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：

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

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

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

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

即：

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

接着：

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

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

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

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

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

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

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

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

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

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

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

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

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

即：

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

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

同理可得：

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

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

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

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