integral
posted by Eric .
Find dy/dx and d2y/dx2 if y= definite integral sign where a= 1 and b= 3x
1/(t^2+t+1) dt
how do i even start. do i integrate and then plug in a and b? plz help.

dy/dx = 1/(x^2+x+1) * 3

can you explain how you got this. thanks

excuse me? Didn't we just go through this about differentiating under the integral sign?

but i don't get how you can just replace the variable t with x so easily. i don't get how you did that. and how does the 3 come into the picture for dy/dx.

In general (see the wikipedia article for proof),
∫[a(x),b(x)] f(t) dt
= f(b(x)) db/dx  f(a(x)) da/dx
Since we have a(x) = 1, da/dx = 0 and we are left with
∫[a(x),b(x)] f(t) dt
= f(b(x)) db/dx
b(x) = 3x, so db/dx = 3
So, oops. I made a mistake. It should be
dy/dx = 1/((3x)^2+3x+1) * 3
Good catch! :) 
Thank you very much! I really didn't understand the form but now I do. :)
Also, would d2y/dx2 be:
9(6x+1)/((9x^2+3x+1)^2)
I just want to make sure that d2y/dx2 is asking for the second derivative? 
haven't checked your work, but yes, they want y"

Oh okay thank you!