求y=x/(根号x^2+1)的微分
问题描述：
求y=x/(根号x^2+1)的微分
答
dy=y'dx
= {x/√(x^2+1)}'dx
= {√(x^2+1) -x^2/√(x^2+1) } / (x^2+1) dx
= {(x^2+1) -x^2/ } / [ (x^2+1) √(x^2+1)] dx
= 1/ (x^2+1)^(3/2) dx