How do I find the derivative of this problem?

**rhcprule3** · October 6th 2013, 07:01 AM

f(x)=x2+2(3x2−3)(x2−2)

Do I just multply out the top to get f(x) and then get the derivative by the quotient rule from there?

I got what I thought was the derivative: (12x^2(X^2+2)-(3x^2-3)(x^2-2)(2x))/(x^2+2)^2 but it said I was incorrect.

**emakarov** · October 6th 2013, 10:23 AM

Originally Posted by rhcprule3

f(x)=x2+2(3x2−3)(x2−2)

Do I just multply out the top to get f(x) and then get the derivative by the quotient rule from there?

There is no top or bottom in what you have written; the whole formula is on one line.

In plain text, it is customary to write x^2 for x².

**HallsofIvy** · October 6th 2013, 11:34 AM

You seem to have forgotten a "/"! Did you mean (x^2+ 2)(3x^2- 3)/(x^2- 2)?
(You also seem to have forgotten the parentheses around x^2+ 2. Or did you really mean "x^2+ [2(3x^2- 3)]= x^2+ 6x^2- 6= 7x^2- 6 as you wrote?)

By the quotient rule, the derivative is [((x^2+2)(3x^2- 3))'(x^2- 2)- (x^2+2)(3x^2-3)(x^2- 2)']/(x^2-2)^2.
You can do the derivative of the numertor, ((x^2+2)(3x^2- 3))' either by first multiplying it out or by using the product rule.

**rhcprule3** · October 6th 2013, 03:38 PM

sorry the problem was supposed to be (3x^2-3)(x^2-2) all over (x^2+2)

**Plato** · October 6th 2013, 04:07 PM

Originally Posted by rhcprule3

sorry the problem was supposed to be (3x^2-3)(x^2-2) all over (x^2+2)

Use logarithmic differentiation: $y' = \frac{{(3{x^2} - 3)({x^2} - 2)}}{{\left( {{x^2} + 2} \right)}}\left[ {\frac{{6x}}{{(3{x^2} - 3)}} + \frac{{2x}}{{({x^2} - 2)}} - \frac{{2x}}{{({x^2} + 2)}}} \right]$

**votan** · October 6th 2013, 04:10 PM

Originally Posted by rhcprule3

sorry the problem was supposed to be (3x^2-3)(x^2-2) all over (x^2+2)

y = (3x^2-3)(x^2-2)/(x^2+2)
This is of the form y = u/v ==> (vu' - v'u)/v^2

Can you continue from there?

**HallsofIvy** · October 6th 2013, 04:23 PM

Plato's point being that if $y= \frac{(3x^2-3)(x^2- 2)}{x^2+ 2}$ then $ln(y)= ln\left(\frac{(3x^2-3)(x^2- 2)}{x^2+ 2}\right)= ln(3x^2- 3)+ ln(x^2- 2)- ln(x^2+ 2)$ so that $d(ln(y))/dx= (1/y)(dy/dx)= 6x/(3x^2- 3)+ 2x/(x^2- 2)- 2x/(x^2+ 2)$ and then
$dy/dx= [(3x^2- 3)(x^2- 2)/(x^2+ 2)]6x/(3x^2- 3)+ 2x/(x^2- 2)- 2x/(x^2+ 2)$

**rhcprule3** · October 6th 2013, 05:22 PM

I've tried using the quotient rule to get the derivative but the webwork still says im incorrect

**votan** · October 6th 2013, 05:37 PM

Originally Posted by rhcprule3

I've tried using the quotient rule to get the derivative but the webwork still says im incorrect

Show us your work we will help you finding what you missed

**SlipEternal** · October 6th 2013, 05:42 PM

Originally Posted by rhcprule3

I got what I thought was the derivative: (12x^2(X^2+2)-(3x^2-3)(x^2-2)(2x))/(x^2+2)^2 but it said I was incorrect.

Originally Posted by rhcprule3

I've tried using the quotient rule to get the derivative but the webwork still says im incorrect

You have now been shown many different ways of finding the solution. If you got the answer wrong before, then try the quotient rule again. You did it wrong when you tried it before.

Break it down. You have $f(x) = \dfrac{u(x)}{v(x)}$ . Write out the functions $u(x)$ and $v(x)$ . Then find the derivatives $u'(x)$ and $v'(x)$ . Write them here for us to check, and we can figure out where you might be going wrong. From your answer in your first post, it appears you think the derivative of $(3x^2-3)(x^2-2)$ is $12x^2$ . Is that the case?

**rhcprule3** · October 6th 2013, 06:00 PM

yes i believe the derivative is 12x^2 from u(x) and I belive the derivative for v(x) is 2x

**SlipEternal** · October 6th 2013, 06:12 PM

Then that is your problem. If $u(x) = (3x^2-3)(x^2-2)$ , then $u'(x) \neq 12x^2$ . Use the product rule (as HallsofIvy said).

**rhcprule3** · October 7th 2013, 05:24 AM

Ive tried everything I got u'(x) to equal 12x^3-18x. So I did the quotient rule: (36x^2-18)(x^2+2)-2x(12x^3-18x) all over (x^2+2)^2. I've tried foiling and product rule on (36x^2-18)(x^2+2) but I still ended up with the wrong anser and its EXTREMELY frustrating!!!

**tom@ballooncalculus** · October 7th 2013, 06:05 AM

You'll feel better when you realise that you've made errors, which is more reassuring than if the maths weren't capable of behaving itself! So re-read the given solutions. Here's another:

... where (key in spoiler) ...

Spoiler:

Full size

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Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

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**rhcprule3** · October 7th 2013, 09:38 AM

Ive tried for f'(2) by plugging 2 for the x values but its still not giving it to me.... even with the product rule youve told me to use...

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