To solve something like:

$\displaystyle f(x)= (8x^2 + 2)^6 * (3x^2-9)^(14) $

Would I use the chain rule twice?

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- Mar 13th 2010, 03:56 PMZanderistTwo Chain-Rules in one?
To solve something like:

$\displaystyle f(x)= (8x^2 + 2)^6 * (3x^2-9)^(14) $

Would I use the chain rule twice? - Mar 13th 2010, 04:04 PMRiyzar
Multiply $\displaystyle 14(3x^2-9) $ out and then use the product rule.

$\displaystyle [f(x)g(x)]' = f(x)g'(x)+g(x)f'(x)$

You'll have to use the chain rule once.

EDIT: Looking at your code did you mean $\displaystyle (3x^2-9)^{14} $? - Mar 13th 2010, 04:07 PMZanderist
- Mar 13th 2010, 04:11 PMRiyzar
Ah ok, then yes you will.

You should use the chain rule to find the derivatives of $\displaystyle (8x^2+2)^6 $ and of $\displaystyle (3x^2-9)^{14}$

And then use the product rule. - Mar 13th 2010, 04:18 PMZanderist
So that ends up being:

$\displaystyle 48(8x^2+2)^5*(3x^2-9)^14 + (8x^2+2)^6 * 42(3x^2-9)^13$ - Mar 13th 2010, 04:27 PMRiyzar
Close, look at this: $\displaystyle [(8x^2+2)^6]' = 6(8x^2+2)^5(16x)$

- Mar 13th 2010, 04:29 PMZanderist
oh god, I added in that 48 by mistake.

- Mar 13th 2010, 04:34 PMRiyzar
Easy mistake (Happy) you did the same thing for the derivative of $\displaystyle (3x^2-9)^{14}$ So just make sure you check that and then you should be good!

- Mar 13th 2010, 05:16 PMZanderist
I enter in:

$\displaystyle (48x^2+12)^5 *(3x^2-9)^{14} + (8x^2+2)^6 * (42x^2-126)^{13}$

And it comes back as incorrect, what do I need to do next?

Wait that's not my question.

I now want to know what to do with the 16x

in

$\displaystyle f'(x)= 6(8x^2+2)^5*(16x)$

Math anxiety - Mar 13th 2010, 05:45 PMione
$\displaystyle f'(x)= 6(8x^2+2)^5*(16x)=96x(8x^2+2)^5$

$\displaystyle g'(x)= 14(3x^2-9)^{13}*(6x)=84x(3x^2-9)^{13}$

$\displaystyle

[f(x)g(x)]' = f(x)g'(x)+g(x)f'(x)$

$\displaystyle =(8x^2+2)^6(84x)(3x^2-9)^{13}+(3x^2-9)^{14}(96x)(8x^2+2)^5

$

$\displaystyle =12x(8x^2+6)^5(3x^2-9)^{13}(7(8x^2+2)+8(3x^2-9))$

$\displaystyle =12x(8x^2+6)^5(3x^2-9)^{13}(56x^2+14+24x^2-72)$

$\displaystyle =12x(8x^2+6)^5(3x^2-9)^{13}(80x^2-58)$

$\displaystyle =24x(8x^2+6)^5(3x^2-9)^{13}(40x^2-29)$ - Mar 13th 2010, 07:00 PMZanderist