# 乘积法则

${\displaystyle (fg)'=f'g+fg'\,}$

## 莱布尼兹的发现

{\displaystyle {\begin{aligned}d(u\cdot v)&{}=(u+du)\cdot (v+dv)-u\cdot v\\&{}=u\cdot dv+v\cdot du+du\cdot dv.\end{aligned}}}

${\displaystyle d(u\cdot v)=v\cdot du+u\cdot dv\,}$

${\displaystyle {\frac {d}{dx}}(u\cdot v)=v\cdot {\frac {du}{dx}}+u\cdot {\frac {dv}{dx}}}$

${\displaystyle (u\cdot v)'=v\cdot u'+u\cdot v'.\,}$

## 例子

• 假设我们要求出f(x) = x2 sin(x)的导数。利用乘积法则，可得f'(x) = 2x sin(x) + x2cos(x)（这是因为x2的导数是2x，sin(x)的导数是cos(x)）。
• 乘积法则的一个特例，是“常数因子法则”，也就是：如果c实数f(x)是可微函数，那么cf(x)也是可微的，其导数为(c × f)'(x) = c × f '(x)。
• 乘积法则可以用来推出分部积分法除法定则

## 证明一：利用面积

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

fgx点可导。那么：

${\displaystyle h'(x)=\lim _{w\to x}{h(w)-h(x) \over w-x}=\lim _{w\to x}{f(w)g(w)-f(x)g(x) \over w-x}.\qquad \qquad (1)}$

${\displaystyle f(w)g(w)-f(x)g(x)\qquad \qquad (2)}$

${\displaystyle f(w){\Bigg (}g(w)-g(x){\Bigg )}+g(w){\Bigg (}f(w)-f(x){\Bigg )}.\qquad \qquad (3)}$

${\displaystyle \lim _{w\to x}\left(f(w)\left({g(w)-g(x) \over w-x}\right)+g(w)\left({f(w)-f(x) \over w-x}\right)\right).\qquad \qquad (4)}$

${\displaystyle \left(\lim _{w\to x}f(w)\right)\left(\lim _{w\to x}{g(w)-g(x) \over w-x}\right)+\left(\lim _{w\to x}g(w)\right)\left(\lim _{w\to x}{f(w)-f(x) \over w-x}\right).\qquad \qquad (5)}$

${\displaystyle \lim _{w\to x}f(w)=f(x)\,}$

${\displaystyle \lim _{w\to x}{g(w)-g(x) \over w-x}=g'(x)}$

${\displaystyle \lim _{w\to x}{f(w)-f(x) \over w-x}=f'(x)}$

${\displaystyle \lim _{w\to x}g(w)=g(x)\,}$

${\displaystyle f(x)g'(x)+g(x)f'(x).\,}$

## 证明二：使用对数

f = uv，并假设uv是正数。那么：

${\displaystyle \ln f=\ln u+\ln v.\,}$

${\displaystyle {1 \over f}{d \over dx}f={1 \over u}{d \over dx}u+{1 \over v}{d \over dx}v}$

${\displaystyle {d \over dx}f=v{d \over dx}u+u{d \over dx}v.}$

## 证明三：使用导数的定义

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

fgx点可导。那么：

${\displaystyle h'(x)=\lim _{\Delta {x}\to 0}{\frac {h(x+\Delta {x})-h(x)}{\Delta {x}}}=\lim _{\Delta {x}\to 0}{\frac {f(x+\Delta {x})g(x+\Delta {x})-f(x)g(x)}{\Delta {x}}}}$

${\displaystyle =\lim _{\Delta {x}\to 0}{\frac {f(x+\Delta {x})g(x+\Delta {x})-f(x)g(x+\Delta {x})+f(x)g(x+\Delta {x})-f(x)g(x)}{\Delta {x}}}}$
${\displaystyle =\lim _{\Delta {x}\to 0}{\frac {[f(x+\Delta {x})-f(x)]\cdot g(x+\Delta {x})+f(x)\cdot [g(x+\Delta {x})-g(x)]}{\Delta {x}}}}$
${\displaystyle =\lim _{\Delta {x}\to 0}{\frac {f(x+\Delta {x})-f(x)}{\Delta {x}}}\cdot \lim _{\Delta {x}\to 0}g(x+\Delta {x})+\lim _{\Delta {x}\to 0}f(x)\cdot \lim _{\Delta {x}\to 0}{\frac {g(x+\Delta {x})-g(x)}{\Delta {x}}}}$
${\displaystyle =f'(x)g(x)+f(x)g'(x)}$.

## 推廣

• 若有${\displaystyle n}$个函数${\displaystyle f_{1},f_{2},...,f_{n}}$，则：
${\displaystyle \left({\prod _{k=1}^{n}{f_{k}}}\right)^{\prime }=\sum _{k=1}^{n}{\left({f'_{k}\cdot \prod _{j=1 \atop j\neq k}^{n}{f_{j}}}\right)}}$
• 萊布尼茲法則）若${\displaystyle f,g}$均為可導${\displaystyle n}$次的函數，則${\displaystyle fg}$${\displaystyle n}$次導數為：
${\displaystyle (f\cdot g)^{(n)}=\sum _{k=0}^{n}{n \choose k}f^{(k)}g^{(n-k)}}$

## 应用

${\displaystyle {d \over dx}x^{n}=nx^{n-1}}$

{\displaystyle {\begin{aligned}{d \over dx}x^{k+1}&{}={d \over dx}\left(x^{k}\cdot x\right)\\\\&{}=x{d \over dx}x^{k}+x^{k}{d \over dx}x\\\\&{}=x\left(kx^{k-1}\right)+x^{k}\cdot 1\\\\&{}=(k+1)x^{k}.\end{aligned}}}