Question

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

Answer 1

I rewrite it first.

Explanation:

`f(x) = ln((x+1)/(x-1)) = ln(x+1)-ln(x-1)` `" "` (property of `ln`)

`f'(x) = 1/(x-1)-1/(x+1)` `" "` (derivative of `ln` and chain rule)

`= ((x+1)-(x-1))/((x+1)(x-1))` `" "` (algebra subtract fractions)

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