Question

What is the derivative of `ln((e^x)/(1+e^x))`?

Answer 1

`1/(1+e^x)`

Explanation:

`y'=1/(e^x/(1+e^x)) * (e^x(1+e^x)-e^xe^x)/(1+e^x)^2=1/e^x * (e^x+e^(2x)-e^(2x))/(1+e^x)=`

`=1/e^x * e^x/(1+e^x)=1/(1+e^x)`

Answer 2

`1/(1+e^x)`

Explanation:

You can bypass using the quoitient rule for derivatives by using the quotient rule for logarithms

`ln(a/b) = lna - lnb`

In your case, you will have

`ln(e^x/(1+e^x)) = lne^x - ln(1+e^x)`

Now all you have to do is use the chain rule twice, once for `lnu`, with `u=e^x`, and once for `lnv`, with `v= 1+e^x`.

`d/dx[lne^x - ln(1+e^x)] = d/dxlne^x - d/dxln(1+e^x)`

You have

`d/dx(lnu) = d/(du)lnu * d/dx(u)`

`d/dx(lnu) = 1/u * d/dxe^x = 1/color(red)(cancel(color(black)(e^x))) * color(red)(cancel(color(black)(e^x))) = 1`

and

`d/dx(lnv) = d/(dv)lnv * d/dx(v)`

`d/dx(lnv) = 1/v * d/dx(1+e^x) = 1/(1+e^x) * e^x`

The target derivative will thus be

`d/dxln(e^x/(1+e^x)) = 1 - e^x/(1+e^x) = (1 + color(red)(cancel(color(black)(e^x))) - color(red)(cancel(color(black)(e^x))))/(1+e^x) = color(green)(1/(1+e^x)`