# Two Chain-Rules in one?

• Mar 13th 2010, 03:56 PM
Zanderist
Two 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 PM
Riyzar
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 PM
Zanderist
Quote:

Originally Posted by Riyzar
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.

That's (3x^2-9) raised to the 14th
• Mar 13th 2010, 04:11 PM
Riyzar
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 PM
Zanderist
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 PM
Riyzar
Close, look at this: \$\displaystyle [(8x^2+2)^6]' = 6(8x^2+2)^5(16x)\$
• Mar 13th 2010, 04:29 PM
Zanderist
oh god, I added in that 48 by mistake.
• Mar 13th 2010, 04:34 PM
Riyzar
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 PM
Zanderist
Quote:

Originally Posted by Riyzar
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!

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 PM
ione
Quote:

Originally Posted by Zanderist
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

\$\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 PM
Zanderist
Quote:

Originally Posted by ione
\$\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)\$

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