答：x=ln(1+t²)，x'(t)=2t/(1+t²)y=t-arctant，y'(t)=1-1/(1+t²)=t²/(1+t²)dy /dx= (dy/dt) /(dx/dt)=t²/(2t)=t/2dy /dx=t/2
dy / dx=(dy/dt) / (dx/dt) =y'(t) / x'(t)
