Question

How do you find f'(x) and f''(x) given `f(x)=x/(3+e^x)`?

Answer 1

Please see below.

Explanation:

`f(x)=x/(3+e^x)`

Using `d/dx(f/g) = (f'g-fg')/g^2` with `f=x` and `g = 3+e^x`

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

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

Using `d/dx(f/g) = (f'g-fg')/g^2` again, but this time with `f=3+e^x-xe^x` and `g = (3+e^x)^2`

`f''(x) = ((e^x-(1e^x+xe^x))(3+e^x)^2 - (3+e^x-xe^x)(2(3+e^x)e^x))/(3+e^x)^4`

`= ((-xe^x)color(red)((3+e^x)^2) - 2e^x(3+e^x-xe^x)(color(red)((3+e^x))))/(3+e^x)^4`

`= (color(red)((3+e^x))[(-xe^x)color(red)((3+e^x)) - 2e^x(3+e^x-xe^x]))/(3+e^x)^4`

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

Rewrite the numerator to taste.