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Math Help - Help with Chain Rule!!!

  1. #1
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    Help with Chain Rule!!!

    I have the very basic of the chain rule understood however the functions

    with powers that contain equation confuse the hell out of me I hope to get some prompt and excellent health let me show an example

    f(x)=e^{(1+3t)^{2}}

    and the chain rule is

     <br />
f\prime(g(x)) * g\prime(x)<br />

    and the answer i was shown and the stops shown dont make sense from what i think i understand

    f\prime(x)=(e^{1+3t^{2}})(2)(1+3x)3

    I can understand f\prime(x)=(e^{1+3t^{2}}) to
    f\prime(g(x)) and (2)(1+3x) to be g\prime however why would you derive (1+3x) as you did all steps of the chain rule why would you need a further derivation

    pretty much anything to power containing a constant confuses me

    hope this posts get a reply unlike my other ones
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  2. #2
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by sk8erboyla2004 View Post
    I have the very basic of the chain rule understood however the functions

    with powers that contain equation confuse the hell out of me I hope to get some prompt and excellent health let me show an example

    f(x)=e^{(1+3t)^{2}}

    and the chain rule is

     <br />
f\prime(g(x)) * g\prime(x)<br />

    and the answer i was shown and the stops shown dont make sense from what i think i understand

    f\prime(x)=(e^{1+3t^{2}})(2)(1+3x)3

    I can understand f\prime(x)=(e^{1+3t^{2}}) to
    f\prime(g(x)) and (2)(1+3x) to be g\prime however why would you derive (1+3x) as you did all steps of the chain rule why would you need a further derivation

    pretty much anything to power containing a constant confuses me

    hope this posts get a reply unlike my other ones
    Chain rule twice:

    f(x)=e^{(1+3t)^{2}}

    Let u=(1+3t)^2
    du =2(1+3t) \cdot 3

    f'(x) = e^u du

    = e^{{1+3t}^2} \cdot 2(1+3t) \cdot 3
    <br />
= 6e^{{1+3t}^2} \cdot (1+3t)
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  3. #3
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by mollymcf2009 View Post
    Chain rule twice:

    f(x)=e^{(1+3t)^{2}}

    Let u=(1+3t)^2
    du =2(1+3t) \cdot 3

    f'(x) = e^u du

    = e^{(1+3t)^2} \cdot 2(1+3t) \cdot 3 Mr F edit: Excuse me for butting in. matheagle has corrected this line (and the next).The original line was = e^{{1+3t}^2} \cdot 2(1+3t) \cdot 3. (So what matheagle has said is perfectly correct).

    <br />
= 6e^{(1+3t)^2} \cdot (1+3t)
    This is what you meant 6(1+3t)e^{(1+3t)^2}.

    The derivative of e^{g(x)} is e^{g(x)}\cdot g'(x).
    Last edited by mr fantastic; March 13th 2009 at 03:42 AM.
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  4. #4
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by matheagle View Post
    This is what you meant 6(1+3t)e^{(1+3t)^2}.

    The derivative of e^{g(x)} is e^{g(x)}\cdot g'(x).

    That's what I did have. Same thing.
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  5. #5
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by mollymcf2009 View Post
    That's what I did have. Same thing.
    You meant that, but look at what you typed.
    You have a typo. It's a TeX error. You used {} instead of (), making the exponent of e, accidentally 1+3t^2 instead of (1+3t)^2.
    I know you know what you're doing, but the typo will cause confusion. That's why I pointed it out.
    Last edited by matheagle; March 13th 2009 at 03:45 PM.
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  6. #6
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by matheagle View Post
    You meant that, but look at what you typed.
    You have a typo. It's a TeX error. You used {} instead of (), making the exponent of e, accidentally 1+3t^2 instead of (1+3t)^2.
    I know you know what you're doing, but the typo will cause confusion. That's why I pointed it out.

    You're totally right. When you quoted me, I didn't realize that you had corrected within my quote. I just read it like it was what I had put, so I couldn't understand what you were talking about. Thanks for catching that. Those parentheses MUST be there. Thanks and sorry for being short with you, I was SO tired!! Have a great weekend!
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  7. #7
    MHF Contributor matheagle's Avatar
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    No problem. You weren't nasty at all.
    I've even been called names here, right mr fancy.
    I debated about fixing your typo or not within the quote.
    In the end I decided to type it correctly.
    My concern here is that students will get confused over typos like that.
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