解答： y=1/x^2+x+1=(x²+x+1)^(-1)y'=-(x²+x+1)^(-2)*(x²+x+1)' =-(x²+x+1)^(-2)*(2x+1) =-(2x+1)/(x²+x+1)²y=1/lnx=(lnx)^(-1)y'=-(lnx)^(-2)*(lnx)' =-(lnx)^(-2)*(1/x) =-1/(x*ln²x)
