高数y=x/(x^2+1)^1/2的微分如题,y等于根号（x的平方加1）分之x的微分,dy=[1/(x^2+1)^(2/3)]dx dy等于{【（x的平方加1）的2分之3次方】分之1}乘以dx吃饭，等等再看。
问题描述：
高数y=x/(x^2+1)^1/2的微分
如题,y等于根号（x的平方加1）分之x的微分,
dy=[1/(x^2+1)^(2/3)]dx dy等于{【（x的平方加1）的2分之3次方】分之1}乘以dx
吃饭，等等再看。
答
∵y=x/(x^2+1)^（1/2）∴dy=d[x/(x^2+1)^(1/2)] =[x/(x^2+1)^(1/2)]′dx ={[(x^2+1)^(1/2)-x^2/(x^2+1)^(1/2)]/(x^2+1)}dx ={[(x^2+1)-x^2]/[(x^2+1)^(3/2)]}dx =[1/(x^2+1)^(3/2)]dx =dx/(x^2+1)^(3...