令t=√((1+x)/(x-1)),则y=ln t
y'=t'/t
t=√((1+x)/(x-1))=(1+ 2/(x-1) )^(1/2)
则t'=(1/2)·(1+ 2/(x-1) )^(-1/2) ·(-2/(x-1)²)
= -1/( t· (x-1)² )
因此
y'= -1/( t²· (x-1)² )
= -((x-1)/(1+x)) / (x-1)²
= 1/(1-x²)