Derivada de $$$\frac{1}{1 + e^{- x}}$$$

A função $$$\frac{1}{1 + e^{- x}}$$$ é a composição $$$f{\left(g{\left(x \right)} \right)}$$$ de duas funções $$$f{\left(u \right)} = \frac{1}{u}$$$ e $$$g{\left(x \right)} = 1 + e^{- x}$$$.

Aplique a regra da cadeia $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:

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

Aplique a regra de poder $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$ com $$$n = -1$$$:

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

Volte para a variável antiga:

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

A derivada de uma soma/diferença é a soma/diferença das derivadas:

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

A função $$$e^{- x}$$$ é a composição $$$f{\left(g{\left(x \right)} \right)}$$$ de duas funções $$$f{\left(u \right)} = e^{u}$$$ e $$$g{\left(x \right)} = - x$$$.

Aplique a regra da cadeia $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:

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

A derivada da exponencial é $$$\frac{d}{du} \left(e^{u}\right) = e^{u}$$$:

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

Volte para a variável antiga:

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

A derivada de uma constante é $$$0$$$:

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

Aplique a regra múltipla constante $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ com $$$c = -1$$$ e $$$f{\left(x \right)} = x$$$:

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

Aplique a regra de potência $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ com $$$n = 1$$$, ou seja, $$$\frac{d}{dx} \left(x\right) = 1$$$:

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

Simplificar:

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

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