f(x) = arctan[2x/(2-x^2)]f'(x) = {2[2-x^2-x(-2x)]/(2-x^2)^2}/[1+4x^2/(2-x^2)^2]= 2(2+x^2)/[(2-x^2)^2+4x^2] = (1+x^2/2)/(1+x^4/4)令 t=x/√2, 则f'(x) = (1+t^2)/(1+t^4) = 1/(1+t^4)+t^2/(1+t^4).
