求导y=tcos(t^2)-∫1到t^2的极限1/(2√u)cosudu设x=cos(t^2)y=tcos(t^2)-∫1到t^21/(2√u)cosudu,t>0dy/dt这是分段函数

Question

求导y=tcos(t^2)-∫1到t^2的极限1/(2√u)cosudu
设x=cos(t^2)
y=tcos(t^2)-∫1到t^2 1/(2√u)cosudu ,t>0
dy/dt
这是分段函数

Answer 1

y=tcos(t^2)-∫1到t^2 1/(2√u)cosudu
dy/dt=cos(t^2)-2t^2sin(t^2)-(1/2t)cos(t^2)2t
=-2t^2sin(t^2)
dx/dt=-2tsin(t^2)
dy/dx=t