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I'm referring specifically to the $\frac{\delta f}{\delta x}|_{y}, \frac{\delta f}{\delta y}|_{x}, \frac{\delta u}{\delta x}|_{t} and \frac{\delta u}{\delta t}|_{x}$. I've included the rest to provide some context.

Edit: They also provide an example which confused me. Could someone also explain how they did the example?:

enter image description here

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This is sometimes used to mean "keeping this variable constant". It emphasizes that while you take the partial derivative with respect to e.g. $x$, you are keeping $y$ constant.

I think it has roots in certain areas of applied math where there may be constraints. For example if you have pressure, volume and temperature of a gas, then they always satisfy a constraint. So when you write the partial derivative with respect to say, pressure, do you let the volume and temperature change with the constraint? (physically natural but not strictly speaking what the usual partial derivative on $\mathbf{R}^3$ is) or just keep them constant? (violates the gas equation but is mathematically natural). Here the notation would be telling you explicitly to do the latter.

I don't like this notation because often it is completely redundant, but sometimes it is absolutely crucial.

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  • $\begingroup$ This makes so much more sense. It is sometimes used as "evaluate here" like I posted, but this is clearly the correct interpretation in this context. +1 $\endgroup$ – The Count Mar 9 '17 at 15:39
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It means the same as "of $t$", I.e., $f|_t=f(t)$.

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  • $\begingroup$ I edited my post, would you mind also explaining the example I added? $\endgroup$ – mrnovice Mar 9 '17 at 15:15
  • $\begingroup$ That would be best as a completely new question, since it is largely unrelated to the first. $\endgroup$ – The Count Mar 9 '17 at 15:22
  • $\begingroup$ I would disagree, since what they have done in the example, would appear to contradict your answer (to my understanding of it). $\endgroup$ – mrnovice Mar 9 '17 at 15:25
  • $\begingroup$ Well maybe I am misunderstanding what the added example says. There are no vertical bars anywhere in it... $\endgroup$ – The Count Mar 9 '17 at 15:27
  • $\begingroup$ They are using the formula in the last line of the picture above it, so by what your answer says, surely the first part should be in terms of $t$ and the second part in terms of $x$? $\endgroup$ – mrnovice Mar 9 '17 at 15:29

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