Question

一道求微分的题目y=x/根号（x平方+1）微分是多少,

Answer 1

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