Question

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

Answer 1

`dy/dx=(e^(-x))/(1+e^(-x))^2,(d^2y)/(dx^2)=(e^(-2x)-e^(-x))/(1+e^(-x))^3`

Explanation:

`color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(d/dx(e^(-x))=-e^(-x))color(white)(2/2)|)))`

There are 2 approaches to differentiating this function.

`(1)" Using the quotient rule"`

`(2)" expressing " y=(1+e^(-x))^-1" and use chain rule"`

I'll use approach (1) you could perhaps try approach (2). The result will be the same.

differentiate using the `color(blue)"quotient rule"`

`" Given " y=(g(x))/(h(x))" then"`

`color(red)(bar(ul(|color(white)(2/2)color(black)(dy/dx=(h(x)g'(x)-g(x)h'(x))/(h(x))^2)color(white)(2/2)|)))`

`g(x)=1rArrg'(x)=0`

`h(x)=1+e^(-x)rArrh'(x)=-e^(-x)`
`color(blue)"------------------------------------------------------------"`

`rArrdy/dx=((1+e^-x).0-1.(-e^-x))/(1+e^-x)^2=(e^-x)/(1+e^-x)^2`

`" To find " (d^2y)/(dx^2)" differentiate " dy/dx`

differentiate using the `color(blue)"quotient rule/chain rule"`

`"here " g(x)=e^-xrArrg'(x)=-e^-x`

`h(x)=(1+e^-x)^2rArrh'(x)=2(1+e^-x).(-e^-x)rarr`

`rArrh'(x)=-2e^-x(1+e^-x)`
`color(blue)"---------------------------------------------------------------"`

`(d^2y)/(dx^2)=((1+e^-x)^2(-e^-x)-(e^-x).(-2e^-x(1+e^-x)))/(1+e^-x)^4`

`=(-e^-x(1+e^-x)^2+2e^(-2x)(1+e^-x))/(1+e^-x)^4`

`=(e^-x(1+e^-x)(2e^-x-1-e^-x))/(1+e^-x)^4larr" factoring"`

`=(e^-xcancel((1+e^-x))(e^-x-1))/(cancel((1+e^-x)^3`

`=(e^(-2x)-e^-x)/(1+e^-x)^3`