求y=arctan[2x⼀(1-x^2）]的导数，请写一下详细的解题过程，万分感谢！

y=arctan[2x/(1-x^2）]
y=arctanu u=2x/(1-x^2) u'=(2(1-x^2)-2x(-2x))/(1-x^2)^2=(2x^2+2)/(1-x^2)^2
那么导数 y'=1/(1+u^2)*u'=1/(1+4x^2/(1-x^2)^2) * ( 2x^2+2)/(1-x^2)^2
=( 2x^2+2)/((1-x^2)^2+4x^2) 底下是个完全平方
=( 2x^2+2)/(1+x^2)^2
=2/(1+x^2)

y=arctanx的导数是 y'=1/(1+x**2) 将2x/(1-x**2)看成整体带入整理得到，y'=(1-x**2)**2/(1+x**2)**2

y=arctan[2x/(1-x^2)]
y'=[2x/(1-x^2)]'*[1/[1+[2x/(1-x^2)]^2]]
=[1/(1-x)-1/(1+x)]' *[1/[1+4x^2/(1-2x^2+x^4)]]
=[1/(1-x)^2+1/(1+x)^2]*[(1-x^2)^2/(1+x^2)^2]
=[(2+2x^2)/[(1-x)^2(1+x)^2]]*(1-x^2)^2/(1+x^2)^2]
=2/(1+x^2)