# Two Chain-Rules in one?

• Mar 13th 2010, 04:56 PM
Zanderist
Two Chain-Rules in one?
To solve something like:

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

Would I use the chain rule twice?
• Mar 13th 2010, 05:04 PM
Riyzar
Multiply $14(3x^2-9)$ out and then use the product rule.

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

You'll have to use the chain rule once.

EDIT: Looking at your code did you mean $(3x^2-9)^{14}$?
• Mar 13th 2010, 05:07 PM
Zanderist
Quote:

Originally Posted by Riyzar
Multiply $14(3x^2-9)$ out and then use the product rule.

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

You'll have to use the chain rule once.

That's (3x^2-9) raised to the 14th
• Mar 13th 2010, 05:11 PM
Riyzar
Ah ok, then yes you will.

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

And then use the product rule.
• Mar 13th 2010, 05:18 PM
Zanderist
So that ends up being:

$48(8x^2+2)^5*(3x^2-9)^14 + (8x^2+2)^6 * 42(3x^2-9)^13$
• Mar 13th 2010, 05:27 PM
Riyzar
Close, look at this: $[(8x^2+2)^6]' = 6(8x^2+2)^5(16x)$
• Mar 13th 2010, 05:29 PM
Zanderist
oh god, I added in that 48 by mistake.
• Mar 13th 2010, 05:34 PM
Riyzar
Easy mistake (Happy) you did the same thing for the derivative of $(3x^2-9)^{14}$ So just make sure you check that and then you should be good!
• Mar 13th 2010, 06:16 PM
Zanderist
Quote:

Originally Posted by Riyzar
Easy mistake (Happy) you did the same thing for the derivative of $(3x^2-9)^{14}$ So just make sure you check that and then you should be good!

I enter in:

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

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

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

Math anxiety
• Mar 13th 2010, 06:45 PM
ione
Quote:

Originally Posted by Zanderist
I enter in:

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

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

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

Math anxiety

$f'(x)= 6(8x^2+2)^5*(16x)=96x(8x^2+2)^5$

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

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

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

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

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

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

$=24x(8x^2+6)^5(3x^2-9)^{13}(40x^2-29)$
• Mar 13th 2010, 08:00 PM
Zanderist
Quote:

Originally Posted by ione
$f'(x)= 6(8x^2+2)^5*(16x)=96x(8x^2+2)^5$

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

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

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

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

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

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

$=24x(8x^2+6)^5(3x^2-9)^{13}(40x^2-29)$

Alright, so what I've learned for this problem, is to treat it as two chain rules, then use the product rule.