设y=(1/6)ln[(x+1)2/(x2-x+1)]+(1/√3)arctan[(2x-1)/√3]，x≠-1，求y′．
y=(1/3)ln(x+1)-(1/6)ln(x2-x+1)+(1/√3)arctan[(2x-1)/√3]y'=1/3(x+1)-(x2-x+1)′/6(x2-x+1)+1/√3{1/[1+(2x-1)2/3]}[(2x-1)/√3]=1/[3(x+1)]-(2x-1)/[6(x2-x+1)]+1/(2x2-2x+2)=1/(1+x3)
设y=(1/6)ln[(x+1)2/(x2-x+1)]+(1/√3)arctan[(2x-1)/√3]，x≠-1，求y′．
y=(1/3)ln(x+1)-(1/6)ln(x2-x+1)+(1/√3)arctan[(2x-1)/√3]y'=1/3(x+1)-(x2-x+1)′/6(x2-x+1)+1/√3{1/[1+(2x-1)2/3]}[(2x-1)/√3]=1/[3(x+1)]-(2x-1)/[6(x2-x+1)]+1/(2x2-2x+2)=1/(1+x3)
