最后更新: 2025-04-02 15:30 查看: 866 次反馈刷题

多元复合函数求导举例 Part3

## 多元复合函数求导举例

> 多元复合函数求导拿到试题后，首先就是画图，标注变量之间的关系，然后找到对应的路径，这在复杂的题目里，尤其有用。阅读本文前最好已经阅读了上一节内容

`例` 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)

从图中可以看到 **$z$到$x$ 有两条路**
所以可得 $\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 $.

而 **$z$到$y$有1条路** ，所以
$\frac{\partial z}{\partial y}=\frac{\partial f}{\partial v} \frac{\partial v}{\partial y}=-2 y f_v$

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，上例的结果可以表示为**

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

同理也可以引入 $f_{11}^{\prime \prime} 、 f_{12}^{\prime \prime} 、 f_{22}^{\prime \prime}$ 等记号

`例`(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.

(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` 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 $\left\{\begin{array}{l}r=\sqrt{x^2+y^2} \\ \theta=\arctan \frac{y}{x}\end{array}\right.$ ， $u=f(x, y)=f(r \cos \theta, r \sin \theta)=F(r, \theta)$,

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

$\frac{\partial r}{\partial x}=\cos \theta, \frac{\partial \theta}{\partial x}=-\frac{x}{r^2}-\frac{\sin \theta}{r}$

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

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

因此平方相加得

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

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

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

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

第(2)问里， $\left(\frac{\partial^2 u}{\partial x^2}\right)=\frac{\partial}{\partial r}\left(\frac{\partial u}{\partial x}\right) \cdot \frac{\partial r}{\partial x}+\frac{\partial}{\partial \theta}\left(\frac{\partial u}{\partial x}\right) \cdot \frac{\partial \theta}{\partial x}$
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

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

开VIP会员

非会员每天6篇，会员每天16篇，VIP会员无限制访问

题库训练自我测评投稿

上一篇：复合求导链式法则 Part2

下一篇：全微分形式的不变性

本文对您是否有用？有用 (1) 无用 (0)