因为复合函数$z = f\left( {u,x,y} \right),u = \varphi \left( {x,y} \right)$可以看作复合函数$z = f\left( {u,v,w} \right)$，$u = \varphi \left( {x,y} \right),v = \phi \left( {x,y} \right),w = \omega \left( {x,y} \right)$中$v = x,w = y$的特殊情形，所以首先给出函数$z = f\left( {u,v,w} \right)$在对应点$\left( {u,v,w} \right)$的偏导数.

设$u = \varphi \left( {x,y} \right),v = \phi \left( {x,y} \right),w = \omega \left( {x,y} \right)$都在点$\left( {x,y} \right)$具有对$x$及对$y$的偏导数，函数$z = f\left( {u,v,w} \right)$在对应点$\left( {u,v,w} \right)$具有连续偏导数，则复合函数

$$z = f\left[ {\varphi \left( {x,y} \right),\phi \left( {x,y} \right),\omega \left( {x,y} \right)} \right]$$

在点$\left( {x,y} \right)$的两个偏导数都存在，且可用下列公式计算：

$$\eqalign{ & \frac{{\partial z}}{{\partial x}} = \frac{{\partial z}}{{\partial u}}\frac{{\partial u}}{{\partial x}} + \frac{{\partial z}}{{\partial v}}\frac{{\partial v}}{{\partial x}} + \frac{{\partial z}}{{\partial w}}\frac{{\partial w}}{{\partial x}} \cr & \frac{{\partial z}}{{\partial y}} = \frac{{\partial z}}{{\partial u}}\frac{{\partial u}}{{\partial y}} + \frac{{\partial z}}{{\partial v}}\frac{{\partial v}}{{\partial y}} + \frac{{\partial z}}{{\partial w}}\frac{{\partial w}}{{\partial y}} \cr & ................\left( 1 \right).............. \cr} $$

还有这样的情形：复合函数的某些中间变量本身又是复合函数的自变量.例如，设$z = f\left( {u,x,y} \right)$具有连续偏导数，而$u = \varphi \left( {x,y} \right)$具有偏导数.我们来求复合函数$z = f\left( {u,x,y} \right)$对$x$及对$y$的偏导数.

我们知道，当二元函数$z = f\left( {x,y} \right)$的偏导数连续时，函数可微，全微分为

$${\text{d}}z = \frac{{\partial z}}{{\partial x}}\Delta x + \frac{{\partial z}}{{\partial y}}\Delta y$$

二元函数$z = f\left( {x,y} \right)$在空间直角坐标系$Oxyz$中的图形是一张曲面，在某点的偏导数表示在该点处曲面被平行于坐标平面$xOy,yOz$的平面所截得的曲线的切线的斜率.

即便$y = \varphi \left( x \right),\left( {y = \varphi \left( x \right)} \right)$可导），即$y$由$x$决定，在$Oxyz$坐标系中已经形成的$z = f\left[ {x,\varphi \left( x \right)} \right]$的曲面上，在对应点$\left( {x,y} \right)$具有连续偏导数，则$Oxyz$坐标系中求偏导数时，依然将$x$和$y$看做独立变量，有全微分

$${\text{d}}z = \frac{{\partial z}}{{\partial x}}\Delta x + \frac{{\partial z}}{{\partial y}}\Delta y$$

有读者会认为此时求对$y$的偏导数时，按照定义，x应保持不变，而x保持不变，则y不会有变化量，从而对y的偏导数不存在.我们再次重复思考一遍即便有$y = \varphi \left( x \right)$，在$Oxyz$坐标系中对$y$的偏导数存在，求偏导数时为什么自变量可以看做相互独立，理由是：尽管$y$由$x$决定($y = \varphi \left( x \right)$)，在$Oxyz$坐标系中，在由所有的点$\left( {x,y,z} \right)$已经形成的曲面$z = f\left[ {x,\varphi \left( x \right)} \right]$上，坐标$x$和$y$可以看做是相互独立的，因为此时在图形上可以任意指定$y$，由任意指定的$y$所形成的平行于坐标平面$Oxyz$的平面截曲面形成曲线，在曲线上任一点可以求出曲线上在该点的切线的斜率，即求偏导数.

