链式法则,台湾地区亦称连锁律(英語:Chain rule),用于求合成函数的導數。
兩函數
和
的定義域 (
和
) 、值域 (
和
) 都包含於實數系
,若可以定義合成函數
(也就是
),且
於
可微分,且
於
可微分,則
![{\displaystyle {(g\circ f)}^{\prime }(a)=g^{\prime }[f(a)]\cdot f^{\prime }(a)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ff614c1ba145d4516a8270f068d65a3e126823c7)
也可以寫成
![{\displaystyle {\frac {dg[f(x)]}{dx}}{\bigg |}_{x=a}={\frac {dg(y)}{dy}}{\bigg |}_{y=f(a)}\cdot {\frac {df}{dx}}{\bigg |}_{x=a}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9610e4356b9ef7dfb28af08470588b52791f5578)
求函数
的导数。
- 设




求函数
的导数。



嚴謹的證明需要以下連續函數的極限定理:
和
都是实函数,若可以定義合成函數
且


則有
![{\displaystyle \lim _{x\to a}g[f(x)]=g(L)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/40ffccc43e9b9a0dbd90e56bab7ea40527f9cd24)
只要展開極限的ε-δ定義,並考慮
等於或不等於
的兩種狀況,這個極限定理就可以得証。
為了證明連鎖律,定義一個函數
,其定義域
, 而對應規則為
![{\displaystyle G(y)={\begin{cases}\displaystyle {\frac {g(y)-g[f(a)]}{y-f(a)}}&y\neq f(a)\\\\g^{\prime }[f(a)]&y=f(a)\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1d34ddec8759f9765572a0a8749d56581c4fd2c4)
和一個函數
,其定義域
, 而對應規則為

這樣,考慮到
於
的導數是以下函數(定義域為
)的極限
![{\displaystyle \lim _{x\to a}{\frac {g(y)-g[f(a)]}{x-a}}=\lim _{x\to a}G[f(x)]\cdot F(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0530f9ebb0ffef5d696b5129b4d06aa43a78c150)
因為可微則必連續(根據乘法的極限性質),所以
於
連續、
於
連續,故根據上面的極限定理有
![{\displaystyle \lim _{x\to a}G[f(x)]=g^{\prime }[f(a)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/41d8817646d5984e27f59a1584f7534ddd40ac21)
而且針對一開始可微的前提有

再根據乘法的極限性質有
![{\displaystyle \lim _{x\to a}{\frac {g(y)-g[f(a)]}{x-a}}=g^{\prime }[f(a)]\cdot f^{\prime }(a)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4837b37ee19118f8877834d4a01535fc6d7c0790)
即為所求。
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:

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:


如果我们考虑

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
的偏导数的数量积:

更一般地,对于从向量到向量的函数,求导法则为:

复合函数的最初几个高阶导数为:



