求y=arctan[2x/(1-x^2)]的导数,

1个回答

  • 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)