Take the definition of the derivative:

$$\frac{d}{dx}f(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$

Using this definition to calculate the derivative of $e^x$ is not the most trivial thing to do, as one ends up with:

$$\frac{d}{dx}e^x=e^x\lim_{h\to 0}\frac{e^h-1}{h}$$

We can finish this off by a change of variables $n=\frac1h$.

$$e^x\lim_{h\to 0}\frac{e^h-1}{h}=e^x\lim_{n\to\infty}n(e^{1/n}-1)=e^x\cdot\ln e=e^x$$ Note: the second to last equality holds because of a limit definition of the natural logarithm: $\ln x=\lim_{n\to\infty}n(x^{1/n}-1)$.

As we see, calculating the derivative of the exponential function is not easy with the usual limit definition of the derivative. It requires calculating a limit that is not obvious without knowing a special limit definition of $\ln x$.

One can wonder then, are there easier ways of proving that $\frac{d}{dx}e^x=e^x$? Indeed, there are easier ways to prove this. But all of the proofs I have ever seen either assume a taylor series or limit definition of the exponential function, or somehow use the derivative of $\ln x$ which itself has similar calculation problems. Finally, the proofs lack deep motivated intuition, and are raw algebraic manipulations for the most part. They prove things well, but they don't explain things well.

Question: is there a way to find the derivative of the exponential function intuitively?


I didn't ask this question without giving it a little thought.

Path A

I figured that one solution to this problem might be by intuituvely explaining how in the world $\ln x$ is equal to $\lim_{n\to\infty}n(x^{1/n}-1)$.

Euler observed, quite unrigorously, that if $\epsilon$ is an arbitrarily small number, then:


Similarly, if we let $n$ be an arbitrarily large number, we can observe that:


Plugging this observation into the first one, we have:

$$\ln(x^{1/n})=x^{1/n}-1$$ $$\frac1n\ln x=x^{1/n}-1$$ $$\ln x=n(x^{1/n}-1)$$


$$\ln x=\lim_{n\to\infty}n(x^{1/n}-1)$$

This would almost work as a solution, except for the fact that here we make observations that work for logarithms of all bases. The observation $\log_b(1+\epsilon)=\epsilon$ is valid for all bases $b$. The second observation we made doesn't even relate specifically to logarithms. Thus, the "intuition" in this case assumes that the limit can be equal to a logarithm of any base. This is obviously false; computations evidently show that this limit holds only for $b=e$. And it is not evident at all why it has to be $e$ and nothing else.

This solution will be complete if it can be shown why base $e$ and none other work.

Path B

Another solution to this problem would be noting that the exponential function grows proportionally to its size. The problem with this intuition is that it is not at all evident why would this function follow such a behavior.

The mystery is, how does one start with simple algebraic properties of exponents, which are trivially defined by multiplication, and arrive the conclusion that this function follows its unique growth behavior. It might help to note that exponentiation turns an arithmetic sequence into a geometric sequence.

Id est, if:

$$\alpha_n=a+\sum^n_1 d$$ $$\gamma_n=b\prod^n_1 r$$


$$e^{\alpha_n}=e^{a+\sum^n_1 d}=e^a\prod^n_1 e^d=b\prod^n_1 r=\gamma_n$$

If there is a way to start with basic algebraic facts about exponents and end up (intuitively) with the fact that exponential growth is proportional to its size, we could then justify the fact that $e^x$ is the solution of $y'=y$, $y(0)=1$. From there, we could automatically say that the derivative of the natural exponential is itself.

Caveat: While solving this ODE, there is still a problem because we need to compute the integral of $\frac1x$. It turns out that we can intuitively solve this task. We can begin by splitting the area under the curve into n rectangles of equal area $A$, situated between corresponding x coordinates: $\{x_0, x_1, ..., x_n\}$. We will then note that:

$$A=y_0(x_1-x_0)=y_1(x_2-x_1)$$ $$\frac{x_1-x_0}{x_0}=\frac{x_2-x_1}{x_1}$$ $$\frac{x_1}{x_0}-1=\frac{x_2}{x_1}-1$$ $$\frac{x_1}{x_0}=\frac{x_2}{x_1}$$

This will generalize to $\frac{x_n}{x_{n-1}}=\frac{x_{n+1}}{x_n}$. What this means is that, if rectangles are the same area, if we increase the x coordinates geometrically (because the ratio between next and current x coordinate is constant), we increase the area arithmetically. This is precisely what logarithms do, they turn geometric sequences into arithmetic sequences (opposite of the exponentials). Thus, the integral of $\frac1x$ will be some kind of logarithm.

