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。

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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。

∵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)^(3/2)