x \to 0} \left(\frac{g(x+\Delta x)-g(x)}{\Delta x}\right)h(x) - g(x) \lim_{\Delta x \to 0} \left(\frac{h(x+\Delta x)-h(x)}{\Delta x}\right)}{h(x)h(\lim_{\Delta x \to 0} (x+\Delta x))}$
$= \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}$

### 从乘积法则推出

$g'(x)=f'(x)h(x) + f(x)h'(x)\mbox{ } \,$
$f'(x)=\frac{g'(x) - f(x)h'(x)}{h(x)} = \frac{g'(x) - \frac{g(x)}{h(x)}\cdot h'(x)}{h(x)}$
$f'(x)=\frac{g'(x)h(x) - g(x)h'(x)}{\left(h(x)\right)^2}$

### 从复合函数求导法则推出

$\frac{u}{v}\; =\; \frac{1}{4}\left[ \left( u+\frac{1}{v} \right)^{2}-\; \left( u-\frac{1}{v} \right)^{2} \right]$

$\frac{d\left( \frac{u}{v} \right)}{dx}\; =\; \frac{d}{dx}\frac{1}{4}\left[ \left( u+\frac{1}{v} \right)^{2}-\; \left( u-\frac{1}{v} \right)^{2} \right]$

$\frac{d\left( \frac{u}{v} \right)}{dx}\; =\; \frac{1}{4}\left[ 2\left( u+\frac{1}{v} \right)\left( \frac{du}{dx}-\frac{dv}{v^{2}dx} \right)-\; 2\left( u-\frac{1}{v} \right)\left( \frac{du}{dx}+\frac{dv}{v^{2}dx} \right) \right]$

$\frac{d\left( \frac{u}{v} \right)}{dx}\; =\; \frac{1}{4}\left[ \frac{4}{v}\frac{du}{dx}-\frac{4u}{v^{2}}\frac{dv}{dx} \right]$

$\frac{d\left( \frac{u}{v} \right)}{dx}\; =\; \frac{\left[ v\frac{du}{dx}-u\frac{dv}{dx} \right]}{v^{2}}$