Derivative of $\arcsin \frac{x-1}{x+1}$ I was looking at a question that asks for the derivative of $\arcsin (\frac {x+1}{x-1}) $.
The solution starts by saying $y = \frac{x+1}{x-1}$, so $1-y^2= \frac{4x}{(x+1)^2}$ and $\frac{1}{\sqrt{1-y^2}}$ and thus $\frac{x+1}{2\sqrt{x}}  $  
However, I don't see why  $1-y^2= \frac{4x}{(x+1)^2} $

$1-y^2=1-\left(\dfrac{x-1}{x+1}\right)^2=1-\dfrac{x^2-2x+1}{x^2+2x+1}=\dfrac{(x^2+2x+1)-(x^2-2x+1)}{x^2+2x+1}=\dfrac{4x}{x^2+2x+1}=\dfrac{4x}{(x+1)^2}$
