Question

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

Answer 1

`(d^2y)/dx^2 = ( 4x ) / ( x^2-1)^2`

Explanation:

Let `y = ln((x+1)/(x-1))`,
Then:

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

Differentiating wrt `x` we get:

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

Differentiating again wrt `x` we get:

`(d^2y)/dx^2 = -1/(x+1)^2 - (-1)/(x-1)^2`
`:. (d^2y)/dx^2 = ( (x+1)^2 - (x-1)^2 ) / ( (x+1)^2(x-1)^2 )`
`:. (d^2y)/dx^2 = ( (x^2+2x+1) - (x^2-2x+1) ) / ( ((x+1)(x-1))^2 )`
`:. (d^2y)/dx^2 = ( 4x ) / ( x^2-1)^2`