# 链式法则

（重定向自連鎖律

${\displaystyle (f\circ g)'(x)=f'(g(x))g'(x).}$

## 例子

${\displaystyle g(x)=x^{2}+1}$
${\displaystyle h(g)=g^{3}\to h(g(x))=g(x)^{3}.}$
${\displaystyle f(x)=h(g(x))}$
${\displaystyle f'(x)=h'(g(x))g'(x)=3(g(x))^{2}(2x)=3(x^{2}+1)^{2}(2x)=6x(x^{2}+1)^{2}.}$

${\displaystyle {\frac {d}{dx}}\arctan \,x\,=\,{\frac {1}{1+x^{2}}}}$
${\displaystyle {\frac {d}{dx}}\arctan \,f(x)\,=\,{\frac {f'(x)}{1+f^{2}(x)}}}$
${\displaystyle {\frac {d}{dx}}\arctan \,\sin \,x\,=\,{\frac {\cos \,x}{1+\sin ^{2}\,x}}}$

## 证明

fg为函数，x为常数，使得fg(x)可导，且gx可导。根据可导的定义，

${\displaystyle g(x+\delta )-g(x)=\delta g'(x)+\epsilon (\delta )\delta \,}$ ，其中当${\displaystyle \delta \to 0}$ 时，${\displaystyle \epsilon (\delta )\to 0\,}$

${\displaystyle f(g(x)+\alpha )-f(g(x))=\alpha f'(g(x))+\eta (\alpha )\alpha \,}$ ，其中当${\displaystyle \alpha \to 0.\,}$ 时，${\displaystyle \eta (\alpha )\to 0\,}$

 ${\displaystyle f(g(x+\delta ))-f(g(x))\,}$ ${\displaystyle =f(g(x)+\delta g'(x)+\epsilon (\delta )\delta )-f(g(x))\,}$ ${\displaystyle =\alpha _{\delta }f'(g(x))+\eta (\alpha _{\delta })\alpha _{\delta }\,}$

${\displaystyle {\frac {f(g(x+\delta ))-f(g(x))}{\delta }}\to g'(x)f'(g(x)).}$

## 多元复合函数求导法则

${\displaystyle {\ dz \over dt}={\partial z \over \partial x}{dx \over dt}+{\partial z \over \partial y}{dy \over dt}.}$

${\displaystyle {\partial z \over \partial x}={\partial z \over \partial u}{\partial u \over \partial x}+{\partial z \over \partial v}{\partial v \over \partial x}}$
${\displaystyle {\partial z \over \partial y}={\partial z \over \partial u}{\partial u \over \partial y}+{\partial z \over \partial v}{\partial v \over \partial y}.}$

${\displaystyle {\vec {r}}=(u,v)}$

${\displaystyle {\frac {\partial f}{\partial x}}={\vec {\nabla }}f\cdot {\frac {\partial {\vec {r}}}{\partial x}}.}$

${\displaystyle {\frac {\partial (z_{1},\ldots ,z_{m})}{\partial (x_{1},\ldots ,x_{p})}}={\frac {\partial (z_{1},\ldots ,z_{m})}{\partial (y_{1},\ldots ,y_{n})}}{\frac {\partial (y_{1},\ldots ,y_{n})}{\partial (x_{1},\ldots ,x_{p})}}.}$

## 高阶导数

${\displaystyle {\frac {d(f\circ g)}{dx}}={\frac {df}{dg}}{\frac {dg}{dx}}}$
${\displaystyle {\frac {d^{2}(f\circ g)}{dx^{2}}}={\frac {d^{2}f}{dg^{2}}}\left({\frac {dg}{dx}}\right)^{2}+{\frac {df}{dg}}{\frac {d^{2}g}{dx^{2}}}}$
${\displaystyle {\frac {d^{3}(f\circ g)}{dx^{3}}}={\frac {d^{3}f}{dg^{3}}}\left({\frac {dg}{dx}}\right)^{3}+3{\frac {d^{2}f}{dg^{2}}}{\frac {dg}{dx}}{\frac {d^{2}g}{dx^{2}}}+{\frac {df}{dg}}{\frac {d^{3}g}{dx^{3}}}}$
${\displaystyle {\frac {d^{4}(f\circ g)}{dx^{4}}}={\frac {d^{4}f}{dg^{4}}}\left({\frac {dg}{dx}}\right)^{4}+6{\frac {d^{3}f}{dg^{3}}}\left({\frac {dg}{dx}}\right)^{2}{\frac {d^{2}g}{dx^{2}}}+{\frac {d^{2}f}{dg^{2}}}\left\{4{\frac {dg}{dx}}{\frac {d^{3}g}{dx^{3}}}+3\left({\frac {d^{2}g}{dx^{2}}}\right)^{2}\right\}+{\frac {df}{dg}}{\frac {d^{4}g}{dx^{4}}}.}$