一道求微分的题目y=x/根号（x平方+1）微分是多少,
问题描述：
一道求微分的题目
y=x/根号（x平方+1）
微分是多少,
答
其实是求导数,dy=y'dx
y'=[x/根号(x^2+1)]'={x'根号(x^2+1)-x[根号(x^2+1)]'}/(x^2+1)=[根号(x^2+1)-x^2/根号(x^2+1)]/(x^2+1)=1/(x^2+1)^(3/2)
答
因为y=x/√[(x^2)+1]
所以：
dy={√[(x^2)+1]-(x^2)/{[(x^2)+1]^(3/2)}}dx/[(x^2)+1]
={{[(x^2)+1]^2-x^2}/[(x^2)+1]^(3/2)}dx/[(x^2)+1]
={(x^4+x^2+1)/[(x^2)+1]^(3/2)}dx/[(x^2)+1]
=(x^4+x^2+1)/{[(x^2)+1]^2}
=[(x^4+x^2+1)/(x^4+2x^2+1)]dx