# Steps to solving the following: f(x) = (2x+1)^3/(3x+1)^4)

• May 4th 2009, 02:30 PM
calcconfused
Steps 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, 02:42 PM
redsoxfan325
Quote:

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

What are we solving for? Are we trying to find the zeroes?

$\frac{(2x+1)^3}{(3x+1)^4} = 0$ ?

If so, we need to solve $(2x+1)^3 = 0 \implies x=-\frac{1}{2}$

Or do you want to integrate that?
• May 4th 2009, 04:48 PM
calcconfused
im sorry...i meant find f'(x) of the above equation
• May 4th 2009, 06:07 PM
redsoxfan325
Quote:

Originally Posted by calcconfused
im sorry...i meant find f'(x) of the above equation

$f'(x) = \frac{(3x+1)^4\cdot 3(2x+1)^2\cdot 2 - (2x+1)^3\cdot 4(3x+1)^3\cdot 3}{((3x+1)^4)^2} =$ $\frac{6(3x+1)^4(2x+1)^2-12(3x+1)^3(2x+1)^3}{(3x+1)^8} = \frac{6(3x+1)(2x+1)^2-12(2x+1)^3}{(3x+1)^5}$ $= \frac{(2x+1)^2(6(3x+1)-12(2x+1))}{(3x+1)^5}$

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

That's probably the nicest way to look at it.
• May 5th 2009, 12:22 AM
calcconfused
Quote:

Originally Posted by redsoxfan325
$f'(x) = \frac{(3x+1)^4\cdot 3(2x+1)^2\cdot 2 - (2x+1)^3\cdot 4(3x+1)^3\cdot 3}{((3x+1)^4)^2} =$ $\frac{6(3x+1)^4(2x+1)^2-12(3x+1)^3(2x+1)^3}{(3x+1)^8} = \frac{6(3x+1)(2x+1)^2-12(2x+1)^3}{(3x+1)^5}$ $= \frac{(2x+1)^2(6(3x+1)-12(2x+1))}{(3x+1)^5}$

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

That's probably the nicest way to look at it.

thanks but the answer i have in here says its (2x+1)^2/(3x+1)^8 * (x+1)
• May 5th 2009, 08:45 AM
redsoxfan325
Quote:

Originally Posted by calcconfused
thanks but the answer i have in here says its (2x+1)^2/(3x+1)^8 * (x+1)

Maple confirms that $\frac{d}{dx}\left[\frac{(2x+1)^3}{(3x+1)^4}\right] = \frac{-6(x+1)(2x+1)^2}{(3x+1)^5}$
• May 5th 2009, 12:11 PM
calcconfused
thanks..im just so confused as my professor told me a different way of getting it/different answer.
• May 5th 2009, 12:58 PM
Jameson
Quote:

Originally Posted by calcconfused
thanks..im just so confused as my professor told me a different way of getting it/different answer.

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).