Sigmoid Function, Derivative and Limits

We have two objectives. First, we need derivate the sigmoid function:

$S(x) = \frac {1}{1 + e^{-px}}$

using very simple techniques.

Second, we must to find the more right and more left function limits:

$ \lim_{x \to +\infty }{S(x)} = 1$ e $\lim_{x \to -\infty }{S(x)} = 0$ 

Derivative sigmoid

We are looking for

$\frac{\mathrm{d} S(x)}{\mathrm{d} x} = S'(x)$.

To deriving it, we'll apply the quocient rule, which is remembered here:

$\left ( \frac{f}{g} \right )' = \frac{f'g  -  g'f}{g^2}$

Said that, taking:

$f(x) = 1 \Rightarrow f'(x) = 0$
and
$g(x) = 1 + e^{-px} \Rightarrow g'(x) = -pe^{-px}$

Then,

$S'(x) = \left ( \frac{f}{g} \right )' = \frac{0g - g'1}{g^2} = \frac{pe^{-px}}{(1  +  e^{-px})^2}$

Objective reached. But if you try Wolfram Alpha to do it, you'll take a different answer:

Ok, no problem. Let's show that both answers are equivalent.

$ = \frac{pe^{-px}}{(1  +  e^{-px})^2}$

$ = \frac{(e^{2px})pe^{-px}}{(e^{2px})(1  +  2e^{-px}  +  e^{-2px})}$

$ = \frac{pe^{px}}{e^{2px}  +  2e^{px}  +  1}$

And finally

$ = \frac{pe^{px}}{(e^{px}  +  1)^2}$ 

Limits in $+\infty$ and $-\infty$

The two limits are easy to do:

$lim_{x \to +\infty }{\frac {1}{1 + e^{-px}}} = \frac {1}{1 + \frac{1}{\infty}} = \frac {1}{1 + 0} = 1$

$lim_{x \to -\infty }{\frac {1}{1 + e^{-px}}} = \frac {1}{1 + e^{p\infty}} = \frac {1}{1 + \infty} = 0$