解:
y=1/根号(x^2+1)
所以
y'=[1/根号(x^2+1)]'=[(x^2+1)^(-1/2)]'
=[(-1/2)(x^2+1)^(-3/2)][2x]
所以dy=[(-1/2)(x^2+1)^(-3/2)][2x]dx