设f'(x)=lnx,令y=f[(2x-1)/(x+1)],求dy/dx

Question

设f'(x)=lnx,令y=f[(2x-1)/(x+1)],求dy/dx

Answer 1

设y=f(u)
dy/dx =dy/du*du/dx=d(lnu)/du *du/dx=1/u *du/dx=1/u *[2(x+1)-(2x-1)]/(x+1)^2
=3/[(2x-1)(x+1)]