Have I done the right derivative?
$\frac{d}{dx}log(2\pi x^2)^{-n/2}=\frac{1}{(2\pi x)^{n/2}} (\frac{-n}{2} (2\pi x)^{3n/2} 2\pi)$
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would you show how you ended with this result, otherwise you just can check on wolfram alpha... – Noa Even Aug 05 '19 at 13:57
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Good idea. However, inserting this to Wolfram gives me $Log[1/((2 \pi)^(n/2) (x^2)^(n/2))]$ and that seems wrong. I don't expect Log to remain. Is Wolfram correct? – Aug 05 '19 at 14:12
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Did you insert $Log$ or $\log$ , as $Log$ is mathematically incorrect for $\log$ and it'll be considered as a constant $Log$ ? – 19aksh Aug 05 '19 at 14:16
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If you mean $y=\log\big[(2\pi x^2)^{-n/2}\big]$
we can simplify it as $y = -\frac{n}{2}\big[\log(2\pi) + 2\log(x)\big]$
$$\implies\frac{dy}{dx} = 0 -\frac{n}{2}\times\frac{2}{x} = -\frac{n}{x}$$
19aksh
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Yes that is the derivative. However, your solution have dropped out a $^2$ in $x^2$. – Aug 05 '19 at 14:14
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