y=arctan(1-x^2)/(1+x^2)y'={[arctan(1-x^2)]'×(1+x^2)-arctan(1-x^2)×(1+x^2)‘}/(1+x^2)^2={1/[1+(1-x^2)^2]×(-2x)×(1+x^2)-arctan(1-x^2)×2x}/(1+x^2)^2dy=-{2x×(1+x^2)/[1+(1-x^2)^2]+2x×arctan(1-x^2)}/...
求函数的微分：y= arctan(1-x^2)/1+x^2 具体算式与答案
y=arctan(1-x^2)/(1+x^2)y'={[arctan(1-x^2)]'×(1+x^2)-arctan(1-x^2)×(1+x^2)‘}/(1+x^2)^2={1/[1+(1-x^2)^2]×(-2x)×(1+x^2)-arctan(1-x^2)×2x}/(1+x^2)^2dy=-{2x×(1+x^2)/[1+(1-x^2)^2]+2x×arctan(1-x^2)}/...