How to find the first and second derivative of y = 1 1 + e...

${\displaystyle \frac{dy}{dx}=\frac{e^{-x}}{{(1+e^{-x})}^2},\frac{d^{2}y}{{dx}^2}=\frac{e^{-2x}-e^{-x}}{{(1+e^{-x})}^3}}$
Solution:
${\displaystyle \text{Reminder}\overline{\underline{\left|\frac{2}{2}\frac{d}{dx}\left(e^{-x}\right)=-e^{-x}\frac{2}{2}\right|}}}$
There are 2 approaches to differentiating this function.
(1) Using the quotient rule
${\displaystyle \left(2\right)\text{expressing}y={(1+e^{-x})}^{-1}\text{and use chain rule}}$
I'll use approach (1), but you could try approach (2). The outcome will be the same.
differentiate using the quotient rule
${\displaystyle \text{Given}y=\frac{g\left(x\right)}{h\left(x\right)}\text{then}}$
${\displaystyle \overline{\underline{\left|\frac{2}{2}\frac{dy}{dx}=\frac{h\left(x\right)g\prime \left(x\right)-g\left(x\right)h\prime \left(x\right)}{{\left(h\left(x\right)\right)}^2}\frac{2}{2}\right|}}}$
${\displaystyle g\left(x\right)=1\Rightarrow g\prime \left(x\right)=0}$
${\displaystyle h\left(x\right)=1+e^{-x}\Rightarrow h\prime \left(x\right)=-e^{-x}}$
${\displaystyle \Rightarrow \frac{dy}{dx}=\frac{(1+e^{-x}).0-1.(-e^{-x})}{{(1+e^{-x})}^2}=\frac{e^{-x}}{{(1+e^{-x})}^2}}$
${\displaystyle \text{To find}\frac{d^{2}y}{{dx}^2}\text{differentiate}\frac{dy}{dx}}$
differentiate using the quotient rule/chain rule
${\displaystyle \text{here}g\left(x\right)=e^{-x}\Rightarrow g\prime \left(x\right)=-e^{-x}}$
${\displaystyle h\left(x\right)={(1+e^{-x})}^2\Rightarrow h\prime \left(x\right)=2(1+e^{-x}).(-e^{-x})\to }$
${\displaystyle \Rightarrow h\prime \left(x\right)=-2e^{-x}(1+e^{-x})}$
${\displaystyle \frac{d^{2}y}{{dx}^2}=\frac{{(1+e^{-x})}^2(-e^{-x})-\left(e^{-x}\right).(-2e^{-x}(1+e^{-x}))}{{(1+e^{-x})}^4}}$
${\displaystyle =\frac{-e^{-x}{(1+e^{-x})}^2+2e^{-2x}(1+e^{-x})}{{(1+e^{-x})}^4}}$
${\displaystyle =\frac{e^{-x}(1+e^{-x})(2e^{-x}-1-e^{-x})}{{(1+e^{-x})}^4}\leftarrow \text{factoring}}$
${\displaystyle =\frac{e^{-x}\cancel{(1+e^{-x})}(e^{-x}-1)}{\cancel{{(1+e^{-x})}^3}}}$
${\displaystyle =\frac{e^{-2x}-e^{-x}}{{(1+e^{-x})}^3}}$