# Math Help - A simple differentiation

1. ## 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?

2. Originally Posted by chella182
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?

3. 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.