高数积分问题

2024-11-18 05:36:04
推荐回答(1个)
回答1:

x=cos(t^2)

dx/dt = -2t.sin(t^2)

y=tcos(t^2) -∫(1->t^2)   [cosu/(2√u) ] du

dy/dt 

=cos(t^2) - 2t^2.sin(t^2) -2t [cos(t^2)/(2t) ]

=-2t^2.sin(t^2) 

dy/dx = (dy/dt)/(dx/dt) = t

d/dt (dy/dt ) =1

d^2y/dx^2

=d/dt (dy/dt ) / (dx/dt)

=-1/[2t.sin(t^2)]