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复合函数的中间变量为多元函数的情形

定理 2 设 $u=u(x, y), v=v(x, y)$ 在点 $(x, y)$ 处都具有偏导数 $\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}$ 及 $\frac{\partial v}{\partial x}, \frac{\partial v}{\partial y}$ ， 函数 $z=f(u, v)$ 在对应点 $(u, v)$ 具有连续的偏导数 $\frac{\partial z}{\partial u}$ 和 $\frac{\partial z}{\partial v} ，$
则复合函数 $z=f[u(x, y), v(x, y)]$ 在 $(x, y)$ 处的两个偏导数存在，并有求导公式

## 复合函数的中间变量为多元函数的情形
$$
\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u} \frac{\partial u}{\partial x}+\frac{\partial z}{\partial v} \frac{\partial v}{\partial x},
$$

$$
\frac{\partial z}{\partial y}=\frac{\partial z}{\partial u} \frac{\partial u}{\partial y}+\frac{\partial z}{\partial v} \frac{\partial v}{\partial y} \text {. }
$$

定理证明从略.
定理 2 中的复合函数的函数的结构图是
![图片](/uploads/2022-12/image_20221231c2c1f85.png)
我们可以借助函数结构图，利用前面分析的方法与结论，能够直接写出公式
(2) 和公式 (3) 的求导公式.
例 3 设 $z=\mathrm{e}^u \sin v$, 而 $u=x y, v=x+y$, 求 $\frac{\partial z}{\partial x}$ 和 $\frac{\partial z}{\partial y}$.
解 由公式 (2) 和 (3) 可得

$$
\begin{aligned}
\frac{\partial z}{\partial x} & =\frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x}+\frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial x}=\mathrm{e}^u \sin v \cdot y+\mathrm{e}^u \cos v \cdot 1 \\
& =\mathrm{e}^u(y \sin v+\cos v)=\mathrm{e}^{x y}[y \sin (x+y)+\cos (x+y)] \\
\frac{\partial z}{\partial y} & =\frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial y}+\frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial y}=\mathrm{e}^u \sin v \cdot x+\mathrm{e}^u \cos v \cdot 1 \\
& =\mathrm{e}^u(x \sin v+\cos v)=\mathrm{e}^{\mathrm{xy}}[x \sin (x+y)+\cos (x+y)]
\end{aligned}
$$

例 4 求 $z=\left(3 x^2+y^2\right)^{4 x+2 y}$ 的偏导数.
解 设 $u=3 x^2+y^2 ， v=4 x+2 y$ ，则 $z=u^v$.
于是 $\frac{\partial z}{\partial u}=v \cdot u^{v-1}, \frac{\partial z}{\partial v}=u^v \cdot \ln u, \frac{\partial u}{\partial x}=6 x ， \frac{\partial u}{\partial y}=2 y ， \frac{\partial v}{\partial x}=4 ， \frac{\partial v}{\partial y}=2$.
则 $\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u} \frac{\partial u}{\partial x}+\frac{\partial z}{\partial v} \frac{\partial v}{\partial x}=v \cdot u^{v-1} \cdot 6 x+u^v \cdot \ln u \cdot 4$

$$
=6 x(4 x+2 y)\left(3 x^2+y^2\right)^{4 x+2 y-1}+4\left(3 x^2+y^2\right)^{4 x+2 y} \ln \left(3 x^2+y^2\right) .
$$

$$
\begin{aligned}
& \frac{\partial z}{\partial y}=\frac{\partial z}{\partial u} \frac{\partial u}{\partial y}+\frac{\partial z}{\partial v} \frac{\partial v}{\partial y}=v \cdot u^{v-1} \cdot 2 y+u^v \cdot \ln u \cdot 2 \\
& =2 y(4 x+2 y)\left(3 x^2+y^2\right)^{4 x+2 y-1}+2\left(3 x^2+y^2\right)^{4 x+2 y} \ln \left(3 x^2+y^2\right)
\end{aligned}
$$

这种类型的题目也可以用全微分的方法来解决，感兴趣的读者可以试一下.

