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Math Help - A simple differentiation

  1. #1
    Senior Member chella182's Avatar
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    A simple differentiation

    Okay, this is part of a bigger question, but you don't need to know that. I just can't quite seem to get this right. I've got this function l...

    l=3n\ln{(\bar{x}_{g})}-n\ln{6}-4n\ln{\theta}-\frac{n\bar{x}}{\theta}

    ...and I want to differentiate with respect to \theta. So collecting the terms that don't depend on \theta together as one general constant I get...

    l=K-4n\ln{\theta}-\frac{n\bar{x}}{\theta}

    Looks simple, and I got an answer, but, ideally, when I find the second derivative I need it to be quite obviously negative (so that the answer I get when solving the first derivative equal to 0 is a maximum of the function), which it isn't, which is making me think that I've done something wrong. Might be worth me pointing out that n and \bar{x} are also constants, so they don't affect the differentiating; just make it look rank.

    Any ideas?
    Last edited by chella182; April 30th 2009 at 07:43 AM.
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  2. #2
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    Quote Originally Posted by chella182 View Post
    Okay, this is part of a bigger question, but you don't need to know that. I just can't quite seem to get this right. I've got this function l...

    l=3n\ln{(\bar{x}_{g})}-n\ln{6}-4n\ln{\theta}-\frac{n\bar{x}}{\theta}

    ...and I want to differentiate with respect to \theta. So collecting the terms that don't depend on \theta together as one general constant I get...

    l=K-4n\ln{\theta}-\frac{n\bar{x}}{\theta}

    Looks simple, and I got an answer, but, ideally, when I find the second derivative I need it to be quite obviously negative (so that the answer I get when solving the first derivative equal to 0 is a maximum of the function), which it isn't, which is making me think that I've done something wrong. Might be worth me pointing out that n and \bar{x} are also constants, so they don't affect the differentiating; just make it look rank.

    Any ideas?
    what did you get for the second derivative?
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  3. #3
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    Hello, chella182!


    L \:=\:3n\ln{(\bar{x}_{g})}-n\ln{6}-4n\ln{\theta}-\frac{n\bar{x}}{\theta}

    Then: . L' \;=\;-\frac{4n}{\theta} + \frac{n\bar{x}}{\theta^2} \:=\:0 \quad\Rightarrow\quad -4n\theta + n\bar{x} \:=\:0\quad\Rightarrow\quad\boxed{ \theta \:=\:\frac{\bar{x}}{4}} \quad\hdots\:\text{critical value}


    We have: . L' \:=\:-4n\theta^{-1} + n\bar{x}\,\theta^{-2}

    . . L'' \;=\;4n\theta^{-2} - 2n\bar{x}\,\theta^{-3} \;=\;\frac{2n}{\theta^3}\left(2\theta - \bar{x}\right)


    \text{If }\,\theta = \frac{\bar{x}}{4}\!:\;\;L'' \;=\;\frac{2n}{\left(\frac{\bar{x}}{4}\right)^3}\l  eft[2\left(\frac{\bar{x}}{4}\right) - x\right] \;=\;\frac{2n\cdot64}{\bar{x}^3}\left[\frac{\bar{x}}{2} - \bar{x}\right] \;=\;\frac{128n}{\bar{x}^3}\left(-\frac{1}{2}\bar{x}\right)

    . . . . . \boxed{L'' \;=\;-\frac{64n}{\bar{x}^2} }\quad\hdots\:\text{The concavity is downward.}


    The critical value is a maximum.

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