Square in denominator of derivative

[autobot.d]

\frac{d}{d \sigma ^2} [log(\sigma ^ 2) - \frac{1}{\sigma ^ 2}]

I think the first part is

\frac{1}{\sigma ^ 2}

but pretty clueless after that. I also want to take the second derivative.
Any help or a reference would be great.
Thanks!

 

[mikeph]

substitute x = sigma^2 and differentiate as you normally would.

 

[autobot.d]

I think this is right.

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

then for the second derivative

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

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

equations 51 and 60

Thanks

 

[HallsofIvy]

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

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

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

Let x= \sigma^2. Then your problem becomes
\frac{d}{dx}(ln(x)- x^{-2}.
Differentiate that and replace x with \sigma^2. You don't need to use the chain rule!

 

[autobot.d]

Awesome, I end up with the same answer as before. Thanks for the help.