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Math Help - derivatives question

  1. #1
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    derivatives question

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  2. #2
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    Re: derivatives question

    please don't ask a question twice in two separate threads. In your last post you were confused by the notation. Are you straight on that now?
    Last edited by romsek; April 21st 2014 at 05:51 AM.
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  3. #3
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    Re: derivatives question

    I'm sorryyyyy!!!!!!

    I was confused about which notation to preserve when manipulating the equation! I still can't do it!
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    Re: derivatives question

    Quote Originally Posted by Applestrudle View Post
    I'm sorryyyyy!!!!!!

    I was confused about which notation to preserve when manipulating the equation! I still can't do it!
    I don't think it's true. I let $f(x)=e^{a x}$ and chugged it through and the difference of the two sides isn't zero.
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  5. #5
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    Re: derivatives question

    Quote Originally Posted by romsek View Post
    I don't think it's true. I let $f(x)=e^{a x}$ and chugged it through and the difference of the two sides isn't zero.
    actually it is looking to be true. I wasn't quite evaluating it correctly. I'll need to toy with this a bit or maybe one of the math powerhouses will chime in with something deviously clever.

    This wiki page has the derivation you need.

    look at the section on higher derivatives.
    Last edited by romsek; April 21st 2014 at 01:23 PM.
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  6. #6
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    Re: derivatives question

    Given any y = f(x), then y' = f'(x). Using implicit differentiation and taking the derivative of both sides with respect to y, you get:

    \begin{align*}\dfrac{d}{dy}(y) & = \dfrac{d}{dy}\left(f(x)\right) \\ 1 & = f'(x)\dfrac{dx}{dy} \\ \dfrac{dx}{dy} & = \dfrac{1}{y'}\end{align*}

    For the second derivative:

    \begin{align*}\dfrac{d}{dy}\left( \dfrac{dx}{dy} \right) & = \dfrac{d}{dy}\left( \dfrac{1}{y'} \right) \\ \dfrac{d^2x}{dy^2} & = -\dfrac{y''\dfrac{dx}{dy}}{(y')^2} \\ & = -\dfrac{y''}{(y')^3}\end{align*}

    One more time:

    \begin{align*}\dfrac{d}{dy}\left(\dfrac{d^2x}{dy^2  }\right) & = \dfrac{d}{dy}\left(-\dfrac{y''}{(y')^3}\right) \\ \dfrac{d^3x}{dy^3} & = -\dfrac{(y')^3y''' \dfrac{dx}{dy} - 3(y'')^2(y')^2\dfrac{dx}{dy}}{(y')^6} \\ & = \dfrac{3(y'')^2(y') - (y')^2y'''}{(y')^6} \\ & = \dfrac{3(y'')^2}{(y')^5} - \dfrac{y'''}{(y')^4}\end{align*}
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