一道求微分的题目 y=x/根号（x平方+1） 微分是多少,
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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
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
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