**复合函数的中间变量既有一元也有为多元函数的情形**
这种情形比较复杂，我们仅以一种情况为例，其他的类似可得.
定理 3 如果函数 $u=u(x, y)$ 在点 $(x, y)$ 具有对 $x$ 及对 $y$ 的偏导数，函数 $v=v(y)$ 在点 $y$ 可导，函数 $z=f(u, v)$ 在对应点 $(u, v)$ 具有连续偏导数，则复合函 数 $z=f[u(x, y), v(y)]$ 在对应点 $(x, y)$ 的两个偏导数存在，且有
定理证明从略. 该复合函数的结构图为

![图片](/uploads/2022-12/image_20221231dec92fc.png)
例 5 设函数 $z=\mathrm{e}^{u^2+v^2}$ ，而 $u=x^2 \sin y, v=\cos y$ ，求 $\frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}$.
解 由 (4)，(5) 可知，

$$
\begin{aligned}
& \frac{\partial z}{\partial x}=\mathrm{e}^{u^2+v^2} \cdot 2 u \cdot 2 x \sin y=4 x^3 \sin ^2 y \mathrm{e}^{x^4 \sin ^2 y+\cos ^2 y} \\
& \frac{\partial z}{\partial y}=\mathrm{e}^{u^2+v^2} \cdot 2 u \cdot x^2 \cos y+\mathrm{e}^{u^2+v^2} \cdot 2 v \cdot(-\sin y)=\mathrm{e}^{x^4 \sin ^2 y+\cos ^2 y}\left(x^4-1\right) \sin 2 y
\end{aligned}
$$

例 6 设 $z=f(x, y, u)=(x-y)^u, u=x y$ ，求 $\frac{\partial z}{\partial x} ， \frac{\partial z}{\partial y}$.
解 函数的结构图为

![图片](/uploads/2022-12/image_20221231ec8f3c9.png)
其中 $x, y$ 既是复合函数的中间变量，又是自变量. 则

$$
\begin{aligned}
& \frac{\partial z}{\partial x}=\frac{\partial f}{\partial x} \frac{\mathrm{d} x}{\mathrm{~d} x}+\frac{\partial f}{\partial u} \frac{\partial u}{\partial x} \\
&=u(x-y)^{u-1} \cdot 1+(x-y)^u \ln (x-y) \cdot y \\
&=x y(x-y)^{x y-1} \cdot 1+y(x-y)^{x y} \ln (x-y), \\
& \frac{\partial z}{\partial y}=\frac{\partial f}{\partial y} \frac{\mathrm{d} y}{\mathrm{~d} y}+\frac{\partial f}{\partial u} \frac{\partial u}{\partial y} \\
&=u(x-y)^{u-1} \cdot(-1) \cdot 1+(x-y)^u \ln (x-y) \cdot x=-x y(x-y)^{x y-1}+x(x-y)^{x y} \ln (x-y) .
\end{aligned}
$$

注 等号两边都有 $\frac{\partial z}{\partial x}$ ，但这两个符号的含义是不一样的，左边的是二元函数 $z=(x-y)^y$ 对 $x$ 的偏导数，右边的是三元函数 $z=f(x, y, u)=(x-y)^u$ 对 $x$ 的偏导数. 为了表示区别, 等号右边的 $\frac{\partial z}{\partial x}$ 常写作 $\frac{\partial f}{\partial x}$. 同理, 等号两边的 $\frac{\hat{\partial z}}{\partial y}$ 的含义也是不一样 的，等号右边的 $\frac{\hat{c}}{\partial y}$ 也常写作 $\frac{\partial}{\partial}$.

例 7 设 $z=f\left(\sin x, x^2-y^2\right) ， f$ 具有一阶连续的偏导数，求 $\frac{\partial z}{\partial x} ， \frac{\partial z}{\partial y}$.
解 设 $u=\sin x, v=x^2-y^2$ ， 则函数 $z=f\left(\sin x, x^2-y^2\right)$ 是由函数 $z=f(u, v)$ ， $u=\sin x, v=x^2-y^2$ 复合而成. 由函数的结构图
![图片](/uploads/2022-12/image_20221231fb818d6.png)

