arctan根号（x^2-1)求导,

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u=x²-1,则u'=2x
v=√u,则v'=1/(2√u)*u'=2x/(2√u)=x/√(x²-1)
所以令y=arctan√(x²-1)=arctanv
则y'=1/(1+v²)*v'
=1/(1+x²-1)*x/√(x²-1)
=1/[x√(x²-1)]