数学永恒定理/乘积、商、复合法则

乘积法则、商法则和复合(或链)法则微积分中最基本的法则或公式。对于函数 $u(x)$ 和 $v(x)$，这些法则为：

1. 乘积法则: $D_{x}(uv)=v\cdot D_{x}u+u\cdot D_{x}v$
2. 商法则: $D_{x}({\frac {u}{v}})={\frac {1}{v^{2}}}(v\cdot D_{x}u-u\cdot D_{x}v)$
3. 复合法则: $D_{x}v=D_{y}v\cdot D_{x}y$

证明

乘积法则

$D_{x}(uv)$ $=\lim _{\Delta x\to 0}{\frac {\Delta (uv)}{\Delta x}}$ $=\lim _{\Delta x\to 0}{\frac {u(x+\Delta x)\cdot v(x+\Delta x)-u(x)\cdot v(x)}{\Delta x}}$ $=\lim _{\Delta x\to 0}{\frac {u(x+\Delta x)\cdot v(x+\Delta x)-u(x)\cdot v(x+\Delta x)+u(x)\cdot v(x+\Delta x)-u(x)\cdot v(x)}{\Delta x}}$ $=\lim _{\Delta x\to 0}{\frac {[u(x+\Delta x)-u(x)]\cdot v(x+\Delta x)+u(x)\cdot [v(x+\Delta x)-v(x)]}{\Delta x}}$ $=\lim _{\Delta x\to 0}{\frac {(\Delta u)\cdot v(x+\Delta x)+u(x)\cdot (\Delta v)}{\Delta x}}$ $=\lim _{\Delta x\to 0}[{\frac {\Delta u}{\Delta x}}\cdot v(x+\Delta x)+{\frac {\Delta v}{\Delta x}}\cdot u(x)]$ $=\lim _{\Delta x\to 0}[D_{x}u\cdot v(x+\Delta x)+u\cdot D_{x}v]$ $=v\cdot D_{x}u+u\cdot D_{x}v$

$\therefore D_{x}(uv)=v\cdot D_{x}u+u\cdot D_{x}v$ [已证明]

商法则

$D_{x}({\frac {u}{v}})$ $=\lim _{\Delta x\to 0}{\frac {\Delta ({\frac {u}{v}})}{\Delta x}}$ $=\lim _{\Delta x\to 0}{\frac {1}{\Delta x}}\cdot [{\frac {u(x+\Delta x)}{v(x+\Delta x)}}-{\frac {u(x)}{v(x)}}]$ $=\lim _{\Delta x\to 0}{\frac {1}{\Delta x}}\cdot [{\frac {u(x+\Delta x)\cdot v(x)-u(x)\cdot v(x+\Delta x)}{v(x)\cdot v(x+\Delta x)}}]$ $=\lim _{\Delta x\to 0}{\frac {1}{\Delta x}}\cdot [{\frac {u(x+\Delta x)\cdot v(x)-v(x)\cdot u(x)+v(x)\cdot u(x)-u(x)\cdot v(x+\Delta x)}{v(x)\cdot v(x+\Delta x)}}]$ $=\lim _{\Delta x\to 0}{\frac {1}{\Delta x}}\cdot [{\frac {v(x)\cdot (u(x+\Delta x)-u(x))+u(x)\cdot (v(x)-v(x+\Delta x)}{v(x)\cdot v(x+\Delta x)}}]$ $=\lim _{\Delta x\to 0}{\frac {1}{\Delta x}}\cdot [{\frac {v(x)\cdot (\Delta u)+u(x)\cdot (\Delta v)}{v(x)\cdot v(x+\Delta x)}}]$ $=\lim _{\Delta x\to 0}{\frac {{\frac {v(x)\cdot (\Delta u)}{\Delta x}}+{\frac {u(x)\cdot (\Delta v)}{\Delta x}}}{v(x)\cdot v(x+\Delta x)}}$ $=\lim _{\Delta x\to 0}{\frac {v(x)\cdot {\frac {\Delta u}{\Delta x}}+u(x)\cdot {\frac {\Delta v}{\Delta x}}}{v(x)\cdot v(x+\Delta x)}}$ $=\lim _{\Delta x\to 0}{\frac {v(x)\cdot D_{x}u+u(x)\cdot D_{x}v}{v(x)\cdot v(x+\Delta x)}}$ $={\frac {v(x)\cdot D_{x}u+u(x)\cdot D_{x}v}{v(x)\cdot v(x)}}$ $={\frac {v\cdot D_{x}u+u\cdot D_{x}v}{v^{2}}}$

$\therefore D_{x}({\frac {u}{v}})={\frac {v\cdot D_{x}u-u\cdot D_{x}v}{v^{2}}}$

复合函数法则

$D_{x}v$ $=\lim _{\Delta x\to 0}{\frac {\Delta v}{\Delta x}}$ $=\lim _{\Delta x\to 0}{\frac {\Delta v\cdot \Delta y}{\Delta x\cdot \Delta y}}$ $=\lim _{\Delta x\to 0}{\frac {\Delta v}{\Delta y}}\cdot \lim _{\Delta x\to 0}{\frac {\Delta y}{\Delta x}}$ $=\lim _{\Delta y\to 0}{\frac {\Delta v}{\Delta y}}\cdot \lim _{\Delta x\to 0}{\frac {\Delta y}{\Delta x}}$ $=D_{y}v\cdot D_{x}y$

$\therefore D_{x}v=D_{y}v\cdot D_{x}y$