$$$\frac{1}{1 + e^{- x}}$$$ 的二阶导数

求一阶导数 $$$\frac{d}{dx} \left(\frac{1}{1 + e^{- x}}\right)$$$

函数$$$\frac{1}{1 + e^{- x}}$$$是两个函数$$$f{\left(u \right)} = \frac{1}{u}$$$和$$$g{\left(x \right)} = 1 + e^{- x}$$$的复合$$$f{\left(g{\left(x \right)} \right)}$$$。

应用链式法则 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$：

$${\color{red}\left(\frac{d}{dx} \left(\frac{1}{1 + e^{- x}}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\frac{1}{u}\right) \frac{d}{dx} \left(1 + e^{- x}\right)\right)}$$

应用幂次法则 $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$，其中 $$$n = -1$$$:

$${\color{red}\left(\frac{d}{du} \left(\frac{1}{u}\right)\right)} \frac{d}{dx} \left(1 + e^{- x}\right) = {\color{red}\left(- \frac{1}{u^{2}}\right)} \frac{d}{dx} \left(1 + e^{- x}\right)$$

返回到原变量：

$$- \frac{\frac{d}{dx} \left(1 + e^{- x}\right)}{{\color{red}\left(u\right)}^{2}} = - \frac{\frac{d}{dx} \left(1 + e^{- x}\right)}{{\color{red}\left(1 + e^{- x}\right)}^{2}}$$

和/差的导数等于导数的和/差：

$$- \frac{{\color{red}\left(\frac{d}{dx} \left(1 + e^{- x}\right)\right)}}{\left(1 + e^{- x}\right)^{2}} = - \frac{{\color{red}\left(\frac{d}{dx} \left(1\right) + \frac{d}{dx} \left(e^{- x}\right)\right)}}{\left(1 + e^{- x}\right)^{2}}$$

常数的导数是$$$0$$$:

$$- \frac{{\color{red}\left(\frac{d}{dx} \left(1\right)\right)} + \frac{d}{dx} \left(e^{- x}\right)}{\left(1 + e^{- x}\right)^{2}} = - \frac{{\color{red}\left(0\right)} + \frac{d}{dx} \left(e^{- x}\right)}{\left(1 + e^{- x}\right)^{2}}$$

函数$$$e^{- x}$$$是两个函数$$$f{\left(u \right)} = e^{u}$$$和$$$g{\left(x \right)} = - x$$$的复合$$$f{\left(g{\left(x \right)} \right)}$$$。

应用链式法则 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$：

$$- \frac{{\color{red}\left(\frac{d}{dx} \left(e^{- x}\right)\right)}}{\left(1 + e^{- x}\right)^{2}} = - \frac{{\color{red}\left(\frac{d}{du} \left(e^{u}\right) \frac{d}{dx} \left(- x\right)\right)}}{\left(1 + e^{- x}\right)^{2}}$$

指数函数的导数为 $$$\frac{d}{du} \left(e^{u}\right) = e^{u}$$$：

$$- \frac{{\color{red}\left(\frac{d}{du} \left(e^{u}\right)\right)} \frac{d}{dx} \left(- x\right)}{\left(1 + e^{- x}\right)^{2}} = - \frac{{\color{red}\left(e^{u}\right)} \frac{d}{dx} \left(- x\right)}{\left(1 + e^{- x}\right)^{2}}$$

返回到原变量：

$$- \frac{e^{{\color{red}\left(u\right)}} \frac{d}{dx} \left(- x\right)}{\left(1 + e^{- x}\right)^{2}} = - \frac{e^{{\color{red}\left(- x\right)}} \frac{d}{dx} \left(- x\right)}{\left(1 + e^{- x}\right)^{2}}$$

对 $$$c = -1$$$ 和 $$$f{\left(x \right)} = x$$$ 应用常数倍法则 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$：

$$- \frac{e^{- x} {\color{red}\left(\frac{d}{dx} \left(- x\right)\right)}}{\left(1 + e^{- x}\right)^{2}} = - \frac{e^{- x} {\color{red}\left(- \frac{d}{dx} \left(x\right)\right)}}{\left(1 + e^{- x}\right)^{2}}$$

应用幂法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，取 $$$n = 1$$$，也就是说，$$$\frac{d}{dx} \left(x\right) = 1$$$：

$$\frac{e^{- x} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)}}{\left(1 + e^{- x}\right)^{2}} = \frac{e^{- x} {\color{red}\left(1\right)}}{\left(1 + e^{- x}\right)^{2}}$$

化简：

$$\frac{e^{- x}}{\left(1 + e^{- x}\right)^{2}} = \frac{1}{4 \cosh^{2}{\left(\frac{x}{2} \right)}}$$

因此，$$$\frac{d}{dx} \left(\frac{1}{1 + e^{- x}}\right) = \frac{1}{4 \cosh^{2}{\left(\frac{x}{2} \right)}}$$$。

接下来，$$$\frac{d^{2}}{dx^{2}} \left(\frac{1}{1 + e^{- x}}\right) = \frac{d}{dx} \left(\frac{1}{4 \cosh^{2}{\left(\frac{x}{2} \right)}}\right)$$$

对 $$$c = \frac{1}{4}$$$ 和 $$$f{\left(x \right)} = \frac{1}{\cosh^{2}{\left(\frac{x}{2} \right)}}$$$ 应用常数倍法则 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$：

$${\color{red}\left(\frac{d}{dx} \left(\frac{1}{4 \cosh^{2}{\left(\frac{x}{2} \right)}}\right)\right)} = {\color{red}\left(\frac{\frac{d}{dx} \left(\frac{1}{\cosh^{2}{\left(\frac{x}{2} \right)}}\right)}{4}\right)}$$

