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April 3rd, 2014, 10:51 AM   #1
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Help understanding this simplification


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.
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April 3rd, 2014, 11:04 AM   #2
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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:

$\displaystyle f'(x)=\frac{6(3-x)^5(x+3)^5+5(x+3)^6(3-x)^4}{(3-x)^{10}}$

Now, if we look at the two terms in the numerator, we see they both contain the factor $(3-x)^4(x+3)^5$, and so we may factor that out to get:

$\displaystyle f'(x)=\frac{(3-x)^4(x+3)^5\left(6(3-x)+5(x+3) \right)}{(3-x)^{10}}$

Next, we see that both the numerator and denominator contain the factor $(3-x)^4$ so we may divide that out:

$\displaystyle f'(x)=\frac{(x+3)^5\left(6(3-x)+5(x+3) \right)}{(3-x)^6}$

Finally, distribute within the second factor in the numerator:

$\displaystyle f'(x)=\frac{(x+3)^5\left(18-6x+5x+15 \right)}{(3-x)^6}$

and collect like terms:

$\displaystyle f'(x)=\frac{(x+3)^5(33-x)}{(3-x)^6}$
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April 3rd, 2014, 11:28 AM   #3
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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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