1. 一元函数与多元函数复合的情形

定理1 如果函数$u = \phi \left( t \right)$及$v = \varphi \left( t \right)$都在点$t$可导，函数$z = f\left( {u,v} \right)$对应点$\left( {u,v} \right)$具有连续偏导数，则复合函数$z = f\left[ {\phi \left( t \right),\varphi \left( t \right)} \right]$在点$t$可导，且有

$$\frac{{{\text{d}}z}}{{{\text{d}}t}} = \frac{{\partial z}}{{\partial u}}\frac{{{\text{d}}u}}{{{\text{d}}t}} + \frac{{\partial z}}{{\partial v}}\frac{{{\text{d}}v}}{{{\text{d}}t}}$$

证明 设$t$有增量$\Delta t$，这时$u = \phi \left( t \right)$、$v = \varphi \left( t \right)$的对应增量为$\Delta u$、$\Delta v$，由此，函数$z = f\left( {u,v} \right)$相应地有增量$\Delta z$.按假定，函数$z = f\left( {u,v} \right)$在点$\left( {u,v} \right)$具有连续偏导数，这时函数的全增量$\Delta z$可表示为

$$\Delta z = \frac{{\partial z}}{{\partial u}}\Delta u + \frac{{\partial z}}{{\partial v}}\Delta v + {\varepsilon _1}\Delta u + {\varepsilon _2}\Delta v$$

#上式的推导如下#

由假定，函数$z = f\left( {u,v} \right)$在点$\left( {u,v} \right)$具有连续偏导数，从而偏导数$\frac{{\partial z}}{{\partial u}},\frac{{\partial z}}{{\partial v}}$在点$\left( {u,v} \right)$的某个邻域内存在，设点$\left( {u + \Delta u,v + \Delta v} \right)$为这邻域内任意一点，考察函数的全增量

\[\begin{gathered} \Delta z = f\left( {u + \Delta u,v + \Delta v} \right) - f\left( {u,v} \right) \hfill \\ = \hfill \\ \left[ {f\left( {u + \Delta u,v + \Delta v} \right) - f\left( {u,v + \Delta v} \right)} \right] \hfill \\ + \left[ {f\left( {u,v + \Delta v} \right) - f\left( {u,v} \right)} \right] \hfill \\ \end{gathered} \]

在第一个方括号内的表达式，由于$\left( {v + \Delta v} \right)$不变，因而可以看做是$u$的一元函数$f\left( {u,v + \Delta v} \right)$的增量，于是，应用拉格朗日中值定理，有

\[\begin{gathered} f\left( {u + \Delta u,v + \Delta v} \right) - f\left( {u,v + \Delta v} \right) \hfill \\ = {f_u}\left( {u + {\theta _1}\Delta u,v + \Delta v} \right)\Delta u, \hfill \\ \left( {0 < {\theta _1} < 1} \right) \hfill \\ \end{gathered} \]

又依假设，$\frac{{\partial z}}{{\partial u}}$在点$\left( {u,v} \right)$连续，所以上式可写为

$$\eqalign{ & f\left( {u + \Delta u,v + \Delta v} \right) - f\left( {u,v + \Delta v} \right) \cr & = {f_u}\left( {u,v} \right)\Delta u + {\varepsilon _1}\Delta u,\left( {0 < {\theta _1} < 1} \right) \cr} $$

#

\[\begin{gathered} \mathop {\lim }\limits_{\Delta u \to 0,\Delta v \to 0} \left[ \begin{gathered} {f_u}\left( {u + {\theta _1}\Delta u,v + \Delta v} \right) \hfill \\ - {f_u}\left( {u,v} \right) \hfill \\ \end{gathered} \right] = 0 \hfill \\ {f_u}\left( {u + {\theta _1}\Delta u,v + \Delta v} \right) - {f_u}\left( {u,v} \right) \hfill \\ = 0 + {\varepsilon _1}, \hfill \\ {f_u}\left( {u + {\theta _1}\Delta u,v + \Delta v} \right) \hfill \\ = {f_u}\left( {u,v} \right) + {\varepsilon _1}, \hfill \\ \end{gathered} \]

其中${\varepsilon _1}$为$\Delta u,\Delta v$的函数，且当$\Delta u \to 0,\Delta v \to 0$时，${\varepsilon _1} \to 0$.

同理可证明第二个方括号内的表达式可写为

\[\begin{gathered} \left[ {f\left( {u,v + \Delta v} \right) - f\left( {u,v} \right)} \right] \hfill \\ = {f_v}\left( {u,v} \right)\Delta v + {\varepsilon _2}\Delta v \hfill \\ \end{gathered} \]

其中${\varepsilon _2}$为$\Delta v$的函数，且当$\Delta v \to 0$时，${\varepsilon _2} \to 0$.

