x=sint,y=tsint+cost,求d^2y/dx^2It=π/4
dx/dt=cost
dy/dt=sint+tcost-sint=tcost(问题是这个!为什么不是dy/dt=tcost-sint请问前面那个sint是怎么来的?)
y'=dy/dx=(dy/dt)/(dx/dt)=t
y"=d^2y/dx^2=d(y')/dx=d(y')/dt/(dx/dt)=1/cost
当t=π/4时,y"=1/cos(π/4)=√2