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April 3rd, 2014, 10:51 AM  #1 
Newbie Joined: Feb 2014 Posts: 16 Thanks: 2  Help understanding this simplification
Hi, I am looking at a problem from a calculus class about finding a derivative of the function $\displaystyle (x + 3)^6/(3  x)^5$. However, since my problem is not in understanding how to get the derivative but in why it is being simplified the way it is (and that part is just algebra), I thought I would post it here. If I was wrong, I apologize in advance and hope a mod will move my post. So this is the derivative: I did the problem up to this point and had no issues. However, I'm afraid I do not see what is happening in either of these two steps that I am showing here. Could someone please help clarify? Last edited by SamSeymour; April 3rd, 2014 at 10:56 AM. 
April 3rd, 2014, 11:04 AM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs 
First, you made a good and conscientious choice about where to post this question, since it is about the algebra involved after completing the differentiation and not the differentiation itself. Okay, we have obtained: $\displaystyle f'(x)=\frac{6(3x)^5(x+3)^5+5(x+3)^6(3x)^4}{(3x)^{10}}$ Now, if we look at the two terms in the numerator, we see they both contain the factor $(3x)^4(x+3)^5$, and so we may factor that out to get: $\displaystyle f'(x)=\frac{(3x)^4(x+3)^5\left(6(3x)+5(x+3) \right)}{(3x)^{10}}$ Next, we see that both the numerator and denominator contain the factor $(3x)^4$ so we may divide that out: $\displaystyle f'(x)=\frac{(x+3)^5\left(6(3x)+5(x+3) \right)}{(3x)^6}$ Finally, distribute within the second factor in the numerator: $\displaystyle f'(x)=\frac{(x+3)^5\left(186x+5x+15 \right)}{(3x)^6}$ and collect like terms: $\displaystyle f'(x)=\frac{(x+3)^5(33x)}{(3x)^6}$ 
April 3rd, 2014, 11:28 AM  #3 
Newbie Joined: Feb 2014 Posts: 16 Thanks: 2 
Aha, factoring! I guess I had a momentary brain freeze there. That's usually my problem. That was crystal clear, thank you. 

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