多元复合函数求导法则

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一元复合函数：$y=f(\varphi(x)) \Leftrightarrow y=f(u), u=\varphi(x)$

其求导有链式法则：$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d} y}{\mathrm{~d} u} \frac{\mathrm{d} u}{\mathrm{~d} x}$

画出函数关系图：$y\rightarrow u\rightarrow x$，可见从$y$到x$$有一条路径,所以结果是1项的和，每一段路径（对应一个导数）乘起来。

全导数

一元函数与多元函数复合的情形

如果函数$u=\varphi(t)$及$v=\psi(t)$都在点$t$可导，函数$z=f(u, v)$在对应 点$(u, v)$具有连续偏导数，那么复合函数$z=f[\varphi(t), \psi(t)]$在点$t$可导，且有

\begin{align}\frac{\mathrm{d} z}{\mathrm{~d} t}=\frac{\partial z}{\partial u}\cdot\frac{\mathrm{d} u}{\mathrm{~d} t}+\frac{\partial z}{\partial v} \cdot\frac{\mathrm{d} v}{\mathrm{~d} t}\end{align}

注意：$\partial$的表示偏导数，$\mathrm{d}$表示一元函数的导数

多元函数与多元函数复合的情形

如果函数$u=\varphi(x,y)$及$v=\psi(x,y)$都在点$(x,y)$具有对$x$及对$y$的偏导数，函数$z=f(u,v)$在对应点$(u,v)$具有连续偏导数，那么复合函数$z=f[\varphi(x,y),\psi(x,y)]$在点$(x,y)$的两个偏导数都存在，且有

\begin{align} \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 y}=\frac{\partial z}{\partial u} \frac{\partial u}{\partial y}+\frac{\partial z}{\partial v} \frac{\partial v}{\partial y} \end{align}

混合情形

如果函数$u=\varphi(x,y)$在点$(x,y)$具有对$x$及对$y$的偏导数，函数$v=\psi(y)$在点$y$可导，函数$z=f(u,v)$在对应点$(u,v)$具有连续偏导数，那么复合函数$z=f[\varphi(x,y),\psi(y)]$在点$(x,y)$的两个偏导数都存在，且有

\begin{align} &\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u} \frac{\partial u}{\partial x} \\& \frac{\partial z}{\partial y}=\frac{\partial z}{\partial u} \frac{\partial u}{\partial y}+\frac{\partial z}{\partial v} \frac{\mathrm{d} v}{\mathrm{~d} y} \end{align}

例题

设$z=u^{2} e^{v}, \quad u=\sin t, \quad v=e^{t}$。求$\frac{d z}{d t}$

$\begin{aligned} \frac{d z}{d t} & =\frac{\partial z}{\partial u} \cdot \frac{d u}{d t}+\frac{\partial z}{\partial v} \cdot \frac{d v}{d t} \\ & =2 u e^{v} \cdot \cos t+u^{2} e^{v} \cdot e^{t} \\ & =2 \sin t \cdot e^{e^{t}} \cdot \cos t+\sin ^{2} t \cdot e^{e^{t}} \cdot e^{t} \end{aligned}$

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设$z=\mathrm{e}^{u}\sin v$，而$u=xy,v=x+y$。求$\frac{\partial z}{\partial x}$和$\frac{\partial z}{\partial y}$

\begin{align} \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}=\mathrm{e}^{u} \sin v \cdot y+\mathrm{e}^{u} \cos v \cdot 1 \\ & =\mathrm{e}^{x y}[y \sin (x+y)+\cos (x+y)] \\ \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}=\mathrm{e}^{u} \sin v \cdot x+\mathrm{e}^{u} \cos v \cdot 1 \\ & =\mathrm{e}^{\mathrm{x} y}[x \sin (x+y)+\cos (x+y)] \end{align}

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设$u=f(x,y,z)=\mathrm{e}^{x^{2}+y^{2}+z^{2}}$，而$z=x^{2}\sin y$。求$\frac{\partial u}{\partial x}$和$\frac{\partial u}{\partial y}$

\begin{align} \frac{\partial u}{\partial x} & =\frac{\partial f}{\partial x}+\frac{\partial f}{\partial z} \frac{\partial z}{\partial x}=2 x \mathrm{e}^{x^{2}+y^{2}+z^{2}}+2 z \mathrm{e}^{x^{2}+y^{2}+z^{2}} \cdot 2 x \sin y \\ & =2 x\left(1+2 x^{2} \sin ^{2} y\right) \mathrm{e}^{x^{2}+y^{2}+x^{4} \sin ^{2} y} \\ \frac{\partial u}{\partial y} & =\frac{\partial f}{\partial y}+\frac{\partial f}{\partial z} \frac{\partial z}{\partial y}=2 y \mathrm{e}^{x^{2}+y^{2}+z^{2}}+2 z \mathrm{e}^{x^{2}+y^{2}+z^{2}} \cdot x^{2} \cos y \\ & =2\left(y+x^{4} \sin y \cos y\right) \mathrm{e}^{x^{2}+y^{2}+x^{4} \sin ^{2} y} \end{align}

