# Square in denominator of derivative

1. Mar 21, 2013

### 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!

2. Mar 21, 2013

### mikeph

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

3. Mar 21, 2013

### 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 [Broken]

equations 51 and 60

Thanks

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

### 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!

Last edited by a moderator: Mar 21, 2013
5. Mar 21, 2013

### autobot.d

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