Question

How do you find the derivative of `[e^x / (1 - e^x)]`?

Answer 1

`e^x/(1-e^x)^2`

Explanation:

Ahh... We gotta use the good ol' quotient rule. Using Lagrange's notation (one that I do not use often) as it would get messy in In Leibniz' notation, we see:

`(f/g)'=(f'g-fg')/(g^2)`

So (back to Leibniz)

Let `f(x)=e^x` and `g(x)=1-e^x`

`d/dx[e^x/(1-e^x)]=d/dx[f(x)/g(x)]=((df)/dx g(x)-(dg)/dxf(x))/(g^2(x))`

So, we've gotta figure out what the derivative of `f(x)` and `g(x)` is

`(df)/dx=d/dx[e^x]=e^x`

`(dg)/dx=d/dx[1]-d/dx[e^x]=-e^x`

Then sub back in

`((df)/dx g(x)-(dg)/dxf(x))/(g^2(x))=(e^x(1-e^x)+e^x(e^x))/(1-e^x)^2`

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

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

And that's the most we can simplify for now. Therefore:

`d/dx[e^x/(1-e^x)]=e^x/(1-e^x)^2`