# [高数]记录一个较复杂的复合函数求偏导过程

## 正文

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

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

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

$$\frac{du}{dr} \frac{\partial r}{\partial x}.$$

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

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

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

$$\frac{x}{r}.$$

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

$$\frac{x}{r} \frac{du}{dr}.$$

$$A = \frac{\partial u}{\partial x} = \frac{x}{r} \frac{du}{dr}.$$

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

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

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

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

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

$$\frac{x}{r} (\frac{x}{r} \frac{du}{dr})_{r}^{‘} =$$

$$\frac{x}{r} [(\frac{x}{r})_{r}^{‘} \frac{du}{dr} + \frac{x}{r} (\frac{du}{dr})_{r}^{‘}].$$

$$\frac{\partial \sqrt{x^{2} + y^{2}}}{\partial x}.$$

$$r(x,y) = \sqrt{x^{2} + y^{2}} \Rightarrow$$

$$x(r,y) = \sqrt{r^{2} – y^{2}} ①$$

$$(\frac{x}{r})_{r}^{‘} =$$

$$(x \cdot \frac{1}{r})_{r}^{‘} =$$

$$(x)_{r}^{‘} \cdot \frac{1}{r} + x \cdot (\frac{1}{r})_{r}^{‘} =$$

$$(\sqrt{r^{2} – y^{2}})_{r}^{‘} \cdot \frac{1}{r} + x \cdot (\frac{1}{r})_{r}^{‘} =$$

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

$$\frac{1}{\sqrt{r^{2} – y^{2}}} – \frac{x}{r^{2}} =$$

$$\frac{1}{x} – \frac{x}{r^{2}} ②$$

$$\frac{x}{r} [(\frac{x}{r})_{r}^{‘} \frac{du}{dr} + \frac{x}{r} (\frac{du}{dr})_{r}^{‘}] =$$

$$\frac{x}{r} [(\frac{1}{x} – \frac{x}{r^{2}}) \frac{du}{dr} + \frac{x}{r} (\frac{du}{dr})_{r}^{‘}] =$$

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

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

$$\frac{\partial^{2} u}{\partial y^{2}} = \frac{du}{dr}(\frac{1}{r} – \frac{y^{2}}{r^{3}}) + \frac{d^{2}u}{dr^{2}} \cdot \frac{y^{2}}{r^{2}}.$$

## 注意

$$\frac{\partial^{2} u}{\partial x^{2}} = \frac{\partial}{\partial x}(\frac{\partial u}{\partial x}).$$

$$\frac{\partial^{2} u}{\partial x^{2}} \neq (\frac{\partial u}{\partial x}) (\frac{\partial u}{\partial x}).$$

$$\frac{\partial^{2} u}{\partial x^{2}} \neq (\frac{\partial u}{\partial x})^{2}.$$

EOF