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设$z=f(u,v,t)=uv+\sin t$，而$u=\mathrm{e}^{t},v=\cos t$。求全导数$\frac{\mathrm{d}z}{\mathrm{~d}t}$

\begin{align} \frac{\mathrm{d} z}{\mathrm{~d} t} & =\frac{\partial f}{\partial u} \frac{\mathrm{d} u}{\mathrm{~d} t}+\frac{\partial f}{\partial v} \frac{\mathrm{d} v}{\mathrm{~d} t}+\frac{\partial f}{\partial t}=v \mathrm{e}^{t}-u \sin t+\cos t \\ & =\mathrm{e}^{t} \cos t-\mathrm{e}^{t} \sin t+\cos t=\mathrm{e}^{t}(\cos t-\sin t)+\cos t \end{align}

抽象复合函数求偏导

为了写法和计算上的方便，引入记导：

\begin{align} f(u, v):\left\{\begin{array}{l} f_{u}^{\prime} \rightarrow f_{1}^{\prime}, f_{v}^{\prime} \rightarrow f_{2}^{\prime} \\ f_{u u}^{\prime \prime} \rightarrow f_{11}^{\prime \prime}, f_{u v}^{\prime \prime} \rightarrow f_{12}^{\prime \prime}, \cdots\end{array}\right. \end{align}

\begin{align}f(u, v, w):\left\{\begin{array}{l} f_{u}^{\prime} \rightarrow f_{1}^{\prime \prime}, f_{v}^{\prime} \rightarrow f_{2}^{\prime}, f_{w}^{\prime} \rightarrow f_{3}^{\prime} \\ f_{u v}^{\prime \prime} \rightarrow f_{12}^{\prime \prime}, f_{v w}^{\prime \prime} \rightarrow f_{23}^{\prime \prime}, \cdots \end{array}\right. \end{align}

全微分形式不变性

全微分形式不变性设函数$z=f(u,v)$具有连续偏导数，则有全微分

\begin{align}\mathrm{d} z=\frac{\partial z}{\partial u} \mathrm{~d} u+\frac{\partial z}{\partial v} \mathrm{~d} v\end{align}

如果$u$和$v$又是中间变量，即$u=\varphi(x,y),v=\psi(x,y)$，且这两个函数也具有连续偏导数，那么复合函数

\begin{align}z=f[\varphi(x, y), \psi(x, y)]\end{align}

的全微分为

\begin{align}\mathrm{d} z=\frac{\partial z}{\partial x} \mathrm{~d} x+\frac{\partial z}{\partial y} \mathrm{~d} y\end{align}

由此可见，无论$u$和$v$是自变量还是中间变量，还是$z=f(u,v)$的全微分形式是一样的。这个性质叫做全微分形式不变性

例题

利用全微分形式不变性解本节的例题2。设$z=\mathrm{e}^{u}\sin v$，而$u=xy,v=x+y$。求$\frac{\partial z}{\partial x}$和$\frac{\partial z}{\partial y}$

$\mathrm{d} z=\mathrm{d}\left(\mathrm{e}^{u} \sin v\right)=\mathrm{e}^{u} \sin v \mathrm{~d} u+\mathrm{e}^{u} \cos v \mathrm{~d} v$

因为

$\mathrm{d} u=\mathrm{d}(x y)=y \mathrm{~d} x+x \mathrm{~d} y, \mathrm{~d} v=\mathrm{d}(x+y)=\mathrm{d} x+\mathrm{d} y$

代人后归并含$\mathrm{d}x$及$\mathrm{d}y$的项，得

$\mathrm{d} z=\left(\mathrm{e}^{u} \sin v \cdot y+\mathrm{e}^{u} \cos v\right) \mathrm{d} x+\left(\mathrm{e}^{u} \sin v \cdot x+\mathrm{e}^{u} \cos v\right) \mathrm{d} y$

即

\begin{align} &\frac{\partial z}{\partial x} \mathrm{~d} x+\frac{\partial z}{\partial y} \mathrm{~d} y \\ &= \mathrm{e}^{x y}[y \sin (x+y)+\cos (x+y)] \mathrm{d} x+\mathrm{e}^{x y}[x \sin (x+y)+\cos (x+y)] \mathrm{d} y \end{align}

比较上式两边的$\mathrm{d}x$和$\mathrm{d}y$的系数,就同时得到两个偏导数$\frac{\partial z}{\partial x}$和$\frac{\partial z}{\partial y}$，它们与上面的结果一样