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$$\frac{(1-α^2)}{2σ}\exp-\frac{|x-μ|-α(χ-μ)}{σ}$$

Could someone explain how to take the first and second partial derivative with respect to μ of this function? I tried taking the log-likelihood first and then taking the derivative, but I got super confused. I also have this question, when there is, for instance, a function $\log2*(|x-μ|-α(x-μ))$, which rule should I use to take the derivative? The logarithm rule and then the chain rule?

My approach was $log(1-α^2)-log2σ(-|x-μ|-α(x-μ)$ Where thé first term Is 0 but then I’m not sure whether this is correct by applying the logarithm rule
$\frac{|x|-αΧ}{|x-μ|-α(x-μ)}$

Thanks :)

Anna
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  • First, eliminate the constant at the front (of course)... – David G. Stork Mar 21 '22 at 17:54
  • Yes but the difficult part comes after .. – Anna Mar 21 '22 at 18:13
  • Of course. But you should learn that you should pare down your question to its core. That way 1) you're more likely to solve it on your own, 2) you'll show that you understand at least some of the concepts, 3) you'll save everyone time, and 4) you'll get more help. Potential helpers think you don't understand the irrelevance of a constant here! Is that what you want? – David G. Stork Mar 21 '22 at 18:15
  • I added my solution now , hopefully it provides a better overview of my mistake – Anna Mar 21 '22 at 18:18
  • After eliminating the irrelevant constant factor at the beginning (which just confuses and complicates the problem), the basic rules of differentiation give: $$\frac{\left(\alpha -\text{Abs}'(x-\mu )\right) e^{\frac{\left| x-\mu \right| -\alpha (\xi -\mu )}{\sigma }}}{\sigma }$$ – David G. Stork Mar 21 '22 at 18:23

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