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Thread: 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

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

    and the chain rule is

    $\displaystyle
    f\prime(g(x)) * g\prime(x)
    $

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

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

    I can understand $\displaystyle f\prime(x)=(e^{1+3t^{2}})$ to
    $\displaystyle f\prime(g(x))$ and $\displaystyle (2)(1+3x)$ to be $\displaystyle g\prime$ however why would you derive $\displaystyle (1+3x)$ 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

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

    and the chain rule is

    $\displaystyle
    f\prime(g(x)) * g\prime(x)
    $

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

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

    I can understand $\displaystyle f\prime(x)=(e^{1+3t^{2}})$ to
    $\displaystyle f\prime(g(x))$ and $\displaystyle (2)(1+3x)$ to be $\displaystyle g\prime$ however why would you derive $\displaystyle (1+3x)$ 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:

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

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

    $\displaystyle f'(x) = e^u du$

    $\displaystyle = e^{{1+3t}^2} \cdot 2(1+3t) \cdot 3$
    $\displaystyle
    = 6e^{{1+3t}^2} \cdot (1+3t)$
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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:

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

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

    $\displaystyle f'(x) = e^u du$

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

    $\displaystyle
    = 6e^{(1+3t)^2} \cdot (1+3t)$
    This is what you meant $\displaystyle 6(1+3t)e^{(1+3t)^2}$.

    The derivative of $\displaystyle e^{g(x)}$ is $\displaystyle e^{g(x)}\cdot g'(x)$.
    Last edited by mr fantastic; Mar 13th 2009 at 02:42 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 $\displaystyle 6(1+3t)e^{(1+3t)^2}$.

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

    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 $\displaystyle 1+3t^2$ instead of $\displaystyle (1+3t)^2$.
    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; Mar 13th 2009 at 02:45 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 $\displaystyle 1+3t^2$ instead of $\displaystyle (1+3t)^2$.
    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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