# How did they simplify this function

I'm currently practicing differentiation. The exercise I currently have is the following

Find the derivative of:

$(x + 6)^3 (9 x^3 - 2)^5$

Okay, well I can do that now. When I do this I uses the chain/product rule to get the following result:

$3(x+6)^2 (9x^3 -2)^5 + 45(x+6)^3 (3x^2) (9x^3 - 2)^4$

However, when I put this in wolfram alpha I get the following result (and it matches the answer from my exercise):

$6(x+6)^2 (2-9 x^3)^4 (27 x^3+135 x^2-1)$

I'm staring at this for an hour now, but I don't get how they get rid of the addition, and how they'd get rid of (for example) $(9x^3-2)^5$

-

I don't know how wolfram alpha's algorithms actually did this, but here's how it could have been done:
$3(x+6)^2(9x^3-2)^5+45(x+6)^3(3x^2)(9x^3-2)^4$
$=3(x+6)^2(9x^3-2)^4(9x^3-2)+3(x+6)^2(9x^3-2)^4(45x^2)(x+6)$
$=3(x+6)^2(9x^3-2)^4((9x^3-2)+45x^2(x+6))$
$=3(x+6)^2(9x^3-2)^4(9x^3-2+45x^3+270x^2)$
$=3(x+6)^2(9x^3-2)^4(54x^3+270x^2-2)$
$=6(x+6)^2(9x^3-2)^4(27x^3+135x^2-1)$

-
Hmm, yeah thanks when I read it like that it seems logically. Aaww, man I hope when I get my test I can just leave the original function I had, because I'm not going to make such thing up ^^ – Timo Willemsen Jan 24 '11 at 10:04
@Timo It doesn't come out of thin air. It is the same process as factoring $3x^2 y^5+45x^3 y^4$. Look for factors that are the same in each thing added together and then pull them out front. If you are confused by the powers being different try writing them out like $(x+6)(x+6)...$. – Brian Jan 24 '11 at 13:32
@Timo Though your answer is correct, it might not be considered a proper form. Just like it is usually not recommended to leave 15/25 instead of 3/5 as answer. If you are concerned about the test, make sure to clarify it with professor, since some would mark such an answer down. – artdanil Jan 24 '11 at 14:02