Question

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

Answer 1

I think it's simplest to rewrite first.

Explanation:

`y = ln((x+1)/(x-1)) = ln(x+1)-ln(x-1)`

Recall that `d/dx(lnu) = 1/u (du)/dx`, so

`d/dx(ln(x+1)) = 1/(x+1)` and `d/dx(ln(x-1)) = 1/(x-1)`

So

`dy/dx = [1/(x+1)-1/(x-1)]`

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