函数$$$\frac{1}{\cosh^{2}{\left(\frac{x}{2} \right)}}$$$是两个函数$$$f{\left(u \right)} = \frac{1}{u^{2}}$$$和$$$g{\left(x \right)} = \cosh{\left(\frac{x}{2} \right)}$$$的复合$$$f{\left(g{\left(x \right)} \right)}$$$。

应用链式法则 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$：

$$\frac{{\color{red}\left(\frac{d}{dx} \left(\frac{1}{\cosh^{2}{\left(\frac{x}{2} \right)}}\right)\right)}}{4} = \frac{{\color{red}\left(\frac{d}{du} \left(\frac{1}{u^{2}}\right) \frac{d}{dx} \left(\cosh{\left(\frac{x}{2} \right)}\right)\right)}}{4}$$

应用幂次法则 $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$，其中 $$$n = -2$$$:

$$\frac{{\color{red}\left(\frac{d}{du} \left(\frac{1}{u^{2}}\right)\right)} \frac{d}{dx} \left(\cosh{\left(\frac{x}{2} \right)}\right)}{4} = \frac{{\color{red}\left(- \frac{2}{u^{3}}\right)} \frac{d}{dx} \left(\cosh{\left(\frac{x}{2} \right)}\right)}{4}$$

返回到原变量：

$$- \frac{\frac{d}{dx} \left(\cosh{\left(\frac{x}{2} \right)}\right)}{2 {\color{red}\left(u\right)}^{3}} = - \frac{\frac{d}{dx} \left(\cosh{\left(\frac{x}{2} \right)}\right)}{2 {\color{red}\left(\cosh{\left(\frac{x}{2} \right)}\right)}^{3}}$$

函数$$$\cosh{\left(\frac{x}{2} \right)}$$$是两个函数$$$f{\left(u \right)} = \cosh{\left(u \right)}$$$和$$$g{\left(x \right)} = \frac{x}{2}$$$的复合$$$f{\left(g{\left(x \right)} \right)}$$$。

应用链式法则 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$：

$$- \frac{{\color{red}\left(\frac{d}{dx} \left(\cosh{\left(\frac{x}{2} \right)}\right)\right)}}{2 \cosh^{3}{\left(\frac{x}{2} \right)}} = - \frac{{\color{red}\left(\frac{d}{du} \left(\cosh{\left(u \right)}\right) \frac{d}{dx} \left(\frac{x}{2}\right)\right)}}{2 \cosh^{3}{\left(\frac{x}{2} \right)}}$$

双曲余弦函数的导数为$$$\frac{d}{du} \left(\cosh{\left(u \right)}\right) = \sinh{\left(u \right)}$$$：

$$- \frac{{\color{red}\left(\frac{d}{du} \left(\cosh{\left(u \right)}\right)\right)} \frac{d}{dx} \left(\frac{x}{2}\right)}{2 \cosh^{3}{\left(\frac{x}{2} \right)}} = - \frac{{\color{red}\left(\sinh{\left(u \right)}\right)} \frac{d}{dx} \left(\frac{x}{2}\right)}{2 \cosh^{3}{\left(\frac{x}{2} \right)}}$$

返回到原变量：

$$- \frac{\sinh{\left({\color{red}\left(u\right)} \right)} \frac{d}{dx} \left(\frac{x}{2}\right)}{2 \cosh^{3}{\left(\frac{x}{2} \right)}} = - \frac{\sinh{\left({\color{red}\left(\frac{x}{2}\right)} \right)} \frac{d}{dx} \left(\frac{x}{2}\right)}{2 \cosh^{3}{\left(\frac{x}{2} \right)}}$$

对 $$$c = \frac{1}{2}$$$ 和 $$$f{\left(x \right)} = x$$$ 应用常数倍法则 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$：

$$- \frac{\sinh{\left(\frac{x}{2} \right)} {\color{red}\left(\frac{d}{dx} \left(\frac{x}{2}\right)\right)}}{2 \cosh^{3}{\left(\frac{x}{2} \right)}} = - \frac{\sinh{\left(\frac{x}{2} \right)} {\color{red}\left(\frac{\frac{d}{dx} \left(x\right)}{2}\right)}}{2 \cosh^{3}{\left(\frac{x}{2} \right)}}$$

应用幂法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，取 $$$n = 1$$$，也就是说，$$$\frac{d}{dx} \left(x\right) = 1$$$：

$$- \frac{\sinh{\left(\frac{x}{2} \right)} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)}}{4 \cosh^{3}{\left(\frac{x}{2} \right)}} = - \frac{\sinh{\left(\frac{x}{2} \right)} {\color{red}\left(1\right)}}{4 \cosh^{3}{\left(\frac{x}{2} \right)}}$$

因此，$$$\frac{d}{dx} \left(\frac{1}{4 \cosh^{2}{\left(\frac{x}{2} \right)}}\right) = - \frac{\sinh{\left(\frac{x}{2} \right)}}{4 \cosh^{3}{\left(\frac{x}{2} \right)}}$$$。

因此，$$$\frac{d^{2}}{dx^{2}} \left(\frac{1}{1 + e^{- x}}\right) = - \frac{\sinh{\left(\frac{x}{2} \right)}}{4 \cosh^{3}{\left(\frac{x}{2} \right)}}$$$。