由此可知，在偏导数连续的假定下，全增量$\Delta z$可以表示为

\[\begin{gathered} \Delta z = {f_u}\left( {u,v} \right)\Delta u + {f_v}\left( {u,v} \right)\Delta v \hfill \\ + {\varepsilon _1}\Delta u + {\varepsilon _2}\Delta v \hfill \\ = \frac{{\partial z}}{{\partial u}}\Delta u + \frac{{\partial z}}{{\partial v}}\Delta v + {\varepsilon _1}\Delta u + {\varepsilon _2}\Delta v \hfill \\ \end{gathered} \]

#推导结束#

将

$$\Delta z = \frac{{\partial z}}{{\partial u}}\Delta u + \frac{{\partial z}}{{\partial v}}\Delta v + {\varepsilon _1}\Delta u + {\varepsilon _2}\Delta v$$

两端各除以$\Delta t$,得

\[\begin{gathered} \frac{{\Delta z}}{{\Delta t}} = \hfill \\ \frac{{\partial z}}{{\partial u}}\frac{{\Delta u}}{{\Delta t}} + \frac{{\partial z}}{{\partial v}}\frac{{\Delta v}}{{\Delta t}} + {\varepsilon _1}\frac{{\Delta u}}{{\Delta t}} + {\varepsilon _2}\frac{{\Delta v}}{{\Delta t}} \hfill \\ \end{gathered} \]

因为当$\Delta t \to 0$时，$\Delta u \to 0,\Delta v \to 0,\frac{{\Delta u}}{{\Delta t}} \to \frac{{{\text{d}}u}}{{{\text{d}}t}},\frac{{\Delta v}}{{\Delta t}} \to \frac{{{\text{d}}v}}{{{\text{d}}t}}$，所以

\[\begin{gathered} \mathop {\lim }\limits_{\Delta t \to 0} \frac{{\Delta z}}{{\Delta t}} = \hfill \\ \frac{{\partial z}}{{\partial u}}\frac{{{\text{d}}u}}{{{\text{d}}t}} + \frac{{\partial z}}{{\partial v}}\frac{{{\text{d}}v}}{{{\text{d}}t}} + 0\frac{{{\text{d}}u}}{{{\text{d}}t}} + 0\frac{{{\text{d}}u}}{{{\text{d}}t}} \hfill \\ \mathop {\lim }\limits_{\Delta t \to 0} \frac{{\Delta z}}{{\Delta t}} = \frac{{\partial z}}{{\partial u}}\frac{{{\text{d}}u}}{{{\text{d}}t}} + \frac{{\partial z}}{{\partial v}}\frac{{{\text{d}}v}}{{{\text{d}}t}} \hfill \\ \frac{{{\text{d}}z}}{{{\text{d}}t}} = \frac{{\partial z}}{{\partial u}}\frac{{{\text{d}}u}}{{{\text{d}}t}} + \frac{{\partial z}}{{\partial v}}\frac{{{\text{d}}v}}{{{\text{d}}t}} \hfill \\ \end{gathered} \]

$\frac{{{\text{d}}z}}{{{\text{d}}t}}$称为全导数.

2. 多元函数与多元函数复合的情形

定理2 如果函数$u = \phi \left( {x,y} \right)$及$v = \varphi \left( {x,y} \right)$都在点$\left( {x,y} \right)$具有对$x$及对$y$的偏导数，函数$z = f\left( {u,v} \right)$在对应点$\left( {u,v} \right)$具有连续偏导数，则复合函数$z = f\left[ {\phi \left( {x,y} \right),\varphi \left( {x,y} \right)} \right]$在点$\left( {x,y} \right)$的两个偏导数存在，且有

$$\eqalign{ & \frac{{\partial z}}{{\partial x}} = \frac{{\partial z}}{{\partial u}}\frac{{\partial u}}{{\partial x}} + \frac{{\partial z}}{{\partial v}}\frac{{\partial v}}{{\partial x}} \cr & \frac{{\partial z}}{{\partial y}} = \frac{{\partial z}}{{\partial u}}\frac{{\partial u}}{{\partial y}} + \frac{{\partial z}}{{\partial v}}\frac{{\partial v}}{{\partial y}} \cr} $$

这里求$\frac{{\partial z}}{{\partial x}}$时，将$y$看做常量，因此$u = \phi \left( {x,y} \right)$及$v = \varphi \left( {x,y} \right)$仍可看做一元函数而应用定理一.但是由于复合函数$z = f\left[ {\phi \left( {x,y} \right),\varphi \left( {x,y} \right)} \right]$以及$u = \phi \left( {x,y} \right)$和$v = \varphi \left( {x,y} \right)$都是$x,y$的二元函数，所以应把定理1中的d改为$\partial $，再把$t$改为$x$.