Derivative of sigmoid function $\sigma (x) = \frac{1}{1+e^{-x}}$

Let's denote the sigmoid function as $\sigma(x) = \dfrac{1}{1 + e^{-x}}$.

The derivative of the sigmoid is $\dfrac{d}{dx}\sigma(x) = \sigma(x)(1 - \sigma(x))$.

Here's a detailed derivation:

$$ \begin{align} \dfrac{d}{dx} \sigma(x) &= \dfrac{d}{dx} \left[ \dfrac{1}{1 + e^{-x}} \right] \\ &= \dfrac{d}{dx} \left( 1 + \mathrm{e}^{-x} \right)^{-1} \\ &= -(1 + e^{-x})^{-2}(-e^{-x}) \\ &= \dfrac{e^{-x}}{\left(1 + e^{-x}\right)^2} \\ &= \dfrac{1}{1 + e^{-x}\ } \cdot \dfrac{e^{-x}}{1 + e^{-x}} \\ &= \dfrac{1}{1 + e^{-x}\ } \cdot \dfrac{(1 + e^{-x}) - 1}{1 + e^{-x}} \\ &= \dfrac{1}{1 + e^{-x}\ } \cdot \left( \dfrac{1 + e^{-x}}{1 + e^{-x}} - \dfrac{1}{1 + e^{-x}} \right) \\ &= \dfrac{1}{1 + e^{-x}\ } \cdot \left( 1 - \dfrac{1}{1 + e^{-x}} \right) \\ &= \sigma(x) \cdot (1 - \sigma(x)) \end{align} $$

Consider $$ f(x)=\dfrac{1}{\sigma(x)} = 1+e^{-x} . $$ Then, on the one hand, the chain rule gives $$ f'(x) = \frac{d}{dx} \biggl( \frac{1}{\sigma(x)} \biggr) = -\frac{\sigma'(x)}{\sigma(x)^2} , $$ and on the other hand, $$ f'(x) = \frac{d}{dx} \bigl( 1+e^{-x} \bigr) = -e^{-x} = 1-f(x) = 1 - \frac{1}{\sigma(x)} = \frac{\sigma(x)-1}{\sigma(x)} . $$ Equate the two expressions, and voilà!

(Cf. also this answer.)

Note that from your given equation,

$(1+e^{-x})\sigma=1$

$\Rightarrow -e^{-x}\sigma+(1+e^{-x})\frac{d\sigma}{dx}=0$ (differentiating using product rule)

$\Rightarrow \frac{d\sigma}{dx}=\sigma.\frac{e^{-x}}{(1+e^{-x})}=\sigma.\frac{(1+e^{-x})-1}{(1+e^{-x})}=\sigma.\left[1-\frac{1}{(1+e^{-x})}\right]=\sigma.(1-\sigma)$