I have an expression of the form:
$\int\limits_{-\infty}^{\infty}\exp \left\{ \frac{-1}{2\sigma^2} (x-\mu)^2 \right\} dx = \sqrt{2\pi\sigma^2}$
and I need to take its derivative with respect to $\sigma^2$. The right hand side seems easy enough:
$\frac{\partial}{\partial\sigma^2}{(2\pi\sigma^2)}^{\frac{1}{2}} = \frac{1}{2}\sqrt{2\pi}(\sigma^2)^{\frac{-1}{2}},$
right? What about the left hand side? I don't know how to take the derivative of something that's inside an integral -- can someone please help? I'm more interested in understanding how it would work then just the answer itself.
EDIT: corrected RHS.
EDIT 2:
Through applying the chain rule, the LHS becomes:
$\int\limits_{-\infty}^\infty \frac{1}{2}(x-\mu)^2\frac{1}{\sigma^4} \exp{\lbrace \frac{-1}{2\sigma^2} (x-\mu)^2 \rbrace } dx $
Does this look right? I'm unsure of how to handle the derivative of a function w.r.t to a squared variable (e.g. $\frac{\partial}{\partial\sigma^2}$ as opposed to $\frac{\partial}{\partial\alpha}$). For example, is this true: $\frac{\partial}{\partial\sigma^2} \sigma^{-2} = -\sigma^{-4}$ ?