一个复合函数求二阶偏导的例题：$u(x, y)$ $=$ $u(\sqrt{x^{2} + y^{2}})$

一、题目

已知，有 $u(x, y)$ $=$ $u(\sqrt{x^{2} + y^{2}})$, $r$ $=$ $\sqrt{x^{2} + y^{2}}$ $>$ $0$.

并且已知函数 $u(x, y)$ 有二阶连续的偏导数，要求计算：

$\frac{\partial u}{\partial x}$、$\frac{\partial ^{2} u}{\partial x^{2}}$、$\frac{\partial u}{\partial y}$、$\frac{\partial ^{2} u}{\partial y^{2}}$.

二、解析

令 $u$ $=$ $u(r)$, $r$ $=$ $\sqrt{x^{2} + y^{2}}$, 则可知，$u(\sqrt{x^{2} + y^{2}})$ 是一元函数 $u$ $=$ $u(r)$ 和二元函数 $r$ $=$ $\sqrt{x^{2} + y^{2}}$ 的复合函数。

于是：

$$
\frac{\partial u}{\partial x} =
$$

$$
\frac{\mathrm{d} u}{\mathrm{d} r} \cdot \frac{\partial r}{\partial x}
$$

又：

$$
\frac{\partial r}{\partial x} =
$$

$$
[(x^{2} + y^{2})^{\frac{1}{2}}]^{\prime}_{x} =
$$

$$
\frac{1}{2} (x^{2} + y^{2})^{\frac{-1}{2}} \cdot 2x =
$$

$$
\frac{x}{\sqrt{x^{2} + y^{2}}} = \frac{x}{r}
$$

即：

$$
\textcolor{purple}{
\frac{\partial u}{\partial x}} = \textcolor{orange}{\frac{\mathrm{d} u}{\mathrm{d} r} \cdot \frac{x}{r}}
$$

接着：

$$
\frac{\partial ^{2} u}{\partial x^{2}} =
$$

$$
\frac{\partial (\frac{\partial u}{\partial x})}{\partial x \partial x} =
$$

$$
\frac{\partial }{\partial x} (\frac{\partial u}{\partial x}) =
$$

$$
\frac{\partial}{\partial x}(\frac{\mathrm{d} u}{\mathrm{d} r} \cdot \frac{\partial r}{\partial x}) =
$$

$$
\frac{\mathrm{d}}{\mathrm{d r}} (\frac{\mathrm{d} u}{\mathrm{d} r} \cdot \frac{\partial r}{\partial x}) \frac{\partial r}{\partial x} =
$$

$$
\Big[ \frac{\mathrm{d}}{\mathrm{d r}} (\textcolor{yellow}{\frac{\mathrm{d} u}{\mathrm{d} r}} \cdot \textcolor{cyan}{\frac{\partial r}{\partial x}}) \Big] \frac{\partial r}{\partial x} =
$$

按照求导法则 $(a \cdot b)^{\prime}$ $=$ $a^{\prime} b$ $+$ $a b^{\prime}$ 可知，$\frac{\mathrm{d}}{\mathrm{d r}} (\textcolor{yellow}{\frac{\mathrm{d} u}{\mathrm{d} r}} \cdot \textcolor{cyan}{\frac{\partial r}{\partial x}})$ 等于先用 $\textcolor{yellow}{\frac{\mathrm{d} u}{\mathrm{d} r}}$ 对 $r$ 求导，再加上 $\textcolor{cyan}{\frac{\partial r}{\partial x}}$ 对 $r$ 求导。

$$
\Big[ \textcolor{blue}{\frac{\mathrm{d}}{\mathrm{d} r}} (\textcolor{yellow}{\frac{\mathrm{d} u}{\mathrm{d} r}}) \cdot \textcolor{cyan}{\frac{\partial r}{\partial x}} + \textcolor{blue}{\frac{\partial}{\partial r}}(\textcolor{cyan}{\frac{\partial r}{\partial x}}) \cdot \textcolor{yellow}{\frac{\mathrm{d} u}{\mathrm{d} r}} \Big] \frac{\partial r}{\partial x} =
$$

$$
\frac{\mathrm{d}^{2} u}{\mathrm{d} r^{2}} \cdot (\frac{\partial r}{\partial x})^{2} + \frac{\partial}{\partial r} (\frac{\partial r}{\partial x}) \frac{\partial r}{\partial x} \cdot \frac{\mathrm{d} u}{\mathrm{d} r} =
$$

$\textcolor{orange}{\frac{\partial}{\partial r} (\frac{\partial r}{\partial x}) \frac{\partial r}{\partial x}}$ $=$ $\frac{\partial}{\partial x} (\frac{\partial r}{\partial x})$ $=$ $\frac{\partial}{\partial x} (\frac{x}{r})$ $=$ $\frac{r – x \cdot \frac{\partial r}{\partial x}}{r^{2}}$ $=$ $\frac{r – \frac{x^{2}}{r}}{r^{2}}$ $\frac{\frac{r^{2} – x^{2}}{r}}{r^{2}}$ $=$ $\frac{r^{2} – x^{2}}{r^{3}}$ $=$ $\textcolor{orange}{\frac{1}{r}}$ $\textcolor{orange}{-}$ $\textcolor{orange}{\frac{x^{2}}{r^{3}}}$

$$
\frac{\mathrm{d}^{2} u}{\mathrm{d} r^{2}} \cdot (\frac{\partial r}{\partial x})^{2} + (\frac{1}{r} – \frac{x^{2}}{r^{3}}) \cdot \frac{\mathrm{d} u}{\mathrm{d} r} =
$$

$$
\frac{\mathrm{d}^{2} u}{\mathrm{d} r^{2}} \cdot (\frac{x}{r})^{2} + (\frac{1}{r} – \frac{x^{2}}{r^{3}}) \cdot \frac{\mathrm{d} u}{\mathrm{d} r} =
$$

$$
\frac{\mathrm{d}^{2} u}{\mathrm{d} r^{2}} \cdot \frac{x^{2}}{r^{2}} + \frac{\mathrm{d} u}{\mathrm{d} r} \cdot (\frac{1}{r} – \frac{x^{2}}{r^{3}})
$$

即：

$$
\textcolor{purple}{\frac{\partial ^{2} u}{\partial x^{2}}} =
$$

$$
\textcolor{orange}{\frac{\mathrm{d}^{2} u}{\mathrm{d} r^{2}} \cdot \frac{x^{2}}{r^{2}} + \frac{\mathrm{d} u}{\mathrm{d} r} \cdot (\frac{1}{r} – \frac{x^{2}}{r^{3}})}
$$

同理可得：

$$
\textcolor{purple}{\frac{\partial u}{\partial y}} = \textcolor{tan}{\frac{\mathrm{d} u}{\mathrm{d} r} \cdot \frac{y}{r}}
$$

$$
\textcolor{purple}{\frac{\partial ^{2} u}{\partial y^{2}}} =
$$

$$
\textcolor{tan}{\frac{\mathrm{d}^{2} u}{\mathrm{d} r^{2}} \cdot \frac{y^{2}}{r^{2}} + \frac{\mathrm{d} u}{\mathrm{d} r} \cdot (\frac{1}{r} – \frac{y^{2}}{r^{3}})}
$$

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