对数曲线y=lnx上哪一点处的曲率半径最小?求出改点处的曲率半径

y'=1/x(x>0)y''=-1/x^2(x>0)ρ=1/K,曲率半径ρ越小,曲率K越大K=|y''/(1+y'^2)^(3/2)|=|-1/x^2/(1+1/x^2)^(3/2)|=x/(x^2+1)^(3/2),x>0令dK/dx=[1*(x^2+1)^(3/2)-x*(3/2)*(x^2+1)^(1/2)*2x]/(x^2+1)^3=[(x^2+1)-3x^2]/...