Question

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

Answer 1

`-2/(x^2-1)`, x > 1.

Explanation:

under the condition x > 1 for log(x-1).
`f(x)=ln((x+1)/(x-1))=ln(x+1)-ln(x-1)`.

Applying function of function rule,

`f'(x)=1/(x+1)(x+1)'-1/(x-1)(x-1)'`

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

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

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