y=arctan(1-x)/(1+x)的微分

y=arctan[(1-x)/(1+x)]=arctan[-1+2/(1+x)]y'=[-1+2/(1+x)]' * (1/(1+(-1+2/(x+1))^2)=-2/(1+x)^2*[1/(1+(1-x)^2/(x+1)^2)=-2/[(1+x)^2+(1-x)^2]=-2/(2+2x^2)=-1/(1+x^2)求微分啊。。是这个么y'=-1/(1+x^2)dy/dx=-1/(1+x^2)dy= -dx/(1+x^2)