The missing bit here, again, is...why is it base e, and not some another base?

Other paths

Those two paths are most likely not the only approaches.


At this point, I overstressed the word "intuition", and I just wanted to explain myself. I just really love to explore things that are proven symbolically, in a natural way. I might be considered weird for trying to do that so deeply for such a simple derivative, but oh well.

Thank you in advance for any good insights into this problem.

  • 5
    $\begingroup$ You could take the view that $e$ is the solution of $$\lim_{h\to 0}\frac{a^h-1}h=1$$ for $a$. $\endgroup$ – Angina Seng Oct 15 '19 at 5:04
  • 10
    $\begingroup$ What exactly is your definition of $\ln$, $\exp$, or $a^x$? If you define $\ln x = \int_1^x \frac{dt}{t}$ and $\exp$ as the inverse function, then $(\ln x)' = \frac 1x$ and $(\exp x)' = \exp x$ follows immediately from basic rules. $\endgroup$ – Martin R Oct 15 '19 at 5:15
  • 4
    $\begingroup$ @KKZiomek You've found the explanation yourself: $\frac d{dx}e^x = e^x\cdot \lim_{h\to 0}\frac{e^h-1}h$. This says, quite literally, that the exponential function grows proportional to its size. The matter of why the proportionality constant is exactly $1$ when the base for the exponential is $e$ is, as Lord Shark points out, often seen as the definition of $e$. In other words, $e$ is chosen to be the number such that the proportionality constant is $1$. $\endgroup$ – Arthur Oct 15 '19 at 13:53
  • 5
    $\begingroup$ KKZ, your comment still doesn't contain a definition of $e$. Just saying $\ln e=1$ is no help, because you don't define $\ln$ properly: $\log_b 1$ is $0$ for all bases $b$, not just for $b=e$. And I think that this missing definition is precisely the source of the problems you are having. $\endgroup$ – TonyK Oct 15 '19 at 14:38
  • 1
    $\begingroup$ Do you know e.g. that $\frac d{dx}\ln f(x)=\frac{f'(x)}{f(x)}$? $\endgroup$ – Hagen von Eitzen Oct 15 '19 at 20:57

This might not exactly be the type of answer you are looking for, but I think it's accurate so I'm writing it here anyway. Often in math, we can encounter objects with multiple (provably equivalent) definitions, some more natural than others, and some can be better in giving an intuitive insight into how the object in question works. Once we've chosen a definition, the others can then be proven as theorems.

If I guess correctly, your definition of the exponential function $\exp$ is given by first defining $$\log(x)=\lim_{n\to\infty}n(x^{1/n}-1),$$ and then defining $\exp$ to be the inverse function of the logarithm. But this is not a very enlightening definition. A much more common definition would literally be that $\exp$ is the unique function $f:\mathbb R\to\mathbb R$ satisfying $f'(x)=f(x)$ for all $x$, and $f(0)=1$. Proving such a function exists and is unique takes some work, then showing that this is indeed equivalent to your definition takes some more. But once this is done, we can then accept, as a fact, that $\exp$ is the function that is meant to satisfy $\exp'=\exp$. We can define this in multiple ways, but this is the core property that is central to the exponential function---some might say, the defining property of the exponential function.

So your question is:

Are there easier ways of proving that $\dfrac{d}{dx}e^x=e^x$?

I would say yes: by assuming it is the case, by taking it to be true by definition. Showing that other definitions are equivalent to this definition is nontrivial and needs to be done of course, but as you observed might not necessarily be very intuitive or give much insight. To understand what the exponential function is deeply, this definition is the way to go.

To show that the definition $\frac{d}{dx}e^x=e^x$ is equivalent to the "arithmetic definition" $e^{a+b}=e^ae^b$ is an interesting problem. To go from the former to the latter, see the link provided by KM101 in the comments. On the other hand, let's try to start from the latter and try to go to the former. Now, differentiation deals with the "local growth rate" of a function with response to small changes in $x$, and we have an additive definition, so we consider $$\exp(x+\Delta x)=\exp(x)\exp(\Delta x)$$ for a small $\Delta x>0$. Now in the definition of the derivative, we consider $$\exp(x)\lim_{\Delta x\to 0}\left(\frac{\exp(\Delta x)-1}{\Delta x}\right).$$ To show that this is indeed $\exp(x)$, all we need to do is to show that the latter limit is $1$. Note that the limit is actually just $\exp'(0)$. Consider what we've done so far: we've reduced the proof of $\exp'=\exp$, which is information regarding the "growth rate" of $\exp$ globally, into the proof of just $\exp'(0)=1$, at a single point!

