Question

高等函数求导y=ln√((1+x)/(x-1))

Answer 1

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)