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Square in denominator of derivative

  1. Mar 21, 2013 #1
    [itex] \frac{d}{d \sigma ^2} [log(\sigma ^ 2) - \frac{1}{\sigma ^ 2}] [/itex]

    I think the first part is

    [itex] \frac{1}{\sigma ^ 2} [/itex]

    but pretty clueless after that. I also want to take the second derivative.
    Any help or a reference would be great.
    Thanks!
     
  2. jcsd
  3. Mar 21, 2013 #2
    substitute x = sigma^2 and differentiate as you normally would.
     
  4. Mar 21, 2013 #3
    I think this is right.

    [itex] \frac{d}{d \sigma ^2} [log(\sigma ^2) - \frac{1}{\sigma ^2}] = \frac{1}{\sigma ^2} + \frac{1}{\sigma ^4} [/itex]

    then for the second derivative

    [itex] \frac{d}{d \sigma ^2} [\frac{1}{\sigma ^2} + \frac{1}{\sigma ^4}] = - \frac{1}{\sigma ^4} - \frac{2}{\sigma ^6} [/itex]

    Yay, nay? How does that look? I used this url as a reference
    https://files.nyu.edu/mrg217/public/mle_introduction1.pdf [Broken]

    equations 51 and 60

    Thanks
     
    Last edited by a moderator: May 6, 2017
  5. Mar 21, 2013 #4

    HallsofIvy

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    You are mistaken. There are two ways to do the first function:
    1) Use the "chain rule". Let [itex]u= x^2[/itex] so that [itex]ln(x^2)= ln(u)[/itex]. Then d ln(u)/du= 1/u and du/dx= 2x. d(ln(x^2)/dx= (d(ln(u))/du)(du/dx).

    2) But it is much simpler to use the "laws of logarithms"- [itex]ln(x^2)= 2ln(x)[/itex] so [itex]d(ln(x^2))/dx= 2 d(ln(x))/dx[/itex].

    For the second, write [tex]1/\sigma^2= \sigma^{-2}[/tex] and use "[tex]d(x^n)/dx= nx^{n-1}[/tex]".

    I just noticed that this was "with respect to [itex]\sigma^2[/itex], NOT with respect to [itex]\sigma[/itex]. That changes everything!

    Let [itex]x= \sigma^2[/itex]. Then your problem becomes
    [tex]\frac{d}{dx}(ln(x)- x^{-2}[/tex].
    Differentiate that and replace x with [itex]\sigma^2[/itex]. You don't need to use the chain rule!
     
    Last edited: Mar 21, 2013
  6. Mar 21, 2013 #5
    Awesome, I end up with the same answer as before. Thanks for the help.
     
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