thanks in advance. you guys have been a huge help with all my other questions!

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- May 4th 2009, 01:30 PMcalcconfusedSteps to solving the following: f(x) = (2x+1)^3/(3x+1)^4)
thanks in advance. you guys have been a huge help with all my other questions!

- May 4th 2009, 01:42 PMredsoxfan325
- May 4th 2009, 03:48 PMcalcconfused
im sorry...i meant find f'(x) of the above equation

- May 4th 2009, 05:07 PMredsoxfan325
$\displaystyle f'(x) = \frac{(3x+1)^4\cdot 3(2x+1)^2\cdot 2 - (2x+1)^3\cdot 4(3x+1)^3\cdot 3}{((3x+1)^4)^2} =$ $\displaystyle \frac{6(3x+1)^4(2x+1)^2-12(3x+1)^3(2x+1)^3}{(3x+1)^8} = \frac{6(3x+1)(2x+1)^2-12(2x+1)^3}{(3x+1)^5}$ $\displaystyle = \frac{(2x+1)^2(6(3x+1)-12(2x+1))}{(3x+1)^5}$

$\displaystyle \boxed{f'(x) =\frac{-6(x+1)(2x+1)^2}{(3x+1)^5}}$

That's probably the nicest way to look at it. - May 4th 2009, 11:22 PMcalcconfused
- May 5th 2009, 07:45 AMredsoxfan325
- May 5th 2009, 11:11 AMcalcconfused
thanks..im just so confused as my professor told me a different way of getting it/different answer.

- May 5th 2009, 11:58 AMJameson
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.