I've been messing around with exponentials lately, and an interesting question I asked myself was, is it possible to calculate the exponential growth at a point?

So, I'm sure you're familiar with derivatives and tangents. With a derivative if you have $f(x)$, and $f'(x)$ is constant from $a$ to $a+1$, then $f(a+1) = f(a)+f'(a)$

Now, let's say we have some function $f(x)=2^x$ where the rate of growth is constant(100% growth every unit), now let's say $g(x)$ will be our equation to describe exponential growth at each point. Since it's constant throughout, $g(x)=2$ at all times. You can then say that, if $g(x)$ is constant from $a$ to $a+1$, then $f(a+1) = f(a)*g(a)$

I've been able to calculate that $$g(x)=exp\left(\frac{f'(x)}{f(x)}\right)$$ Now, this is very interesting, and even somewhat useful to me especially for regression, since exponential growth rates do in fact change a lot in the real world. And you can either solve $g(x)=exp\left(\frac{y'}{y}\right)$ as a differential equation on a case by case basis to get the exponential for a specific growth function, or use the more general integral, particularly if you want to compute it numerically: $$f(x)=c_1exp\left(\int_1^xln(g(t))dt\right)$$

Explanation aside, my question is does $g(x)$ have a more formal name, and where are some places where I can learn more about, or see such math being used?

Edit: After a brief discussion in the comments, I have realized I need to make more clear what exactly g(x) is doing, so I will write, in order, that $$g(x)=\lim_{h\to 0} \sqrt[h]{\frac{f(x+h)}{f(x)}} = exp\left(\frac{f'(x)}{f(x)}\right)$$ And that in a similar to which $f(x+h)\approx f(x)+f'(x)*h$, $$f(x+h)\approx f(x)g(x)^h$$

  • $\begingroup$ I'm not entirely clear on what exactly you mean by $g$, or even what you mean by growth rate. Because $f$ is not growing at a constant rate if it is exponential. Do you mean something like "if $f$ grows then $f'(x) = g(x) f(x)$ for some function $g(x)$ and $g$ is constant?? $\endgroup$
    – memerson
    Commented Apr 7, 2021 at 20:55
  • $\begingroup$ @memerson I guess I could have been more clear, but it was pretty hard to articulate from the beginning. Pretty much with g(x) it's a derivative, but instead of describing linear growth, it describes multiplicative growth. The basis is that for any equation $f(x)=a^x$, the growth rate is constant, hence: $g(x)=a$, and then expanding this out to the idea of "what if g(x) wasn't constant, but instead different at each point?" $\endgroup$ Commented Apr 7, 2021 at 21:04
  • $\begingroup$ It seems like the best way to make this concept of "multiplicative growth" is that it is somehow related to $f'(x)/f(x)$? If we reverse engineer from the equation $g(x) = \operatorname{exp}(f'(x)/f(x))$ then $g(x) = \ln(f'(x)/f(x)$ which seems to be something like what you want. Would you be willing to take this as what you mean by "multiplicative growth"? $\endgroup$
    – memerson
    Commented Apr 7, 2021 at 21:34
  • $\begingroup$ @memerson I think you're still missing the picture. Think back to basic exponential models where you have $a * b^t$, b is your rate of growth, which is constant. It's just that, but dynamic. Also, $g(x) \ne ln(f'(x)/f(x))$ as you mentioned. I believe what you meant to do was $ln(g(x))=f'(x)/f(x)$. Think of g(x) like this: with a normal derivative, you add, with g(x), you multiply. It's the same exact concept, but for two different operations, to represent important information about specific functions. $\endgroup$ Commented Apr 7, 2021 at 22:25
  • $\begingroup$ It's still not clear what the "rate of growth" is for a general function. We have the definition of something for exponential functions. We say that the rate of growth of some exponential function $f(x) = b^x$ is defined to be $b$. Ok, that's great. Then what is the exponential growth of $g(x) = e^{\sin(x)|\tan(1/x) + 5x^2$. The answer isn't well defined because we haven't defined what we mean by growth rate for non-exponential functions. It also isn't clear what special property the growth rate has that you are trying to preserve so how should we know how to extend. $\endgroup$
    – memerson
    Commented Apr 8, 2021 at 2:01

1 Answer 1


After discussing what the operation I was doing really meant in the comments, a new name hit me, multiplicative derivative, and after a quick google search, it turns out the exact thing I have described does in fact already exist and has been written about! It has been coined the term "Multiplicative calculus" by others, and many papers seem to have been written with it. I will post just one, below. The symbol commonly used for it is $$f^*(x)=\lim_{h\to 0}\left(\frac{f(x+h)}{f(x)}\right)^{\frac{1}{h}}=e^{\frac{f'(x)}{f(x)}}$$

similar to how I had already described. Regardless of ending up just answering my own question, hopefully this can serve as a gateway for anyone who was also curious of similar things to find the term they may have been looking for, and thanks to @memerson in particular for our discussion in the comments, which led to me discovering the name.



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