Two Chain-Rules in one?

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March 13th 2010, 03:56 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?
March 13th 2010, 04:04 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 $(3x^2-9)^{14}$ ?
March 13th 2010, 04:07 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
March 13th 2010, 04:11 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.
March 13th 2010, 04:18 PM
Zanderist

So that ends up being:

$48(8x^2+2)^5*(3x^2-9)^14 + (8x^2+2)^6 * 42(3x^2-9)^13$
March 13th 2010, 04:27 PM
Riyzar

Close, look at this: $[(8x^2+2)^6]' = 6(8x^2+2)^5(16x)$
March 13th 2010, 04:29 PM
Zanderist

oh god, I added in that 48 by mistake.
March 13th 2010, 04:34 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!
March 13th 2010, 05:16 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
March 13th 2010, 05:45 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

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

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

$<br /> [f(x)g(x)]' = f(x)g'(x)+g(x)f'(x)$

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

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

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

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

$=24x(8x^2+6)^5(3x^2-9)^{13}(40x^2-29)$
March 13th 2010, 07:00 PM
Zanderist

Quote:

Originally Posted by ione

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

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

$<br /> [f(x)g(x)]' = f(x)g'(x)+g(x)f'(x)$

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

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

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

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

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

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