x=ln(1+t^2),y=arctant+π 求dy/dx和d2y/dx2

网友回答

dx/dt=2t/(1+t²)
dy/dt=1/(1+t²)
dy/dx=1/(2t)
d(dx/dt)/dt=(2-4t²)/(1+t²)²
d(dy/dt)/dt=(-2t)/(1+t²)²
d²y/d²x=[d(dy/dt)/dt]/[d(dx/dt)/dt]
=t/(2t²-1)