Question

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

Answer 1

Simplify using natural log properties, take the derivative, and add some fractions to get `d/dxln((x+1)/(x-1))=-2/(x^2-1)`

Explanation:

It helps to use natural log properties to simplify `ln((x+1)/(x-1))` into something a little less complicated. We can use the property `ln(a/b)=lna-lnb` to change this expression to:
`ln(x+1)-ln(x-1)`

Taking the derivative of this will be a lot easier now. The sum rule says we can break this up into two parts:
`d/dxln(x+1)-d/dxln(x-1)`

We know the derivative of `lnx=1/x`, so the derivative of `ln(x+1)=1/(x+1)` and the derivative of `ln(x-1)=1/(x-1)`:
`d/dxln(x+1)-d/dxln(x-1)=1/(x+1)-1/(x-1)`

Subtracting the fractions yields:
`(x-1)/((x+1)(x-1))-(x+1)/((x-1)(x+1))`
`=((x-1)-(x+1))/(x^2-1)`
`=(x-1-x-1)/(x^2-1)`
`=-2/(x^2-1)`