So how can we prove this, knowing just $\exp(a+b)=\exp(a)\exp(b)$? Well... unfortunately we can't. See, if we define $f'(0)=k$ for any arbitrary real number $k$, where $f$ satisfies $f(a+b)=f(a)f(b)$, we will get a perfectly well-defined function $f$ (you can try to show this). But when we make the choice that $f'(0)=1$ (or in other words $k=1$), then we end up with a function which is its own derivative. Indeed, this is the property that motivates the choice $f'(0)=1$.

In hindsight, if $f'(0)=k$ and $f(a+b)=f(a)f(b)$ for all $a,b$, we can describe $f$ in general: it is simply $f(x)=e^{kx}$.

  • $\begingroup$ That's the thing I'm debating with in my head. I use the diff eq definition of the exponential all the time when I try to justify things intuitively in my head. But then, thinking deeper I realized there's a big abyss between the standard arithmetic definition of exponentials and the fact that y'=y. The only thing shared between those two approaches is that y(0)=1, which is trivial. What is not trivial though, is the question, why are the two definitions equivalent? Why does a function that grows proportionally to itself have the algebraic property of $e^{a+b}=e^ae^b$ and/or vice versa? $\endgroup$ – KKZiomek Oct 15 '19 at 5:33
  • $\begingroup$ The main answer here gives a great explanation of why the property $\exp(a)\exp(b) = \exp(a+b)$ (in addition to the other arithmetic properties) follows from the definition $\frac{\mathrm d}{\mathrm dx} \exp(x) = \exp(x)$. $\endgroup$ – KM101 Oct 15 '19 at 5:45
  • $\begingroup$ @KKZiomek I've added a bit into my answer, please check if that answers your question. $\endgroup$ – YiFan Oct 15 '19 at 5:48

One fairly elementary way to approach this -- elementary enough that it's the way it was approached at the secondary school (= high school) I went to -- is to begin by defining the logarithm rather than the exponential. You say $\log x=\int_1^x1/t\ dt$ from which it immediately follows that the derivative of $\log x$ is $1/x$, you prove that $\log(ab)=\log a+\log b$ by a simple geometrical argument and $\log 1=0$ by inspection, and now you've got a nice log function and know its derivative. Then you define exp to be the inverse of log; you know it behaves like an exponential because you know log behaves like a logarithm; and now the derivative of exp is a trivial calculation using the already-known derivative of log.

(Personally I think I prefer to take YiFan's approach and begin by defining exp in terms of the ODE it satisfies. But the above works pretty well.)

  • 1
    $\begingroup$ Actually, your and YiFan's approach are equivalent. It is almost trivial from the integral definition to show that $\log x$ is 1-1, and thus it follows by the FTC that the inverse function $\exp$ satisfies the ODE. The advantage of the integral definition is that existence of $\log$ is obvious to students as soon as they've studied the basic properties of integration. The existence of a solution for the ODE takes much more development. $\endgroup$ – Paul Sinclair Oct 15 '19 at 16:30
  • 3
    $\begingroup$ Obviously any two approaches are equivalent in some sense, given that there's only one exp and only one log :-). $\endgroup$ – Gareth McCaughan Oct 15 '19 at 16:49
  • $\begingroup$ True - perhaps I should have said "almost the same". $\endgroup$ – Paul Sinclair Oct 15 '19 at 16:51
  • $\begingroup$ I really like this answer too. If I could, I would make both of those best answers $\endgroup$ – KKZiomek Oct 16 '19 at 3:41

If you can accept the derivative of $\ln x$ then this might provide "some" intuition.

$$ y=e^x \Rightarrow \\ \ln y = x \Rightarrow \\ \frac{y '}{y} = 1 \Rightarrow \\ y' = y = e^x $$


Another intuitive approch is to consider $g(x)=2^x$, $h(x)=3^x$ and since by convexity of the exponential function

  • $\frac12<g'(0)<1$
  • $\frac23<h'(0)<2$

we have that there exists $2<e<3$ such that for $f(x)=e^x$

$$ f'(0)=\lim_{x\to 0}\frac{e^x-1}{x}=1$$

and therefore $\forall x_0\in\mathbb R$

$$ f'(x_0)=\lim_{x\to x_0}\frac{e^x-e^{x_0}}{x-x_0}=e^{x_0}\cdot \lim_{(x-x_0)\to 0}\frac{e^{(x-x_0)}-1}{x-x_0}=e^{x_0}$$


The exponential function of base e is defined as the (monovariable real) function equal to the value of its first derivative for any value of x.

You are trying to prove the definition.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.