复合函数的求导法则

接下来介绍复合函数的求导方法，比如：

$$(\ln \tan x)'，{\text{(}}{{\text{e}}^{{x^2}}})'，(\sin \frac{{2x}}{{1 + {x^2}}})'.$$

定理3 如果$u=g(x)$在点$x$可导，而$y=f(u)$在点$u=g(x)$可导，则复合函数$y=f[g(x)]$在点$x$可导，且其导数为

$$\frac{{{\bf{d}}y}}{{{\bf{d}}x}} = f'\left( u \right)\cdot g'(x)$$

或

$$\frac{{{\bf{d}}y}}{{{\bf{d}}x}} = \frac{{{\bf{d}}y}}{{{\bf{d}}u}}\cdot \frac{{{\bf{d}}u}}{{{\bf{d}}x}}$$

证明:由于$y=f(u)$在点$u$可导，因此

$$\mathop {\lim }\limits_{\Delta u \to 0} \frac{{\Delta y}}{{\Delta u}} = f'(u)$$

存在，于是根据极限与无穷小的关系有

$$\frac{{\Delta y}}{{\Delta u}} = f'(u) + {\alpha }$$

其中$α$是$\Delta u \to 0$时的无穷小.上式中$\Delta u \ne 0$ [注意，由于$u=g(x)$,$\Delta u \ne 0$意味着函数$u=g(x)$不能是常值函数（例如$u=3,u=2^5...$或在定义域的某一段区间上有$u$是常数，对于一对多函数，尤其是偶函数对称点附近的点比如$y=\text{sin }x^2$在$x = 0$附近，$\Delta x ≠ 0$时会有一个点使得$u= g(x + \Delta x)-g(x) = 0$,此时只要保证$\Delta x$足够小即可去掉使$u= g(x + \Delta x)-g(x) = 0$的点.所以$u = g(x)$不能是常值函数(这会导致$\frac{{{\bf{d}}y}}{{{\bf{d}}u}}$是对常数u求导)或在定义域内某区间上是常数 ]，用$\Delta u$乘上式两边，得到

$$ \begin{equation} \Delta y = f'\left( u \right)\Delta u + {\alpha }\Delta u \end{equation} $$

当$\Delta u=0$时，规定$α= 0$ ①，这时因左端$\Delta y = f\left( {u + \Delta u} \right) - f\left( u \right) = 0,$而(1)式右端也为零，故(1)式对$\Delta u=0$也成立（如果$α= k ≠ 0$，那么当$\Delta u=0$时，右端也为0，所以当$\Delta u=0$时，$α$可以等于任意实数）.用$\Delta x \ne 0$除(1)式两边，得

$$\frac{{\Delta y}}{{\Delta x}} = \frac{{{f'}\left( u \right)\Delta u}}{{\Delta x}} + \frac{{{\alpha }\Delta u}}{{\Delta x}}$$

于是

$$\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \mathop {{\text{lim}}}\limits_{\Delta x \to 0} [\frac{{{f'}\left( u \right)\Delta u}}{{\Delta x}} + \frac{{\alpha \Delta u}}{{\Delta x}}]$$

根据函数在某点可导必在该点连续的性质可知，当${\Delta x \to 0}$时，${\Delta u \to 0},$从而可以推知

$$\mathop {{\text{lim}}}\limits_{\Delta x \to 0} \alpha = \mathop {{\text{lim}}}\limits_{\Delta u \to 0} \alpha = 0$$

又因$u=g(x)$在点$x$处可导，有

$$\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta u}}{{\Delta x}} = g'(x)$$

故

$$\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = f'\left( u \right)\mathop {{\text{lim}}}\limits_{\Delta x \to 0} \frac{{\Delta u}}{{\Delta x}}$$

即

$$\frac{{\text{d}y}}{{\text{d}x}} = f'\left( u \right)g'(x)$$

讨论

当${\Delta u \to 0}$时，规定$α$=0的作用是什么？

根据极限与无穷小的关系，当

$$\mathop {\lim }\limits_{\Delta u \to 0} \frac{{\Delta y}}{{\Delta u}} = f'(u)$$

有

$$\frac{{\Delta y}}{{\Delta u}} = f'(u) + \alpha $$

此时$\Delta u \ne 0$，用$\Delta u$乘以上式两边，得

\begin{equation} \Delta y = f'\left( u \right)\Delta u + \alpha \Delta u \end{equation}

