thanks in advance. you guys have been a huge help with all my other questions!
$\displaystyle f'(x) = \frac{(3x+1)^4\cdot 3(2x+1)^2\cdot 2 - (2x+1)^3\cdot 4(3x+1)^3\cdot 3}{((3x+1)^4)^2} =$ $\displaystyle \frac{6(3x+1)^4(2x+1)^2-12(3x+1)^3(2x+1)^3}{(3x+1)^8} = \frac{6(3x+1)(2x+1)^2-12(2x+1)^3}{(3x+1)^5}$ $\displaystyle = \frac{(2x+1)^2(6(3x+1)-12(2x+1))}{(3x+1)^5}$
$\displaystyle \boxed{f'(x) =\frac{-6(x+1)(2x+1)^2}{(3x+1)^5}}$
That's probably the nicest way to look at it.
Ok you need to give us more information before this thread gets out of control and too long. You've been shown the way to calculate the derivative and if that's confusing, that's ok. Tell us why it is and what you have been told to do. We can't guess what you know and what you're learning in class.
There are only a few ways to do this problem. The quotient rule like redsoxfan325 did, the product rule (by rewriting one term with a negative exponent) or expanding everything out and then simplifying (which is dumb).
Tell us more and we can help you.