Question

求高数大神解答😭

这个dy/dt，求dy怎么算的过程😭😭😭

Answer 1

变上限积分公式如下。

Answer 2

x=1+2t^2

dx/dt =4t
y=∫(1->1+2lnt) (e^u/u) du
dy/dt
= [e^(1+2lnt)/(1+2lnt)].(1+2lnt)'
= [et^2/(1+2lnt)].(2/t)
=2et/(1+2lnt)
dy/dx
=(dy/dt)/(dx/dt)
=[ 2et/(1+2lnt)] /(4t)
=e/[2(1+2lnt)]