<s>[INST] How to find the derivative with respect to $x^2$? I would like to find the derivative of this function with respect to x^2.
$-0.5\ln({2\pi x^{2}}) - \frac{1}{2 x^{2}}$
The answer I got is:
$-(0.5)(\frac{1}{2\pi x^{2}})(2\pi)(2) - (0.5)(-1)(\frac{1}{x^{4}})(2)$
$=-\frac{1}{x^{2}} + \frac{1}{x^{4}}$
Is it correct?
 [/INST] If in some problems you are having difficulties then you can use chain rule.
y = $-0.5\ln({2\pi x^{2}}) - \frac{1}{2 x^{2}}$ =  $-0.5ln(2{\pi}) - ln(x) -  \frac{x^{-2}}{2}       $
To find: $\frac{dy}{dx^2}$
$\frac{dy}{dx}\frac{dx}{dx^2}$ = $\frac{dy}{dx}\left({\frac{dx^2}{dx}}\right)^{-1}$ = $(2x)^{-1}\frac{dy}{dx}$
$\frac{dy}{dx^2}$ = $\frac{1}{2x}$$(-\frac{1}{x}+x^{-3})$
Or Simply:
$-0.5\frac{1}{2\pi{x^2}}2{\pi}$ - $\frac{1}{2}(-\frac{1}{x^4})$ = $-\frac{0.5}{x^2}$ +$\frac{1}{2x^4}$
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