求导去积分符号，积分去求导符号

一、题目

已知 $f(x)$ 为连续函数，且 $f(x)$ $=$ $\int_{0}^{x} \mathrm{e}^{ -f(t) } \mathrm{~d} t$, 则当 $n$ $\geqslant$ $2$ 时：

$$
f^{(n)}(0) = ?
$$

难度评级：

二、解析

首先要知道，本题隐藏着一个已知条件：

$$
\textcolor{yellow}{
\begin{aligned}
f(0) \\ \\
& = \int_{0}^{x} \mathrm{e}^{ -f(t) } \mathrm{~d} t \Bigg|_{x = 0} \\ \\
& = 0
\end{aligned}
}
$$

于是：

$$
\begin{aligned}
& f(\textcolor{springgreen}{x}) = \int_{0}^{\textcolor{springgreen}{x}} \mathrm{e}^{ -f(\textcolor{orangered}{t}) } \mathrm{ ~d } \textcolor{orangered}{t} \\ \\
\Rightarrow & \ \text{等号两边同时对} \ \textcolor{springgreen}{x} \ \text{求导} \\ \\
\Rightarrow & \ f^{ \prime }(\textcolor{springgreen}{x}) = \mathrm{e}^{ -f(\textcolor{springgreen}{x}) } \\ \\
\Rightarrow & \ \textcolor{pink}{e^{ f(x) }} \cdot f^{ \prime }(\textcolor{springgreen}{x}) = \textcolor{pink}{e^{ f(x) }} \cdot \mathrm{e}^{ -f(\textcolor{springgreen}{x}) } \\ \\
\Rightarrow & \ e^{ f(\textcolor{springgreen}{x}) } \cdot f ^ { \prime } (\textcolor{springgreen}{x} ) = 1 \\ \\
\Rightarrow & \ \text{等号两边同时对} \ \textcolor{springgreen}{x} \ \text{积分} \\ \\
\Rightarrow & \ \int e^{ f(\textcolor{springgreen}{x}) } \cdot f ^ { \prime } (\textcolor{springgreen}{x} ) \mathrm{~d} x = \int 1 \mathrm{~d} x \\ \\
\Rightarrow & \ \mathrm { e } ^ { f ( \textcolor{springgreen}{x} ) } = \textcolor{springgreen}{x} + C \\ \\
\Rightarrow & \ \textcolor{yellow}{ f(0) = 0 \rightarrow \mathrm{e}^{0} = C \rightarrow C = 1 } \\ \\
\Rightarrow & \ \mathrm { e } ^ { f ( x ) } = x + 1 \\ \\
\Rightarrow & \ \textcolor{springgreen}{ f ( x ) = \ln ( x + 1 ) }
\end{aligned}
$$

由于 $\ln ( 1+x )$ 在 $x = 0$ 处的泰勒展开式（麦克劳林展开式）为：

$$
\textcolor{yellow}{
\begin{aligned}
\ln (1+x) & = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n+1} x^{n+1} \\ \\
& = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} x^{n}
\end{aligned}
}
$$

接着，根据泰勒公式可知，泰勒展开式的第 $n$ 项为：

$$
\begin{aligned}
& \frac{f^{(n)} (x_{0})}{n!} \cdot (x – x_{0})^{n} \\ \\
\Rightarrow & \ x_{0} = 0 \\ \\
\Rightarrow & \ \frac{f^{(n)}(0)}{n!} \cdot x^{n}
\end{aligned}
$$

Tip

补充内容：《考研数学常用的泰勒公式》

zhaokaifeng.com

所以：

$$
\begin{aligned}
& \frac{f^{(n)}(0)}{n!} \cdot x^{n} = \frac{(-1)^{n-1}}{n} x^{n} \\ \\
\Rightarrow & \ \frac{f^{(n)}(0)}{n!} = \frac{(-1)^{n-1}}{n} \\ \\
\Rightarrow & \ f^{(n)}(0) = \frac{(-1)^{n-1}}{n} \cdot \textcolor{orangered}{n !} \\ \\
\Rightarrow & \ f^{(n)}(0) = (-1)^{n-1} \cdot \textcolor{orangered}{ \frac{n (n-1) !}{n} } \\ \\
\Rightarrow & \ \textcolor{springgreen}{\boldsymbol{ f^{(n)}(0) = (-1)^{n-1}(n-1)! }}
\end{aligned}
$$

相关例题

[01]. 使用麦克劳林公式（泰勒公式）求解函数 n 阶导

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