可得 $\quad \frac{\partial z}{\partial x}=\frac{\partial f}{\partial u} \frac{\mathrm{d} u}{\mathrm{~d} x}+\frac{\partial f}{\partial v} \frac{\mathrm{d} v}{\mathrm{~d} x}=\cos x \cdot f_u+2 x f_v ， \frac{\partial z}{\partial y}=\frac{\partial f}{\partial v} \frac{\partial v}{\partial y}=-2 y f_v$.
注 有时，为表达简便起见，引入以下记号:

$$
f_1^{\prime}(u, v)=f_u(u, v), \quad f_2^{\prime}(u, v)=f_v(u, v),
$$

这里下标 1 表示对第一个变量求偏导数，下标 2 表示对第二个变量求偏导 数，利用这样的记号，例 7 的结果可以表示为

$$
\frac{\partial z}{\partial x}=\cos x \cdot f_1^{\prime}+2 x f_2^{\prime}, \quad \frac{\partial z}{\partial y}=-2 y f_2^{\prime}
$$

同理也可以引入 $f_{11}^{\prime \prime} 、 f_{12}^{\prime \prime} 、 f_{22}^{\prime \prime}$ 等记号
例 8 (1) 设 $u=f(x+y+z, x y z)$ ，其中 $f$ 具有二阶偏导数，求 $\frac{\partial u}{\partial x} ， \frac{\partial^2 u}{\partial x \partial z}$;
（2）设 $u=f(x, x y, x y z)$ ，其中 $f$ 具有二阶连续偏导数，求 $\frac{\partial^2 u}{\partial x \partial y}$.
解 (1)

$$
\begin{aligned}
\frac{\partial u}{\partial x}=f_1^{\prime}+f_2^{\prime} y z, \frac{\partial^2 u}{\partial x \partial z} & =\frac{\partial}{\partial x}\left(f_1^{\prime}\right)+\frac{\partial}{\partial z}\left(f_2^{\prime} y z\right) \\
& =f_{11}^{\prime \prime}+f_{12}^{\prime \prime} x y+\left(f_{21}^{\prime \prime}+f_{22}^{\prime \prime} x y\right) y z+f_2^{\prime} y
\end{aligned}
$$

注 本题条件中并没有二阶偏导数连续，因此 $f_{12}^{\prime \prime}$ 与 $f_{21}^{\prime \prime}$ 末必相等，因此不要将 其合并.
例 8 (2) 设 $u=f(x, x y, x y z)$ ，其中 $f$ 具有二阶连续偏导数，求 $\frac{\partial^2 u}{\partial x \partial y}$.
(2) $\frac{\partial u}{\partial x}=f_1^{\prime}+y f_2^{\prime}+y z f_3^{\prime}$,

$$
\begin{aligned}
\frac{\partial^2 u}{\partial x \partial y} & =\frac{\partial}{\partial x}\left(f_1^{\prime}\right)+\frac{\partial}{\partial x}\left(y f_2^{\prime}\right)+\frac{\partial}{\partial x}\left(y z f_3^{\prime}\right) \\
& =x f_{12}^{\prime \prime}+x z f_{13}^{\prime \prime}+y\left(x f_{22}^{\prime \prime}+x z f_{23}^{\prime \prime}\right)+f_2^{\prime}+y z\left(x f_{32}^{\prime \prime}+x z f_{33}^{\prime \prime}\right)+z f_3^{\prime} \\
& =f_2^{\prime}+z f_3^{\prime}+x f_{12}^{\prime \prime}+x z f_{13}^{\prime \prime}+x y f_{22}^{\prime \prime}+2 x y z f_{23}^{\prime \prime}+x z f_{33}^{\prime \prime} .
\end{aligned}
$$

注 本题条件中有二阶偏导数连续，因此 $f_{23}^{\prime \prime}=f_{32}^{\prime \prime}$ ， 因此需要将其合并.
例 9 设函数 $u=f(x, y)$ 具有二阶连续偏导数，将下列表达式转换为极坐标 的形式:
(1) $\left(\frac{\partial u}{\partial x}\right)^2+\left(\frac{\partial u}{\partial y}\right)^2$;
(2) $\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}$.
解 (1) 直角坐标系与极坐标系的关系 : $\left\{\begin{array}{l}x=r \cos \theta \\ y=r \sin \theta\end{array}\right.$ ，或 $\left\{\begin{array}{l}r=\sqrt{x^2+y^2} \\ \theta=\arctan \frac{y}{x}\end{array}\right.$ ， $u=f(x, y)=f(r \cos \theta, r \sin \theta)=F(r, \theta)$,
例 9 设函数 $u=f(x, y)$ 具有二阶连续偏导数，将下列表达式转换为极坐标
的形式: $u=f(x, y)=f(r \cos \theta, r \sin \theta)=F(r, \theta)$ ，

