I'm currently studying Elementary differential equations, and I came across a confusion that I had that I think arises from notation, but I would like to clarify with someone. The example problem said this: $$\frac{\frac{df(t)}{dt}}{f(t)} \equiv \frac{d}{dt} (ln|f(t)|)$$ The way I would usually attain this (Assuming that the above is equal to 1) is by splitting the derivative into the differentials and obtaining something like this: $$\frac{d(f(t))}{f(t)} = dt$$ and then integrating on both sides to obtain $$ln|f(t)| = t +c$$ and then taking the derivative to confirm the equivalence. But I feel like my approach is somewhat wrong because the book I am using seems to do this using the derivative operator and integral operator. I think my understanding is somewhat flawed. Can someone please clarify this and explain how the first equivalence is true?


2 Answers 2


First, the equation of interest is not really differential equation in the usual sense. It is merely an identity.

Now, the right-hand side of the second equation in the posted question is not correct. It should have read

$$\frac{df(t)}{f(t)}=d\ln |f(t)|$$

Then, integration leads to the self identity $\ln |f(t)| = \ln |f(t)|$.

To proceed in another way, simply let $y=f(t)$ and use the chain rule. We then have

$$\frac{d\ln f(t)}{dt}=\frac{d\ln y}{dt}=\frac{d\ln y}{dy}\frac{dy}{dt}$$

And proceeding, we obtain

$$\bbox[5px,border:2px solid #C0A000]{\frac{d\ln f(t)}{dt}=\frac{1}{f(t)}\frac{d f(t)}{dt}}$$

  • $\begingroup$ I had commented, but apparently it didn't get submitted. There is nothing wrong with his equation (other than a missing parenthesis on the lh side), and your version ignores the central cause of his hesitation, which is the use of differentials. $\endgroup$ Aug 15, 2015 at 3:45
  • $\begingroup$ Ah. mea culpa. I thought it was the first equation you were complaining about, which was correct, so i downvoted I thought the complaint was wrong. But the fault was mine for not reading carefully enough. I have changed the downvote to an upvote. $\endgroup$ Aug 15, 2015 at 3:52
  • $\begingroup$ @PaulSinclair Thank you!! Much appreciated. $\endgroup$
    – Mark Viola
    Aug 15, 2015 at 4:01
  • $\begingroup$ Thank you! The hint made a lot of sense. I missed that. $\endgroup$ Aug 15, 2015 at 6:37
  • $\begingroup$ You're quite welcome. My pleasure. And please let me know how I can improve my answer. I really want to give you the best answer I can. $\endgroup$
    – Mark Viola
    Aug 15, 2015 at 6:45

Your understanding is flawed, but then again it isn't flawed. You are dealing with something that has previously been a subject of great controversy in mathematics. But now it has been pretty much settled. The naive concept of free differentials has been a part of calculus since it was invented, but is fundamentally flawed. In a great quote, Berkeley castigated them as being "the ghosts of departed quantities". However, they work. Perfectly. This has resulted in the development of rigorous (but highly technical) means of defining differentials that do work (though never as well as the naive concept did). So yes, it is perfectly okay to solve this with the means that you used. But until one sees a rigorous development of differentials, it has to be considered as lacking rigor.

The rigorous solution uses the identity $\int g(u(x))u'(x)dx = \int g(u) du$ with $u = f$ and $g(x) = \ln|x|$

  • $\begingroup$ You likely have heard of Non-Standard Calculus, but I thought that I would share this link since the focus of your answer is around "differentials" and their rigorous (or lack thereof) interpretation. In this beautiful theory they are given a rigorous interpretation using hyperreal numbers. Kind of cool. $\endgroup$
    – Mark Viola
    Aug 15, 2015 at 4:00

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