x=f(t),y=t-arctgt,且dy/dx=t/2,则[1-1/(1+t^2)]/f'(t)=t/2,f'(t)=2t/(1+t^2),所以d∧2y/dx∧2=d(dy/dx)/dt*dt/dx=(1/2)/f'(t)=(1+t^2)/(4t).