$$
\begin{aligned}
\frac{\partial r}{\partial x}= & \frac{x}{\sqrt{x^2+y^2}}=\frac{x}{r}=\cos \theta, \frac{\partial \theta}{\partial x}=\frac{1}{1+\left(\frac{y}{x}\right)^2} \cdot\left(-\frac{y}{x^2}\right)=-\frac{y}{x^2+y^2}=-\frac{x}{r^2}-\frac{\sin t}{r} \\
& \frac{\partial r}{\partial y}=\frac{y}{\sqrt{x^2+y^2}}=\frac{y}{r}=\sin \theta, \frac{\partial \theta}{\partial y}=\frac{1}{1+\left(\frac{y}{x}\right)^2} \cdot \frac{1}{x}=\frac{x}{x^2+y^2}=\frac{\cos }{r}
\end{aligned}
$$

例 9 设函数 $u=f(x, y)$ 具有二阶连续偏导数，将下列表达式转换为极坐标 的形式: $\frac{\partial r}{\partial x}=\cos \theta, \frac{\partial \theta}{\partial x}=-\frac{x}{r^2}-\frac{\sin \theta}{r}$

$$
\frac{\partial u}{\partial x}=\frac{\partial u}{\partial r} \cdot \frac{\partial r}{\partial x}+\frac{\partial u}{\partial \theta} \cdot \frac{\partial \theta}{\partial x}=\frac{\partial u}{\partial r} \cdot \frac{x}{r}-\frac{\partial u}{\partial \theta} \cdot \frac{y}{r^2}=\frac{\partial u}{\partial r} \cos \theta-\frac{\partial u}{\partial \theta} \cdot \frac{\sin \theta}{r}
$$

$$
\frac{\partial r}{\partial y}=\sin \theta, \frac{\partial \theta}{\partial y}=\frac{\cos \theta}{r}
$$

因此

$$
\begin{aligned}
\left(\frac{\partial u}{\partial x}\right)^2+\left(\frac{\partial u}{\partial y}\right)^2 & =\left(\frac{\partial u}{\partial r} \cos \theta-\frac{\partial u}{\partial \theta} \cdot \frac{\sin \theta}{r}\right)^2+\left(\frac{\partial u}{\partial r} \sin \theta+\frac{\partial u}{\partial \theta} \cdot \frac{\cos \theta}{r}\right)^2 \\
& =\left(\frac{\partial u}{\partial r}\right)^2+\frac{1}{r^2}\left(\frac{\partial u}{\partial \theta}\right)^2
\end{aligned}
$$

例 9 设函数 $u=f(x, y)$ 具有二阶连续偏导数，将下列表达式转换为极坐标 的形式: $\frac{\partial u}{\partial x}=\frac{\partial u}{\partial r} \cos \theta-\frac{\partial u}{\partial \theta} \cdot \frac{\sin \theta}{r}$

$$
\text { ( 2 ) } \begin{aligned}
\left(\frac{\partial^2 u}{\partial x^2}\right)= & \frac{\partial}{\partial r}\left(\frac{\partial u}{\partial x}\right) \cdot \frac{\partial r}{\partial x}+\frac{\partial}{\partial \theta}\left(\frac{\partial u}{\partial x}\right) \cdot \frac{\partial \theta}{\partial x} \\
= & \frac{\partial}{\partial r}\left(\frac{\partial u}{\partial r} \cos \theta-\frac{\partial u}{\partial \theta} \cdot \frac{\sin \theta}{r}\right) \cos \theta-\frac{\partial}{\partial \theta}\left(\frac{\partial u}{\partial r} \cos \theta-\frac{\partial u}{\partial \theta} \cdot \frac{\sin \theta}{r}\right) \frac{\sin \theta}{r} \\
= & \left(\frac{\partial^2 u}{\partial r^2} \cos \theta-\frac{\partial^2 u}{\partial \theta \partial r} \sin \theta+\frac{\partial u}{\partial \theta} \cdot \frac{\sin \theta}{r^2}\right) \cos \theta \\
& -\left(\frac{\partial^2 u}{\partial r \partial \theta} \cos \theta-\frac{\partial u}{\partial r} \sin \theta-\frac{\partial^2 u}{\partial \theta^2} \cdot \frac{\sin \theta}{r}-\frac{\partial u}{\partial \theta} \frac{\cos \theta}{r}\right) \frac{\sin \theta}{r} \\
= & \frac{\partial^2 u}{\partial r^2} \cos ^2 \theta-2 \frac{\partial^2 u}{\partial \theta \partial r} \cdot \frac{\sin \theta \cos \theta}{r}+2 \frac{\partial u}{\partial \theta} \frac{\sin \theta \cos \theta}{r^2}+\frac{\partial u}{\partial r} \frac{\sin ^2 \theta}{r}+\frac{\partial^2 u}{\partial \theta^2} \cdot \frac{\sin ^2 \theta}{r^2}
\end{aligned}
$$

