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

dx/dt=2t/(1+t²)dy/dt=1/(1+t²)dy/dx=1/(2t)d(dx/dt)/dt=(2-4t²)/(1+t²)²d(dy/dt)/dt=(-2t)/(1+t²)²d²y/d²x=[d(dy/dt)/dt]/[d(dx/dt)/dt]=t/(2t²-1)跟答案不一样啊 好像错了d(dx/dt)/dt=(2+2t²-4t²)/(1+t²)² =(2-2t²)/(1+t²)²d(dy/dt)/dt=(-2t)/(1+t²)²d²y/d²x=[d(dy/dt)/dt]/[d(dx/dt)/dt]=t/(t²-1)