How do you find the first and second derivative of `y=1/(1+e^x)`?

First derivative is just applying chain rule to `(1+e^x)^-1`
which gives:
`d/dx((1+e^x)^-1)=-1*(1+e^x)^-2*e^x=-e^x/(1+e^x)^2`
For the second derivative we need the quotient rule which states:
`d/dx((u(x))/(v(x)))=(u'(x)*v(x)-u(x)*v'(x))/(v(x)^2)`

`u(x)=e^x` ; `u'(x)=e^x`
`v(x)=(1+e^x)^2` ; `v'(x)=2*(1+e^x)*e^x`
Plugging this in we get:

`d^2/dx^2(1/1+e^x)=(e^x*(1+e^x)^2-e^x*e^x*2*(1+e^x))/((1+e^x)^2)^2`
This can be simplified to:
`=(e^x*(1+e^x)((1+e^x)-2e^x))/((1+e^x)^4)=(e^x*((1+e^x)-2e^x))/((1+e^x)^3)=(e^x*(1-e^x))/((1+e^x)^3)`

Given that

`y=\frac{1}{1+e^x}`
Differentiating w.r.t. `x` as follows

`\frac{dy}{dx}=\frac{d}{dx}(1+e^x)^{-1}`

`\frac{dy}{dx}=-(1+e^x)^{-2}\frac{d}{dx}(1+e^x)`

`\frac{dy}{dx}=-\frac{1}{(1+e^x)^{2}}e^x`

`\frac{dy}{dx}=-\frac{e^x}{(1+e^x)^{2}}`
Differentiating w.r.t. `x` as follows

`\frac{d}{dx}\frac{dy}{dx}=-\frac{d}{dx}\frac{e^x}{(1+e^x)^{2}}`

`\frac{d^2y}{dx^2}=-\frac{(1+e^x)^{2}\frac{d}{dx}e^x-e^x\frac{d}{dx}(1+e^x)^2}{((1+e^x)^{2})^2}`

`\frac{d^2y}{dx^2}=-\frac{(1+e^x)^{2}e^x-e^x2(1+e^x)e^x}{(1+e^x)^{4}}`

`\frac{d^2y}{dx^2}=\frac{2e^{2x}-e^x(1+e^x)}{(1+e^x)^{3}}`

`\frac{d^2y}{dx^2}=\frac{e^x(2e^{x}-(1+e^x))}{(1+e^x)^{3}}`

`\frac{d^2y}{dx^2}=\frac{e^x(e^{x}-1)}{(1+e^x)^{3}}`