# Sigmoid derivative using quotient rule

Sigmoid function defined by $f(x)=\frac{1}{1+e^{-x}}$ can be derived easily with derivative of a composed function like here: Derivative of sigmoid function. However I was asking myself how this could be achieved with quotient ratio.

What I tried is:

$$f'(x)=\lim_{\Delta x \to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x} = \lim_{\Delta x \to 0} \frac{1}{\Delta x}\Big[ \frac{1}{1+e^{-x- \Delta x}} - \frac{1}{1+e^{-x}} \Big]$$ $$= \lim_{\Delta x \to 0} \frac{1}{\Delta x}\Big[ \frac{1+e^{-x}-(1+e^{-x-\Delta x})}{(1+e^{-x- \Delta x})(1+e^{-x})}\Big] = \lim_{\Delta x \to 0} \frac{1}{\Delta x} \frac{e^{-x}-e^{-x-\Delta x}}{(1+e^{-x- \Delta x})(1+e^{-x})}$$ $$= \lim_{\Delta x \to 0} \frac{1-e^{-\Delta x}}{\Delta x} \frac{e^{-x}}{(1+e^{-x- \Delta x})(1+e^{-x})}$$

Now I can't simplify. So I was thinking if is correct to assert that

If $\lim_{x \to 0} f_1(x)-f_2(x)=0$, than $\lim_{x \to 0} \frac{f_1(x)}{f_2(x)}=1$ ?

If is possible than I can show easily that the first fraction goes to $1$, and we can plug in $0$ instead of the remaining $\Delta x$ and get the formula for derivative function. I think it would be necessary to show that the sign is the same for both $f_1$ and $f_2$, which is trivial also.

However I am wondering if this kind of reasoning is mathematically correct.

Use $$\lim_{h\to0}\frac{e^h-1}h=1$$
Observe that the exponent of $e,$ the numerator and the limit variable must be same or equivalent