例 9 设函数 $u=f(x, y)$ 具有二阶连续偏导数，将下列表达式转换为极坐标 的形式: $\frac{\partial u}{\partial y}=\frac{\partial u}{\partial r} \sin \theta-\frac{\partial u}{\partial \theta} \cdot \frac{\cos \theta}{r}$
同理，

$$
\frac{\partial^2 u}{\partial y^2}=\frac{\partial^2 u}{\partial r^2} \sin ^2 \theta+2 \frac{\partial^2 u}{\partial \theta \partial r} \cdot \frac{\sin \theta \cos \theta}{r}-2 \frac{\partial u}{\partial \theta} \frac{\sin \theta \cos \theta}{r^2}+\frac{\partial u}{\partial r} \frac{\cos ^2 \theta}{r}+\frac{\partial^2 u}{\partial \theta^2} \cdot \frac{\sin ^2 \theta}{r^2}
$$

因此，

$$
\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=\frac{\partial^2 u}{\partial r^2}+\frac{1}{r} \frac{\partial u}{\partial r}+\frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2}=\frac{1}{r^2}\left(r \frac{\partial}{\partial r}\left(r \frac{\partial u}{\partial r}\right)+\frac{\partial^2 u}{\partial \theta^2}\right)
$$

例 9 设函数 $u=f(x, y)$ 具有二阶连续偏导数，将下列表达式转换为极坐标
的形式: (2 ) $\left(\frac{\partial^2 u}{\partial x^2}\right)=\frac{\partial}{\partial r}\left(\frac{\partial u}{\partial x}\right) \cdot \frac{\partial r}{\partial x}+\frac{\partial}{\partial \theta}\left(\frac{\partial u}{\partial x}\right) \cdot \frac{\partial \theta}{\partial x}$
注 也可将 $r 、 \theta$ 看作是自变量， $x 、 y$ 看作是中间变量.

$$
\begin{aligned}
& \frac{\partial u}{\partial r}=\frac{\partial u}{\partial x} \cdot \frac{\partial x}{\partial r}+\frac{\partial u}{\partial y} \cdot \frac{\partial y}{\partial r}=\frac{\partial u}{\partial x} \cos \theta+\frac{\partial u}{\partial y} \sin \theta \\
& -=-{ }^2-{ }^{\frac{\partial u}{\partial \theta}}=\frac{\partial u}{\partial x} \cdot \frac{\partial x}{\partial \theta}+\frac{\partial u}{\partial y} \cdot \frac{\partial y}{\partial \theta}=\frac{\partial u}{\partial x} \cdot(-\sin \theta)+\frac{\partial u}{\partial y} r \cos \theta
\end{aligned}
$$

以上两式联立，解出

$$
\frac{\partial u}{\partial x}=\frac{\partial u}{\partial r} \cos \theta-\frac{\partial u}{\partial \theta} \cdot \frac{\sin \theta}{r}, \frac{\partial u}{\partial y}=\frac{\partial u}{\partial r} \cdot \sin \theta+\frac{\partial u}{\partial \theta} \cdot \frac{\cos \theta}{r}
$$

## 本章在线教程

https://www.bilibili.com/video/BV1Eb411u7Fw?p=101

https://www.bilibili.com/video/BV1Eb411u7Fw?p=102

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