Question

@Psymon :)

Answer 1

More questions :P I took a quiz and got a 95 and I want to retake to get a 100

Answer 2

So what was the issue?

Answer 3

I don't know, the last question on one of them I just didn't get correct for some reason.

Answer 4

Well, if you got the question, then yeah.

Answer 5

Derivative of G(x)=(x^3-125)*(x^2+9)/(x^2-9)

Answer 6

\[G(x)=(x^3-125)*((x^2+9)/(x^2-9))\]

Answer 7

http://gyazo.com/4b1d9c6befa1bf9d339361c1cf5a7ea6

Answer 8

:(

Answer 9

Awkward how they threw in things that could be factored but accomplish nothing when you factor them, haha. Alright, so its a product rule and a quotient rule at the same time. So for quotient rule, we would do
(derivative of top)*(bottom) - (derivative of bottom)*(top)/ all over the bottom squared. Its just the difference here is that the derivative of the top is a product rule. So we definitely have to be careful. So let's work with the derivative of the top first. So the derivative of the top is a product rule, meaning we do (derivative of first term)*(second term) + (derivative of 2nd term)*(first term). So if I do that, I get:
\[(3x)(x^{2}+9)+ (x^{3}-125)(2x)\]Now this ENTIRE result, as per the quotient rule, is multiplied by the bottom, giving us:

\[(x^{2}-9)[(3x)(x^{2}+9)+(x^{3}-125)(2x)]\] So now we have to do the other part of the quotient rule, the whole minus (derivative of bottom)*(top) So Ill do that and tackit on to the current derivaitve, making sure I have the bottomsquared. In total that gives me:

\[\frac{ (x^{2}-9)[(3x)(x^{2}+9)+(x^{3}-125)(2x)] - (2x)(x^{3}-125)(x^{2}+9) }{ (x^{2}-9)^{2} }\]

quite the mess. I think I did it wrong, so I was just trying to check over myself. This does take forever xD Imma just solve it real fast pen and paper and then try to explain, haha.

Answer 10

I GOT IT RIGHT

Answer 11

Haha, nice xD Yeah, I was gettingit right, too, I just needed to check myself. Too many big numbers in the end -_-