# 混合偏导数与次序无关的前提是：混合偏导数连续

## 二、解析

### 第 (1) 问

\begin{aligned} \frac{\partial u}{\partial x} & = y\left[\mathrm{e}^x+f^{\prime}(x)\right] \\ \\ & \Rightarrow \frac{\partial^2 u}{\partial x \partial y}=\mathrm{e}^x+f^{\prime}(x) \end{aligned}

\begin{aligned} \frac{\partial u}{\partial y} & = f^{\prime}(x) \\ \\ & \Rightarrow \frac{\partial^2 u}{\partial y \partial x} = f^{\prime \prime}(x) \end{aligned}

$$\begin{cases} f^{\prime \prime}(x) \\ f^{\prime}(x) \end{cases}$$

$$\begin{cases} \frac{\partial^2 u}{\partial x \partial y} & = \mathrm{e}^x+f^{\prime}(x) \\ \\ \frac{\partial^2 u}{\partial y \partial x} & = f^{\prime \prime}(x) \end{cases}$$

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

\begin{aligned} f^{\prime}(x) + \mathrm{e}^x = f^{\prime \prime}(x) \\ & \Rightarrow f^{\prime \prime}(x)-f^{\prime}(x) = e^x \end{aligned}

$$f(0)=f^{\prime}(0)=1$$

$$\begin{cases} C_1=1 \\ C_2=0 \end{cases}$$

$$\textcolor{springgreen}{ \boldsymbol{ f(x) = x e^{x} + 1 } }$$

### 第 (2) 问

$$f(x) = x e^{x} + 1$$

$$f^{\prime}(x)=e^{x} (x+1)$$

$$f^{\prime}(x) = 0$$

$$x=-1$$

$$f^{\prime \prime}(x)=e^x(x+2)$$

$$f^{\prime \prime}(-1)=e^{-1} > 0$$

*当 $x$ $<$ $-1$ 时, $f^{\prime}(x)$ $=$ $e^{x} (x+1)$ $<0$;

**当 $x$ $>$ $-1$ 时, $f^{\prime}(x)$ $=$ $e^{x} (x+1)$ $>0$, 故 $(-\infty,-1)$ 为单调递减区间, $(-1,+\infty)$ 为单调递增区间