f(x)=(x+根号(1+x^2)/(1+x^2)=x/(1+x^2)+1/根号(1+x^2)
f'(x)=[x'(1+x^2)-x*(1+x^2)']/(1+x^2)^2-1/2*(1+x^2)^(-3/2)*(1+x^2)'
=[1+x^2-x*2x]/(1+x^2)^2-1/2*(1+x^2)^(-3/2)*2x
=(1-x^2)/(1+x^2)^2-x*(1+x^2)^(-3/2)