Calculus help for exam

solve by natural log:
differentiate
y=ln [(x-1) / (x+1)]

y = ln (1 - (2/(x+1)))
dy/dx = 1/[(x-1) / (x+1)] * d/dx (1 - (2/(x+1)))
dy/dx = [(x+1) / (x-1)] * 2 / (x+1)^2
dy/dx = 2 / (x^2 - 1)

Oh - or the easier way is:

y = ln (x-1) - ln (x+1)
dy/dx = 1/(x-1) - 1/(x+1) = 2/(x^2 - 1)

One of two
y = ln (x-1) - Ln (x+1)

dy/dx= 1/(x-1) - 1/(x+1) = ((x+1- (x-1)) /(x-1) (x+1) = 2/ (x^2-1)

OR
y= Ln u being u=u(x) =[(x-1) / (x+1)]

dy/dx=( dy/du) (du/dx)
dy/dx=(1/u) du/dx
du/dx= ((x+1) -(x-1)) / (x+1)^2
du/dx= 2/(x+1)^2
(1/u) = (x+1)/(x-1)
dy/dx = (x+1)/(x-1) (2/(x+1)^2)
dy/dx= 2/ (x^2-1)

Let u = (x - 1) / (x + 1)
du/dx is given by :-

(x + 1) - (x - 1)
---------------------
(x + 1)²

2
---------
(x + 1)²

y = ln u
dy/du = 1/u

dy/dx = 2 / [ u (x + 1)² ]

dy/dx = 2 (x + 1) / [ (x - 1) (x + 1)² ]

dy/dx = 2 / [ (x - 1)(x + 1) ]