Question

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

Answer 1

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

Explanation:

Let `y= lnu` and `u = (1 + e^x)/(1 - e^x)`.

Then `dy/(du) = 1/u` and `(du)/dx = (e^x(1 - e^x) - (-e^x(1 + e^x)))/(1 - e^x)^2`

`dy/dx = dy/(du) xx (du)/dx`

`dy/dx = 1/u xx (e^x - e^(2x) + e^x + e^(2x))/(1 - e^x)^2`

`dy/dx= 1/u xx (2e^x)/(1 - e^x)^2`

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

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

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

Hopefully this helps!