I am given the following partial differential of a Gaussian where the variance is a function of $t$, $\Xi(t)$ as follows:

$$\frac{\partial}{\partial t} \frac{1}{\sqrt{2\pi \Xi(t)}}\cdot\exp\bigg[\frac{-z^2}{2 \Xi(t)} \bigg]$$

Apparently this can be rewritten as:

$$=\bigg[- \frac{1}{2 \Xi(t)} + \frac{z^2}{2 \Xi(t)^2} \bigg]\bigg(\frac{d \Xi}{dt}\bigg) \frac{1}{\sqrt{2\pi \Xi(t)}}\cdot\exp\bigg[\frac{-z^2}{2 \Xi(t)} \bigg]$$

I can see that the term in front looks it stems from differentiating the exponential but I am not entirely sure how, and I do not get how the $\frac{d \Xi}{dt}$ term appears.

Any help is highly appreciated :-)

  • 1
    $\begingroup$ Its the chain rule, if you had a function $F(\Xi)$ with $\Xi = \Xi(t)$ then $\frac d{dt} F(\Xi(t)) = \frac d{d\Xi} F(\Xi) \cdot \frac d{dt}\Xi$ $\endgroup$ Commented Jul 27, 2018 at 11:01
  • $\begingroup$ that makes sense, I guess I was also worried about the $\pi$ not appearing in the front but I see now that it is basically just multiplied out of the expression where it would otherwise be $\endgroup$
    – user469216
    Commented Jul 27, 2018 at 11:12
  • $\begingroup$ you're welcome :) $\endgroup$ Commented Jul 27, 2018 at 11:29


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