# [SOLVED]having a problem working through an equation

#### DeusAbscondus

##### Active member
I am finding product rule problems difficult.Trying to find the second derivative makes them explode into unweildly equations that are hard to deal with and factorize without making errors. Here is one I have got wrong; can someone spot where:

$72x^2(2x^4-1)^2+24x^3.2(2x^4-1)$
$=72x^2(2x^4-1)2+48x^3(2x^4-1)$
$=24x^2(2x^4-1)[3(2x^4-1)+2]$
$=24x^2(2x^4-1)(6x^4-1)$

$\text{But answer given in my text is: }$
$24x^2(2x^4-1)(22x^4-3)$

Can anyone help see the point where I went astray?
thanks kindly,
DeusAbs

#### SuperSonic4

##### Well-known member
MHB Math Helper
I am finding product rule problems difficult.Trying to find the second derivative makes them explode into unweildly equations that are hard to deal with and factorize without making errors. Here is one I have got wrong; can someone spot where:

$72x^2(2x^4-1)^2+24x^3.2(2x^4-1)$
$=72x^2(2x^4-1)^2+48x^3(2x^4-1)$ SS4: Is that two a typo, I'm guessing it should be to the power 2
$=24x^2(2x^4-1)[3(2x^4-1)+2]$
$=24x^2(2x^4-1)(6x^4-1)$

$\text{But answer given in my text is: }$
$24x^2(2x^4-1)(22x^4-3)$

Can anyone help see the point where I went astray?
thanks kindly,
DeusAbs
I get the same answer as you. Did you take the derivative correctly? If you want post the function you need to take the second derivative of with any working you've done.

#### DeusAbscondus

##### Active member
I get the same answer as you. Did you take the derivative correctly? If you want post the function you need to take the second derivative of with any working you've done.
Hey, you are really nice man! Thanks for speedy response: i actually got it!
Still, nice to have confirmed that i was on the right track.

Best,
D'Abs

(PS you can be sure you'll never run out of questions from me: keen as mustard and not greatly skilled: I am no natural at this, but i LOVE it!)

#### soroban

##### Well-known member
Hello, DeusAbscondus!

$$\text{Differentiate }\:h(x) \:=\: 24x^3(2x^4-1)^2$$

$$h'(x) \;=\;72x^2(2x^4-1)^2+24x^3\cdot2(2x^4-1)$$ . Here!

$$\text{Answer: }\:24x^2(2x^4-1)(22x^4-3)$$

$$\text{We have: }\:h(x) \;=\;\overbrace{24x^3}^{f(x)}\cdot\overbrace{(2x^4-1)^2}^{g(x)}$$

$$\text{Product Rule: }\;h'(x) \;=\;\overbrace{72x^2}^{f'(x)}\cdot\overbrace{(2x^4-1)^2}^{g(x)} + \overbrace{24x^3}^{f(x)}\cdot \overbrace{2(2x^4-1)\color{red}{(8x^3)}}^{g'(x)}$$

. . . . . . $$h'(x) \;=\;72x^2(2x^4-1)^2 + 384x^6(2x^4-1)$$

$$\text{Factor: }\;h'(x) \;=\;24x^2(2x^4-1)\cdot\left(3[2x^4-1] + 16x^4\right)$$

. . . . . . $$h'(x) \;=\;24x^2(2x^4-1)\,(6x^4-3 + 16x^4)$$

. . . . . . $$h'(x) \;=\;24x^2(2x^4-1)(22x^4-3)$$

LaTeX isn't dependable?

#### DeusAbscondus

##### Active member
Hello, DeusAbscondus!

$$\text{We have: }\:h(x) \;=\;\overbrace{24x^3}^{f(x)}\cdot\overbrace{(2x^4-1)^2}^{g(x)}$$

$$\text{Product Rule: }\;h'(x) \;=\;\overbrace{72x^2}^{f'(x)}\cdot\overbrace{(2x^4-1)^2}^{g(x)} + \overbrace{24x^3}^{f(x)}\cdot \overbrace{2(2x^4-1)\color{red}{(8x^3)}}^{g'(x)}$$

. . . . . . $$h'(x) \;=\;72x^2(2x^4-1)^2 + 384x^6(2x^4-1)$$

$$\text{Factor: }\;h'(x) \;=\;24x^2(2x^4-1)\cdot\left(3[2x^4-1] + 16x^4\right)$$

. . . . . . $$h'(x) \;=\;24x^2(2x^4-1)\,(6x^4-3 + 16x^4)$$

. . . . . . $$h'(x) \;=\;24x^2(2x^4-1)(22x^4-3)$$

LaTeX isn't dependable?
soroban, I don't know how you managed to reconstitute the problem, but you got it...!
And so did I, not 30 mins after making initial post.
You were correct of course: (as was SuperSonic4) I hadn't derived the second properly.

I've been floating on a cushion of relief and feel my self-confidence seeping back deep into my maths bones.
Thanks heaps for input!
D'abs
PS the significance of your cryptic comment re. Latex intrigues while it escapes me... please explain some time?

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#### Ackbach

##### Indicium Physicus
Staff member
(PS you can be sure you'll never run out of questions from me: keen as mustard and not greatly skilled: I am no natural at this, but i LOVE it!)
Well, MHB is all about people like you, so anything you think could improve MHB would be welcome to hear. Please post such things in the Questions, Comments, and Feedback forum, and keep asking math questions!