This is just confusing me. I have encountered in thermodynamics the following:

$$d(\ln V) = \frac1VdV$$

Which I think comes from knowing that:

$$\frac d{dV} (\ln V) = \frac1V$$

But I never thought you could just rearrange derivatives in that manner.

I would appreciate any help! Thanks!

  • 1
    $\begingroup$ I never liked thinking of things that way. And sometimes, manipulating those ds can get you into trouble. Rigorously, you can manipulate them with the math of differential forms. But texts on thermodynamics shouldn’t require that unless they say so explicitly. If f(x) = ln(x) then f’(x) = 1/x. That’s all we should need. $\endgroup$
    – NicNic8
    Commented Nov 21, 2020 at 17:46

1 Answer 1


It depends how formal you want to make things, but usually the differential $df$ of a function is defined as $$df = f'(x)\,dx,$$ so in your case, $$d(\ln V) = (\ln V)'\,dV = \tfrac1V\,dV.$$ There are various other questions on this site related to yours.

  • $\begingroup$ Thanks a lot for that definition. I will go through the other questions as well. $\endgroup$ Commented Nov 21, 2020 at 17:58

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