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Math Help - Two Chain-Rules in one?

  1. #1
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    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?
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  2. #2
    Newbie Riyzar's Avatar
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    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} ?
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  3. #3
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    Quote Originally Posted by Riyzar View Post
    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
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  4. #4
    Newbie Riyzar's Avatar
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    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.
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  5. #5
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    So that ends up being:

    48(8x^2+2)^5*(3x^2-9)^14 + (8x^2+2)^6 * 42(3x^2-9)^13
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  6. #6
    Newbie Riyzar's Avatar
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    Close, look at this:  [(8x^2+2)^6]' = 6(8x^2+2)^5(16x)
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  7. #7
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    oh god, I added in that 48 by mistake.
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  8. #8
    Newbie Riyzar's Avatar
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    Easy mistake 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!
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  9. #9
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    Quote Originally Posted by Riyzar View Post
    Easy mistake 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
    Last edited by Zanderist; March 13th 2010 at 06:31 PM.
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  10. #10
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    Quote Originally Posted by Zanderist View Post
    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)
    Last edited by ione; March 13th 2010 at 07:32 PM.
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  11. #11
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    Quote Originally Posted by ione View Post
    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.
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