急设x=2t^(2)-1,y=根号(1+t^2).求dy/dx和d^2y/dx^2
问题描述：
急设x=2t^(2)-1,y=根号(1+t^2).求dy/dx和d^2y/dx^2
答
dy/dx=(dy/dt)/(dx/dt)=[t/√(1+t^2)]/(4t)=1/[4√(1+t^2)]
d^2y/dx^2=[d(dy/dx)/dt]/dx/dt
=1/4*(-1/2)*(1+t^2)^(-3/2)/(4t)
=-1/32*(1+t^2)^(-3/2)/t