链式法则

两函数 ${\displaystyle f}$  和 ${\displaystyle g}$  的定义域 (${\displaystyle D_{f}}$  和 ${\displaystyle D_{g}}$ ) 、值域 (${\displaystyle I_{f}}$  和 ${\displaystyle I_{g}}$ ) 都包含于实数系 ${\displaystyle \mathbb {R} }$  ，若可以定义合成函数 ${\displaystyle g\circ f}$  (也就是 ${\displaystyle I_{f}\cap D_{g}\neq \varnothing }$  )，且 ${\displaystyle f}$  于 ${\displaystyle a\in D_{f}}$  可微分，且 ${\displaystyle g}$  于 ${\displaystyle f(a)\in I_{f}\cap D_{g}}$  可微分，则

${\displaystyle {(g\circ f)}^{\prime }(a)=g^{\prime }[f(a)]\cdot f^{\prime }(a)}$ 

也可以写成

${\displaystyle {\frac {dg[f(x)]}{dx}}{\bigg |}_{x=a}={\frac {dg(y)}{dy}}{\bigg |}_{y=f(a)}\cdot {\frac {df}{dx}}{\bigg |}_{x=a}}$ 

求函数 ${\displaystyle f(x)=(x^{2}+1)^{3}}$ 的导数。

设 ${\displaystyle g(x)=x^{2}+1}$ 

${\displaystyle h(g)=g^{3}\to h(g(x))=g(x)^{3}.}$ 

${\displaystyle f(x)=h(g(x))}$ 

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

求函数 ${\displaystyle \arctan \,\sin \,x}$ 的导数。

${\displaystyle {\frac {d}{dx}}\arctan \,x\,=\,{\frac {1}{1+x^{2}}}}$ 

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

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

严谨的证明需要以下连续函数的极限定理：

${\displaystyle f}$  和 ${\displaystyle g}$  都是实函数，若可以定义合成函数 ${\displaystyle g\circ f}$  且

• ${\displaystyle \lim _{x\to a}f(x)=L}$ 
• ${\displaystyle \lim _{y\to L}g(y)=g(L)}$ 

则有

${\displaystyle \lim _{x\to a}g[f(x)]=g(L)}$ 

只要展开极限的δ-ε定义，并考虑 ${\displaystyle f(x)}$  等于或不等于 ${\displaystyle L}$  的两种状况，这个极限定理就可以得证。

为了证明链式法则，定义一个函数 ${\displaystyle G}$  ，其定义域 ${\displaystyle D_{G}=D_{g}}$  ， 而对应规则为

${\displaystyle G(y)={\begin{cases}\displaystyle {\frac {g(y)-g[f(a)]}{y-g[f(a)]}}&y\neq f(a)\\\\g^{\prime }[f(a)]&y=f(a)\end{cases}}}$ 

和一个函数 ${\displaystyle F}$  ，其定义域 ${\displaystyle D_{F}=D_{f}}$  ， 而对应规则为

${\displaystyle F(x)={\begin{cases}\displaystyle {\frac {f(x)-f(a)}{x-a}}&x\neq a\\\\f^{\prime }(a)&x=a\end{cases}}}$ 

这样，考虑到 ${\displaystyle g\circ f}$  于 ${\displaystyle a}$  的导数是以下函数（定义域为${\displaystyle D_{g\,\circ f}}$ ）的极限

${\displaystyle \lim _{x\to a}{\frac {g(y)-g[f(a)]}{x-a}}=\lim _{x\to a}G[f(x)]\cdot F(x)}$ 

因为可微则必连续(根据乘法的极限性质)，所以 ${\displaystyle f}$  于 ${\displaystyle a}$  连续、 ${\displaystyle G}$  于 ${\displaystyle f(a)}$  连续，故根据上面的极限定理有

${\displaystyle \lim _{x\to a}G[f(x)]=g^{\prime }[f(a)]}$ 

而且针对一开始可微的前提有

${\displaystyle \lim _{x\to a}F(x)=f^{\prime }(a)}$ 

再根据乘法的极限性质有

${\displaystyle \lim _{x\to a}{\frac {g(y)-g[f(a)]}{x-a}}=g^{\prime }[f(a)]\cdot f^{\prime }(a)}$ 

即为所求。 ${\displaystyle \Box }$ 

考虑函数z = f(x, y)，其中x = g(t)，y = h(t)，g(t)和h(t)是可微函数，那么：

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

假设z = f(u, v)的每一个自变量都是二元函数，也就是说，u = h(x, y)，v = g(x, y)，且这些函数都是可微的。那么，z的偏导数为：

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

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

如果我们考虑

${\displaystyle {\vec {r}}=(u,v)}$ 

为一个向量函数，我们可以用向量的表示法把以上的公式写成f的梯度与${\displaystyle {\vec {r}}}$ 的偏导数的数量积：

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

更一般地，对于从向量到向量的函数，求导法则为：

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

复合函数的最初几个高阶导数为：

${\displaystyle {\frac {d(f\circ g)}{dx}}={\frac {df}{dg}}{\frac {dg}{dx}}}$ 

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

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

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

• 乘积法则
• 除法定则