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.


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: $\frac{\partial}{\partial\sigma^2} \sigma^{-2} = -\sigma^{-4}$ ?

  • 1
    $\begingroup$ RHS is still not correct. You forgot to take the $1/2$ down from the exponent into the front (the power rule is $(x^n)'=nx^{n-1}$ in case you forgot). Additionally, $2\pi$ isn't part of the variable so it doesn't go under the negative exponent; it stays as-is. $\endgroup$ – anon Jan 30 '12 at 10:30
  • $\begingroup$ Thank you. It's been a long time since I've had to do any calculus... $\endgroup$ – mpenkov Jan 30 '12 at 10:45
  • $\begingroup$ You omitted a minus sign in the argument of the exponential. $\endgroup$ – John Bentin Jan 30 '12 at 11:44

First off, the derivative you do have is incorrect. Where'd you get the extra $\pi$? Use the power rule:


Now to take the derivative of the integral expression you simply put the differential operator inside:


One can justify this by putting the integral into the limit definition of the partial derivative and then using the linearity of integration to pull the quotient inside the $\int$. To evaluate from here you need to use the chain rule, power rule, and know the derivative of the exponential function. Good luck...

  • $\begingroup$ thank you for clarifying this point. I've updated my question to include my answer for the integral. Could you please let me know if it is correct? $\endgroup$ – mpenkov Jan 30 '12 at 13:21

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