Second derivative of $$$\frac{1}{1 + e^{- x}}$$$

Find the first derivative $$$\frac{d}{dx} \left(\frac{1}{1 + e^{- x}}\right)$$$

The function $$$\frac{1}{1 + e^{- x}}$$$ is the composition $$$f{\left(g{\left(x \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = \frac{1}{u}$$$ and $$$g{\left(x \right)} = 1 + e^{- x}$$$.

Apply the chain rule $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:

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

Apply the power rule $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$ with $$$n = -1$$$:

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

Return to the old variable:

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

The derivative of a sum/difference is the sum/difference of derivatives:

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

The function $$$e^{- x}$$$ is the composition $$$f{\left(g{\left(x \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = e^{u}$$$ and $$$g{\left(x \right)} = - x$$$.

Apply the chain rule $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:

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

The derivative of the exponential is $$$\frac{d}{du} \left(e^{u}\right) = e^{u}$$$:

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

Return to the old variable:

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

The derivative of a constant is $$$0$$$:

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

Apply the constant multiple rule $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ with $$$c = -1$$$ and $$$f{\left(x \right)} = x$$$:

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

Apply the power rule $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ with $$$n = 1$$$, in other words, $$$\frac{d}{dx} \left(x\right) = 1$$$:

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

Simplify:

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

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

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

Apply the constant multiple rule $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ with $$$c = \frac{1}{4}$$$ and $$$f{\left(x \right)} = \frac{1}{\cosh^{2}{\left(\frac{x}{2} \right)}}$$$:

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

The function $$$\frac{1}{\cosh^{2}{\left(\frac{x}{2} \right)}}$$$ is the composition $$$f{\left(g{\left(x \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = \frac{1}{u^{2}}$$$ and $$$g{\left(x \right)} = \cosh{\left(\frac{x}{2} \right)}$$$.

Apply the chain rule $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:

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

Apply the power rule $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$ with $$$n = -2$$$:

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

Return to the old variable:

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

The function $$$\cosh{\left(\frac{x}{2} \right)}$$$ is the composition $$$f{\left(g{\left(x \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = \cosh{\left(u \right)}$$$ and $$$g{\left(x \right)} = \frac{x}{2}$$$.

Apply the chain rule $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:

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

The derivative of the hyperbolic cosine is $$$\frac{d}{du} \left(\cosh{\left(u \right)}\right) = \sinh{\left(u \right)}$$$:

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

Return to the old variable:

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

Apply the constant multiple rule $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ with $$$c = \frac{1}{2}$$$ and $$$f{\left(x \right)} = x$$$:

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

Apply the power rule $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ with $$$n = 1$$$, in other words, $$$\frac{d}{dx} \left(x\right) = 1$$$:

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

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

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