除法定则

除法定则或商定则（英语：Quotient rule）是数学中关于两个函数的商的导数的一个计算定则。

若已知两个可导函数g,h及其导数g',h'，且h(x)≠0，则它们的商

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

的导数为：

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

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

设${\displaystyle f(x)={\tfrac {g(x)}{h(x)}}}$，${\displaystyle h(x)\neq 0}$，且${\displaystyle g}$和${\displaystyle h}$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

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

那么${\displaystyle g(x)=f(x)h(x){\mbox{ }}\,}$

${\displaystyle g'(x)=f'(x)h(x)+f(x)h'(x){\mbox{ }}\,}$

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

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

从复合函数求导法则推出

考虑恒等式，v≠0

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

那么：

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

于是：

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

展开，得：

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

最后，把分子和分母同除以4，便得：

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

参见

• 乘法定则
• 链式法则