用$\Delta x \ne 0$除(2)式两边，得

$$\frac{{\Delta y}}{{\Delta x}} = \frac{{f'\left( u \right)\Delta u}}{{\Delta x}} + \frac{{\alpha \Delta u}}{{\Delta x}}$$

于是

\begin{equation} \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \mathop {{\text{lim}}}\limits_{\Delta x \to 0} \left[ {\frac{{f'\left( u \right)\Delta u}}{{\Delta x}} + \frac{{\alpha \Delta u}}{{\Delta x}}} \right] \end{equation}

(2)式中，$\Delta u$为变量，且不为0，我们将其改为

$$\alpha = \frac{{\Delta y}}{{\Delta u}} - f'(u)$$

此式更能反映$\Delta u ≠ 0$，且$α$是$\Delta u$的函数$α=α(\Delta u)$，由于$α$是$\Delta u \to 0$时的无穷小，上式两端求极限得

$$\mathop {\lim }\limits_{\Delta u \to 0} \alpha = \mathop {\lim }\limits_{\Delta u \to 0} \left[ {\frac{{\Delta y}}{{\Delta u}} - f'\left( u \right)} \right] = f'\left( u \right) - f'\left( u \right) = 0$$

极限是$\Delta u$趋于0，即趋近并等于0时的值（注意极限同时描述了趋近过程和等于两者），那么我们若规定$\Delta u = 0$时$α= 0$，则$α$在$\Delta u = 0$时连续，也满足上述推导.

有些书上说，“若$\Delta u = 0$时，规定$α$=0，此时$\alpha = \frac{{\Delta y}}{{\Delta u}} - f'(u)$在$\Delta u = 0$处连续”，说的就是上述过程.

我们再来看另一种证明方法

$$ {{\rm{d}} \over {{\rm{d}}x}}f\left[ {g\left( x \right)} \right] = \mathop {\lim }\limits_{h \to 0} {{f\left[ {g\left( {x + h} \right)} \right] - f\left[ {g\left( x \right)} \right]} \over h} $$

因为$g(x)$在点$x$可导，令

$$ v = {{g\left( {x + h} \right) - g\left( x \right)} \over h} - g'\left( x \right) $$

则

$$ g\left( {x + h} \right) = g\left( x \right) + \left[ {v + g'\left( x \right)} \right]h $$

且有

$$ \mathop {\lim }\limits_{h \to 0} v = 0 $$

同理，对函数$f$有

$$ f\left( {y + k} \right) = f\left( y \right) + \left[ {w + f'\left( y \right)} \right]k \\ \mathop {\lim }\limits_{h \to 0} w = 0, $$

特别地，令$y=g(x),k=[v+g'(x)]h$，则

$$ f\left( {g\left( x \right) + \left[ {v + g'\left( x \right)} \right]h} \right) = f\left[ {g\left( x \right)} \right] + \left[ {w + f'\left( y \right)} \right]\left[ {v + g'\left( x \right)} \right]h $$ $$ \eqalign{ & f\left( {\left[ {g\left( {x + h} \right)} \right]} \right) - f\left[ {g\left( x \right)} \right] \cr & = f\left( {g\left( x \right) + \left[ {v + g'\left( x \right)} \right]h} \right) - f\left[ {g\left( x \right)} \right] \cr & = f\left[ {g\left( x \right)} \right] + \left[ {w + f'\left( y \right)} \right]\left[ {v + g'\left( x \right)} \right]h - f\left[ {g\left( x \right)} \right] \cr & = \left[ {w + f'\left( y \right)} \right]\left[ {v + g'\left( x \right)} \right]h \cr} $$ $$ \eqalign{ & {{\rm{d}} \over {{\rm{d}}x}}f\left[ {g\left( x \right)} \right] \cr & = \mathop {\lim }\limits_{h \to 0} {{f\left[ {g\left( {x + h} \right)} \right] - f\left[ {g\left( x \right)} \right]} \over h} \cr & = \mathop {\lim }\limits_{h \to 0} \left[ {w + f'\left( y \right)} \right]\left[ {v + g'\left( x \right)} \right] \cr & = f'\left( y \right)g'\left( x \right) \cr & = f'\left[ {g\left( x \right)} \right]g'\left( x \right) \cr & = f'\left( u \right)g'\left( x \right) \cr} $$

对于这种证明方法，我们没有看到对函数$u=g(x)$有不能是常值函数等的要求.

