How is it possible that $d(\ln V) = \frac1VdV$?

Question

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!

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. — NicNic8, Nov 21 '20 at 17:46

score 2 · Accepted Answer · answered Nov 21 '20 at 17:46

2

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.

answered Nov 21 '20 at 17:46

Luke Collins

8,748

Thanks a lot for that definition. I will go through the other questions as well. – dareToDiffer07 Nov 21 '20 at 17:58

How is it possible that $d(\ln V) = \frac1VdV$?

1 Answers1