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Is there a physical meaning for derivative of the function to function ratio? That is, this quantity, $$ Q(x) = \frac{1}{f(x)}\frac{df(x)}{dx} $$ Like for instance, if $f$ is the potential energy, this would be work to potential energy ratio. Or even what can we say about $Q(x)$, say when $Q(x)<0$.

user2167741
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  • This is the scaling factor of the derivitive to the function. This quantity gives an intuition of how the function relates to its gradient. – Nick Aug 05 '18 at 15:12
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    I also learned that this is called 'logarithmic derivative'. https://en.wikipedia.org/wiki/Logarithmic_derivative – user2167741 Aug 05 '18 at 15:34

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This would give $$ \int\limits_{x_0}^x \!Q(\xi) \, d\xi = \ln \frac{f(x)}{f(x_0)} \iff \\ f(x) = f(x_0) \exp {\int\limits_{x_0}^x \! Q(\xi) \, d\xi} $$ An example would be a radioactive decay process ($Q$: decay constant)

Another: light absorption ($Q$: attenuation coefficient).

mvw
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It is also $\frac {d(\ln(f(x))}{dx}$. Note that it has units of the units of $f$ divided by length, so it can be a scale length that shows how quickly $f$ changes.

Ross Millikan
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I know in Physics they use that $\frac{dN}{dt}=Nk$ for various uses so you could use it to show whether a relationship is exponential or not

Henry Lee
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  • I get that - I know the relationship is exponential but wanted to know the physical significance of it. Like mvw says, attenuation coefficient gives a good meaning to it. – user2167741 Aug 05 '18 at 15:33
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    Hmmm I think it is quite hard to give a general meaning to something like that, you could also see it as some kind of relative stability. – Henry Lee Aug 05 '18 at 15:36
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If $f(t)$ measures the size of the population, and $t$ is time, then $Q(t)=f'(t)/f(t)$ is the rate of change of the population per capita.

For example, if each individual reproduces at a constant rate $r$, then $f'(t)/f(t)=r$, so we get exponential population growth $f(t)=f(0) \, e^{rt}$.

But if the per capita growth rate decreases linearly with the size of the population (due to limited resources, for example), $f'(t)/f(t)=r(1-f(t)/K)$, we get logistic growth.

Hans Lundmark
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  • Thanks! Yes, I can see that in case of basic exponential/logistic model. What if $f(t)$ is the rate $r(t)$, where the growth rate $r$ depends on time? How do we interpret it? – user2167741 Aug 05 '18 at 18:21
  • You mean $Q(t)$? Then you get something which I don't know a name for, but which can be solved by separation of variables: $f'/f=r$ gives $\int df/f=\int r(t) , dt$, etc. – Hans Lundmark Aug 05 '18 at 18:33
  • No, then I mean the quantity $r'(t)/r(t)$. I know $r$ follows exponential as well, but I don't know what the quantity would mean. It is some sort of relative rate of change of growth over time. – user2167741 Aug 05 '18 at 18:55
  • Well, if $r(t)$ is the per capita growth rate, then $r'(t)/r(t)$ is rate of change of the per capita growth rate per per capita growth rate... :-) – Hans Lundmark Aug 05 '18 at 19:10