Question

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

Answer 1

I think that this was not the function that you intended to write (probably you omitted parentheses, which are essential when one writes in-line fractions). In fact, you can rewrite that function as follows:
`c(x)=ln(1+e^x/1-e^x)=ln(1+e^x-e^x)=ln(1)=0`
The derivative of a constant function is `0`, so
`c'(x)=0`

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I'll try to infer the function that you wanted to write:
`f(x)=ln((1+e^x)/(1-e^x))`
The code to render this properly is f(x)=ln((1+e^x)/(1-e^x)) (notice the usage of parentheses).

This is a function composition. Given two functions `y=g(x)` and `z=h(y)` you can compute the derivative of the composition `f(x)=h(g(x))` as follows:
`f'(x)=h'(g(x))*g'(x)`
In this particular case `g(x)=(1+e^x)/(1-e^x)` and `h(y)=ln(y)`.

The derivative of `g` we is the derivative of a quotient
`g'(x)=(e^x(1-e^x)-(1+e^x)(-e^x))/(1-e^x)^2=(2e^x)/(1-e^x)^2`
and the derivative of `h` is the derivative of the logarithmic function
`h'(y)=1/y`
So
`f'(x)=h'(g(x)) * g'(x)=1/((1+e^x)/(1-e^x)) * (2e^x)/(1-e^x)^2=(2e^x)/((1+e^x)(1-e^x))=(2e^x)/(1-e^(2x))`

Answer 2

I assume you want to differentiate:

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

You could use `d/dx (lnu) = 1/u (du)/dx` with `u = (1+e^x)/(1-e^x)` and use the quotient rule to find `(du)/dx`, but it's a little quicker to use `ln(a/b) = lna - lnb` to rewrite the function as:

`y=ln(1+e^x) - ln(1-e^x)`. before differentiating:

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

Or, simplifying:

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