例 $y = {{\bf{e}}^{{x^3}}}$，求$\frac{{{\text{d}}y}}{{{\text{d}}x}}.$

解 $y = {{\bf{e}}^{{x^3}}}$是由$y = {{\bf{e}}^{{u}}}$,$u = x^3$复合而成的函数，因此

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{\text{d}}y}}{{{\text{d}}u}} \cdot \frac{{{\text{d}}u}}{{{\text{d}}x}} = {{\text{e}}^u} \cdot 3{x^2} = 3{x^2}{{\text{e}}^{{x^3}}}$$

例$y = \sin \frac{{2x}}{{1 + {x^2}}}$

解$y = \sin \frac{{2x}}{{1 + {x^2}}}$可以看作由$y = \sin u,u = \frac{{2x}}{{1 + {x^2}}}$复合而成，

$$\frac{{{\text{d}}y}}{{{\text{d}}u}} = \cos u$$ $$\frac{{{\text{d}}u}}{{{\text{d}}x}} = \frac{{2\left( {1 + {x^2}} \right) - 2x \cdot 2x}}{{{{(1 + {x^2})}^2}}} = \frac{{2\left( {1 - {x^2}} \right)}}{{{{(1 + {x^2})}^2}}}$$

所以

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \cos u \cdot \frac{{2\left( {1 - {x^2}} \right)}}{{{{(1 + {x^2})}^2}}} = \frac{{2\left( {1 - {x^2}} \right)}}{{{{(1 + {x^2})}^2}}}\cos \frac{{2x}}{{1 + {x^2}}}$$

复合函数的求导法则可以推广到多个中间变量的情形.以两个中间变量为例，设$y = f\left( u \right),u = \varphi \left( v \right),v = \psi (x)$在相应点处可导，则

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{\text{d}}y}}{{{\text{d}}u}} \cdot \frac{{{\text{d}}u}}{{{\text{d}}x}}$$

而

$$\frac{{{\text{d}}u}}{{{\text{d}}x}} = \frac{{{\text{d}}u}}{{{\text{d}}v}} \cdot \frac{{{\text{d}}v}}{{{\text{d}}x}}$$

所以

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{\text{d}}y}}{{{\text{d}}u}}\frac{{{\text{d}}u}}{{{\text{d}}v}} \cdot \frac{{{\text{d}}v}}{{{\text{d}}x}}$$

例 $y = \ln \cos ({{{\bf{e}}}^x})$，求$\frac{{{\text{d}}y}}{{{\text{d}}x}}.$

解$y = \ln u,u = \cos v,v = {{\bf{e}}^x}$

$$\frac{{{\text{d}}y}}{{{\text{d}}u}} = \frac{1}{u},\frac{{{\text{d}}u}}{{{\text{d}}v}} = - \sin v,\frac{{{\text{d}}v}}{{{\text{d}}x}} = {{\bf{e}}^x}$$

所以

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{1}{u} \cdot \left( { - \sin v} \right) \cdot {{\bf{e}}^x} =- \frac{1}{{\cos {{\bf{e}}^x}}} \cdot \sin {{\bf{e}}^x} \cdot {{\bf{e}}^x} = - {{\bf{e}}^x}\tan {{\bf{e}}^x}$$

例设$x>0$，证明幂函数的导数公式

$$\left( {{x^\mu }} \right)' = \mu {x^{\mu - 1}}$$

证明：设$y = {x^\mu }$，两边取对数得：

$$\ln y = \ln {x^\mu } \to {{\bf{e}}^{\ln y}} = {x^\mu } \to {x^\mu } = {{\bf{e}}^{\ln y}} = {{\bf{e}}^{\ln {x^\mu }}} = {{\bf{e}}^{\mu \ln x}}$$

分解：

$$y = {x^\mu } = {{\bf{e}}^{\mu \ln x}} = {{\bf{e}}^u},u = \mu \ln x$$

所以

$${\left( {{x^\mu }} \right)^\prime } = {\left( {{{\bf{e}}^{\mu \ln x}}} \right)^\prime } = {{\bf{e}}^u}\cdot\left( {\mu \ln x} \right)' = {{\bf{e}}^{{\text{ln}}{x^\mu }}}\cdot\frac{\mu }{{{x^\mu }}} = \mu {x^{\mu - 1}}.$$