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)