$$\eqalign{ & {\text{d}}z = \frac{{\partial z}}{{\partial x}}\Delta x + \frac{{\partial z}}{{\partial y}}\Delta y \cr & \frac{{\partial z}}{{\partial x}} = \frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}}\frac{{{\text{d}}y}}{{{\text{d}}x}} \cr} $$

同理，对三元函数$z = f\left( {u,x,y} \right)$，可以在坐标系$Oxyuz$中求偏导数，此时自变量$x,y,u$可以看做相互独立.

证明

由于$z = f\left( {u,x,y} \right)$具有连续偏导数，从而可微，全增量

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

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

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

根据假设，$z = f\left( {u,x,y} \right)$具有连续偏导数，所以上式可写为

$$\eqalign{ & f\left( {u + \Delta u,x + \Delta x,y + \Delta y} \right) \cr & - f\left( {u,x + \Delta x,y + \Delta y} \right) \cr & = {f_u}\left( {u,x,y} \right)\Delta u + {\varepsilon _1}\Delta u \cr} $$

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

同理可得第二、第三个方括号内的表达式可写为

$$\eqalign{ & f\left( {u,x + \Delta x,y + \Delta y} \right) - f\left( {u,x,y + \Delta y} \right) \cr & = {f_x}\left( {u,x,y} \right)\Delta x + {\varepsilon _2}\Delta x \cr & f\left( {u,x,y + \Delta y} \right) - f\left( {u,x,y} \right) \cr & = {f_y}\left( {u,x,y} \right)\Delta y + {\varepsilon _3}\Delta y \cr} $$

其中${\varepsilon _2}$为$\Delta x,\Delta y$的函数，且当$\Delta x \to 0,\Delta y \to 0$时，${\varepsilon _2} \to 0$；${\varepsilon _3}$为$\Delta y$的函数，且当$\Delta y \to 0$时，${\varepsilon _3} \to 0$.

由此可见，在偏导数连续的假定下，全增量可以表示为

$$\eqalign{ & \Delta z = \cr & f\left( {u + \Delta u,x + \Delta x,y + \Delta y} \right) - f\left( {u,x,y} \right) \cr & = {f_u}\left( {u,x,y} \right)\Delta u + {f_x}\left( {u,x,y} \right)\Delta x + \cr & {f_y}\left( {u,x,y} \right)\Delta y + {\varepsilon _1}\Delta u + {\varepsilon _2}\Delta x + {\varepsilon _3}\Delta y \cr} $$

且${\varepsilon _1}\Delta u + {\varepsilon _2}\Delta x + {\varepsilon _3}\Delta y$是随着$\left( {\Delta u,\Delta x,\Delta y} \right) \to \left( {0,0,0} \right)$.这就证明了$z = f\left( {u,x,y} \right)$的全微分为

\[\begin{gathered} {\text{d}}z = \hfill \\ {f_u}\left( {u,x,y} \right)\Delta u + {f_x}\left( {u,x,y} \right)\Delta x \hfill \\ + {f_y}\left( {u,x,y} \right)\Delta y \hfill \\ \end{gathered} \]

两端除以$\Delta x,\Delta y$求极限，得偏导数为

\[\begin{gathered} \frac{{\partial z}}{{\partial x}} = {f_u}\left( {u,x,y} \right)\frac{{\partial u}}{{\partial x}} + {f_x}\left( {u,x,y} \right) \hfill \\ = \frac{{\partial f}}{{\partial u}}\frac{{\partial u}}{{\partial x}} + \frac{{\partial f}}{{\partial x}} \hfill \\ \frac{{\partial z}}{{\partial y}} = {f_u}\left( {u,x,y} \right)\frac{{\partial u}}{{\partial y}} + {f_y}\left( {u,x,y} \right) \hfill \\ = \frac{{\partial f}}{{\partial u}}\frac{{\partial u}}{{\partial y}} + \frac{{\partial f}}{{\partial y}} \hfill \\ \end{gathered} \]