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。

笔记下载:复合函数的偏导数

1,若 \(z=f(u,v),u=\phi(x,y), v=\psi(x,y)\),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

\[z=f(x,y,t), x=\phi(t),y=\psi(t)\]

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,

\[z=f(u,v,x,y), u=\phi(x,y), v=\psi(x,y)\]

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

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

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{align*}\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+2tx^2\cos y\end{align*}

\begin{align*}\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+2sx^2\cos y\end{align*}

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

解:\begin{align*}\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)+2xe^{x^2+2xy+2xt}\cdot\cos t+2xe^{x^2+2xy+2xt}\\ &=e^{x^2+2xy+2xt}(-2\sin t(x+y+z)+2x\cos t+2x)\end{align*}

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{align*}\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(2xe^{x^2+y})\\ &=y\sin z+2x^2y\cos z\cdot e^{x^2+y}\end{align*}

\begin{align*}\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{align*}

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{align*}\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’\cdot 2x+f_2’\cdot ye^{xy}\end{align*}

\begin{align*}\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’\cdot (-2y)+f_2’\cdot xe^{xy}\end{align*}

这里我们记 \(f_1’=\frac{\partial f}{\partial s}, f_2’=\frac{\partial f}{\partial t}\)。