一、前言 
在本文中,我们主要来讨论一下对下面这个式子怎么求导:
$$
y = x^{x}
$$
二、正文 
在基本求导公式中,有关于 $x^{\mu}$ 和 $a^{x}$ 的求导公式(其中 $\mu$ 和 $a$ 都是常数):
$$
\begin{aligned}
\textcolor{yellow}{\left( x^{\mu} \right) ^{\prime} } & \textcolor{yellow}{=} \textcolor{yellow}{ \mu x^{\mu – 1} } \\
\textcolor{yellow}{ \left( a^{x} \right) ^{\prime} } & \textcolor{yellow}{=} \textcolor{yellow}{ a^{x} \ln a }
\end{aligned}
$$
但是,对于 $x^{x}$, 却并没有直接的求导公式可以使用。
其实,对于类似 $x^{x}$ 这样的式子,需要借助对数函数 $\ln$ 才能完成求导:
$$
\begin{aligned}
& y = x^{x} \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \ln y = \ln \left( x^{x} \right) \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \ln y = x \ln x \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \frac{\mathrm{d}}{\mathrm{d} x} \left( \ln y \right) = \frac{\mathrm{d}}{\mathrm{d} x} \left( x \ln x \right) \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \frac{1}{y} \cdot \frac{\mathrm{d} y}{\mathrm{d} x} = \ln x + 1 \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \frac{\mathrm{d} y}{\mathrm{d} x} = y \left( \ln x + 1 \right) \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \frac{\mathrm{d} y}{\mathrm{d} x} = x^{x} \left( \ln x + 1 \right) \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \textcolor{lightgreen}{ \left( x^{x} \right) ^{\prime} = x^{x} \left( \ln x + 1 \right) }
\end{aligned}
$$
如果 $x^{\textcolor{orange}{f(x)}}$ 的幂指数是一个关于 $x$ 的函数 $f(x)$, 则求导过程如下:
$$
\begin{aligned}
& y = x^{\textcolor{orange}{f(x)}} \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \ln y = \ln \left[ x^{\textcolor{orange}{f(x)}} \right] \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \ln y = \textcolor{orange}{f(x)} \ln x \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \frac{\mathrm{d}}{\mathrm{d} x} \left( \ln y \right) = \frac{\mathrm{d}}{\mathrm{d} x} \left[ \textcolor{orange}{f(x)} \ln x \right] \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \frac{1}{y} \cdot \frac{\mathrm{d} y}{\mathrm{d} x} = \textcolor{orange}{f ^{\prime} (x)} \ln x + \frac{\textcolor{orange}{f(x)}}{x} \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \frac{\mathrm{d} y}{\mathrm{d} x} = y \left[ \textcolor{orange}{f ^{\prime} (x)} \ln x + \frac{\textcolor{orange}{f(x)}}{x} \right] \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \frac{\mathrm{d} y}{\mathrm{d} x} = x^{\textcolor{orange}{f(x)}} \left[ \textcolor{orange}{f ^{\prime} (x)} \ln x + \frac{\textcolor{orange}{f(x)}}{x} \right] \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \textcolor{lightgreen}{ \left[ x^{\textcolor{orange}{f(x)}} \right] ^{\prime} = \left[ \textcolor{orange}{f ^{\prime} (x)} \ln x + \frac{\textcolor{orange}{f(x)}}{x} \right] \cdot x^{\textcolor{orange}{f(x)}} }
\end{aligned}
$$
接下来我们就利用上面的到的有关 $x^{f(x)}$ 的求导公式,计算一道题目:
例题
$$
\lim_{x \rightarrow +\infty} (x^{\frac{1}{x}} – 1)^{\frac{1}{\ln x}} = ?
$$
难度评级:
解析
根据在本文中得到的求导公式 $\textcolor{orange}{ \left[ x^{f(x)} \right] ^{\prime} }$ $\textcolor{orange}{=}$ $\textcolor{orange}{ \left[ f ^{\prime} (x) \ln x + \frac{f(x)}{x} \right] \cdot x^{f(x)} }$, 可得:
$$
\lim_{x \rightarrow +\infty} (x^{\frac{1}{x}} – 1)^{\frac{1}{\ln x}} = \lim_{x \rightarrow +\infty} \mathrm{e}^{\frac{\ln(x^{\frac{1}{x}} – 1)}{\ln x}} = \mathrm{e}^{ \lim_{x \rightarrow +\infty} \frac{\ln(x^{\frac{1}{x}} – 1)}{\ln x}}
$$
又因为,根据洛必达法则,可得:
$$
\begin{aligned}
\lim_{x \rightarrow +\infty} \frac{\ln(x^{\frac{1}{x}} – 1)}{\ln x} \\ \\
= \ & \lim_{x \rightarrow +\infty} \frac{\frac{1}{x^{\frac{1}{x}} – 1} \cdot \textcolor{orange}{ \left( x^{\frac{1}{x}} – 1 \right) ^{\prime} } }{\frac{1}{x}} \\ \\
= \ & \lim_{x \rightarrow +\infty} \frac{\frac{1}{x^{\frac{1}{x}} – 1} \cdot \textcolor{orange}{ \left( x^{\frac{1}{x}} \right) ^{\prime} } }{\frac{1}{x}} \\ \\
= \ & \lim_{x \rightarrow +\infty} \frac{\frac{1}{x^{\frac{1}{x}} – 1} \cdot \textcolor{orange}{\left[ x^{\frac{1}{x}} \cdot \frac{(1 – \ln x)}{x^{2}} \right]}}{\frac{1}{x}} \\ \\
= \ & \lim_{x \rightarrow +\infty} \frac{1}{x^{\frac{1}{x}} – 1} \cdot \left[ x^{\frac{1}{x}} \cdot \frac{(1 – \ln x)}{x^{2}} \right] \cdot x \\ \\
= \ & \lim_{x \rightarrow +\infty} \frac{1}{\mathrm{e}^{\frac{1}{x} \ln x} – 1} \cdot x^{\frac{1}{x}} \cdot \frac{(1 – \ln x)}{x} \\ \\
= \ & \lim_{x \rightarrow +\infty} x^{\frac{1}{x}} \cdot \lim_{x \rightarrow +\infty} \frac{1 – \ln x}{x(\mathrm{e}^{\frac{\ln x}{x}} – 1)} \\ \\
= \ & \lim_{x \rightarrow +\infty} \mathrm{e}^{\frac{\ln x}{x}} \cdot \lim_{x \rightarrow +\infty} \frac{1 – \ln x}{x \cdot \frac{\ln x}{x}} \\ \\
= \ & 1 \cdot \lim_{x \rightarrow +\infty} \frac{1 – \ln x}{\ln x} \\ \\
= \ & \lim_{x \rightarrow +\infty} \frac{- \ln x}{\ln x} \\ \\
= \ & \textcolor{lightgreen}{ -1 }
\end{aligned}
$$
综上可知:
$$
\textcolor{lightgreen}{
\lim_{x \rightarrow +\infty} (x^{\frac{1}{x}} – 1) ^{\frac{1}{\ln x}} = \mathrm{e}^{-1}
}
$$
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