复合函数的偏导数

多元复合函数的求导法则，也称为链式法则。对于多元函数的链式法则，比较让人困惑的情形，就是函数里即含有自变量，又含有中间变量的情形。这种情形，我们可以将函数里的自变量当成中间变量来处理即可。所以对于复合函数的偏导数，只需要记住一点：有多少个中间变量，偏导数就有多少项，这样的话，问题就变得容易多了。

1，若 $z=f(u,v),u=\varphi (x,y),v=\psi (x,y)$，则

$$\begin{array}{l}{\displaystyle \frac{\partial z}{\partial x}=\frac{\partial f}{\partial u}\cdot \frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\cdot \frac{\partial v}{\partial x}}\\ {\displaystyle \frac{\partial z}{\partial y}=\frac{\partial f}{\partial u}\cdot \frac{\partial u}{\partial y}+\frac{\partial f}{\partial v}\cdot \frac{\partial v}{\partial y}}\end{array}$$

要注意的是，这里只是两个中间变量，如果有多个中间变量，则有几个中间变量，应该有几项。这是一般的情况。

2，如果函数中即有自变量，也有中间变量，这会比较麻烦一点，如果只有一个自变量

$$z=f(x,y,t),x=\varphi (t),y=\psi (t)$$

则 $$\frac{dz}{dt}=\frac{\partial f}{\partial x}\cdot \frac{dx}{dt}+\frac{\partial f}{\partial y}\cdot \frac{dy}{dt}+\frac{\partial f}{\partial t}$$

3，若函数中即有自变量，也有中间变量，中间变量和自变量都多于一个，

$$z=f(u,v,x,y),u=\varphi (x,y),v=\psi (x,y)$$

则

$$\frac{\partial z}{\partial x}=\frac{\partial f}{\partial u}\cdot \frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\cdot \frac{\partial v}{\partial x}+\frac{\partial f}{\partial x}$$

$$\frac{\partial z}{\partial y}=\frac{\partial f}{\partial u}\cdot \frac{\partial u}{\partial y}+\frac{\partial f}{\partial v}\cdot \frac{\partial v}{\partial y}+\frac{\partial f}{\partial y}$$

例1，设 $z=x^2\sin y,x=s^2+t^2,y=2st$， 求${\displaystyle \frac{\partial z}{\partial s},\frac{\partial z}{\partial t}}$。

解：$$\begin{array}{rl}\frac{\partial z}{\partial s}& =\frac{\partial f}{\partial x}\cdot \frac{\partial x}{\partial s}+\frac{\partial f}{\partial y}\cdot \frac{\partial y}{\partial s}\\ & =2x\sin y\cdot (2s)+x^2\cos y\cdot 2t\\ & =4sx\sin y+2t{x}^2\cos y\end{array}$$

$$\begin{array}{rl}\frac{\partial z}{\partial t}& =\frac{\partial f}{\partial x}\cdot \frac{\partial x}{\partial t}+\frac{\partial f}{\partial y}\cdot \frac{\partial y}{\partial t}\\ & =2x\sin y\cdot (2t)+x^2\cos y\cdot 2s\\ & =4tx\sin y+2s{x}^2\cos y\end{array}$$

例2，设 ${\displaystyle z=e^{x^2+2xy+2xt},x=\cos t,y=sint}$，求 ${\displaystyle \frac{dz}{dt}}$。

解：$$\begin{array}{rl}\frac{dz}{dt}& =\frac{\partial f}{\partial x}\cdot \frac{dx}{dt}+\frac{\partial f}{\partial y}\cdot \frac{dy}{dt}d+\frac{\partial f}{\partial t}\\ & =(2x+2y+2t)e^{x^2+2xy+2xt}\cdot (-\sin t)+2x{e}^{x^2+2xy+2xt}\cdot \cos t+2x{e}^{x^2+2xy+2xt}\\ & =e^{x^2+2xy+2xt}(-2\sin t(x+y+z)+2x\cos t+2x)\end{array}$$

例3，设 $u=xy\sin z,z=e^{x^2+y}$，求 ${\displaystyle \frac{\partial u}{\partial x},\frac{\partial u}{\partial y}}$。

解：$$\begin{array}{rl}\frac{\partial u}{\partial x}& =\frac{\partial f}{\partial x}+\frac{\partial f}{\partial z}\cdot \frac{\partial z}{\partial x}\\ & =y\sin z+xy\cos z\cdot (2x{e}^{x^2+y})\\ & =y\sin z+2x^{2}y\cos z\cdot e^{x^2+y}\end{array}$$

$$\begin{array}{rl}\frac{\partial u}{\partial y}& =\frac{\partial f}{\partial y}+\frac{\partial f}{\partial z}\cdot \frac{\partial z}{\partial y}\\ & =x\sin z+xy\cos z\cdot e^{x^2+y}\end{array}$$

例4，设 $u=f(x^2-y^2,e^{xy})$，设 $f$ 具有一阶连续偏导数，求 ${\displaystyle \frac{\partial u}{\partial x},\frac{\partial u}{\partial y}}$。

解：令 $u=f(s,t),s=x^2-y^2,t=e^{xy}$，则

$$\begin{array}{rl}\frac{\partial u}{\partial x}& =\frac{\partial f}{\partial s}\cdot \frac{\partial s}{\partial x}+\frac{\partial f}{\partial t}\cdot \frac{\partial t}{\partial x}\\ & =f_1^{\prime }\cdot 2x+f_2^{\prime }\cdot y{e}^{xy}\end{array}$$

$$\begin{array}{rl}\frac{\partial u}{\partial y}& =\frac{\partial f}{\partial s}\cdot \frac{\partial s}{\partial y}+\frac{\partial f}{\partial t}\cdot \frac{\partial t}{\partial y}\\ & =f_1^{\prime }\cdot (-2y)+f_2^{\prime }\cdot x{e}^{xy}\end{array}$$

这里我们记 $f_1^{\prime }=\frac{\partial f}{\partial s},f_2^{\prime }=\frac{\partial f}{\partial t}$。