Question

y=1+lnx分之1-lnx 求导数

Answer 1

y=(1+x-x^2)/(1-x+x^2)

y'=[(1+x-x^2)'*(1-x+x^2)-(1+x-x^2)*(1-x+x^2)']/(1-x+x^2)^2

=[(-2x+1)*(1-x+x^2)-(1+x-x^2)*(2x-1)]/(1-x+x^2)^2

=[-(2x-1)*(1-x+x^2)-(1+x-x^2)*(2x-1)]/(1-x+x^2)^2

=(2x-1)*(-1+x-x^2-1-x+x^2)/(1-x+x^2)^2

=(2x-1)*(-2)/(1-x+x^2)^2

=-2(2x-1)/(1-x+x^2)^2