# 除法定则

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

${\displaystyle f'(x)={\frac {g'(x)h(x)-g(x)h'(x)}{[h(x)]^{2}}}}$

## 例子

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

## 证明

### 从牛顿差商推出

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

### 从乘积法则推出

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

### 从复合函数求导法则推出

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

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

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

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

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