I'm working through the proof of $\frac{d}{dx}e^x = e^x$, and trying to understand it, but my mind has gotten stuck at the last step.

Starting with the definition of a derivative, we can formulate it like so:

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

After some algebra, we arrive at:

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

As $h\to0$, the expression approaches $\frac{0}{0}$, which makes it indeterminate. And, this is where my understanding ends. I've tried looking at wikipedia and other descriptions of the proof, but couldn't understand those explanations. It has usually been something along the lines of, "plot $\frac{e^x - 1}{x}$ and see the function's behavior at $0$," which ends up approaching $1$, which can substitute the limit to give the result of the derivative:

$$\frac{d}{dx} e^x = e^x \cdot 1 = e^x$$

I vaguely understand the concept of indeterminate forms, and why it is difficult to know what is happening with the function. But is there a better explanation of how the result of $1$ is obtained?

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    $\begingroup$ The first thing you have to establish is, what does $e$ mean? How to establish the limit depends on how you define $e$. $\endgroup$ Sep 4, 2012 at 2:58
  • $\begingroup$ @GerryMyerson: What do you mean? $e$ is the base of the natural logarithm. Am I missing something? $\endgroup$
    – voithos
    Sep 4, 2012 at 3:02
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    $\begingroup$ At the risk of sounding circular, what's a natural logarithm? $\endgroup$
    – user22805
    Sep 4, 2012 at 3:05
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    $\begingroup$ @voithos : There are many definitions for $e$, and yours is probably the most meaningless, because then one has to define "what is the natural logarithm", and that is just as hard as defining $e$. For instance, $$ \lim_{n \to \infty} \left( 1 + \frac xn \right)^n, \sum_{k=0}^{\infty} \frac{x^k}{k!} $$ are both valid definitions to $e^x$, and other definitions may be found by defining $e^x$ as the inverse function of the logarithmic function. Each definition has its pros and cons. $\endgroup$ Sep 4, 2012 at 3:09
  • $\begingroup$ You need this definition $\lim_{n \rightarrow \infty}(1+\frac{h}{n})^n$ $\endgroup$ Sep 4, 2012 at 3:10

12 Answers 12


Everything in the "proof" will depend on your definition of the function $e^x$. I will choose the definition $$ e^h \overset{\text{def}}{=} \sum_{k=0}^\infty \frac{h^k}{k!}. $$ Using this, one sees that $$ \frac{e^h - 1}{h} = \frac{\sum_{k=0}^\infty \frac{h^k}{k!} - 1}{h} = \sum_{k=1}^\infty \frac{h^{k-1}}{k!} = 1 + h \sum_{k=0}^\infty \frac{h^k}{(k+2)!}. $$ If you have studied convergence tests, you know that the last series on the right converges for all $h \in \mathbb R$, hence taking the limit when $h \to 0$, the RHS goes to $1$ because the series will converge to something and the $h$ factored out will make the product go to zero.

Another way to do this would be to show that this series is an analytic function and is its own Taylor expansion around zero (this is not hard to do using convergence tests), so to differentiate it you can go term by term and readily see that its derivative is itself.

A third approach, which will sound a little stupid and meaningless but is nonetheless funny, is choosing another definition for $e^x$ : consider the differential equation $$ f'(x) = f(x), \quad f(0) = 1. $$ Using differential equation theory it is really not hard at all to show that the solutions to the equation (without the initial condition) is a one-dimensional vector space and there exists an unique element of this vector space which satisfies the initial condition $f(0) = 1$, because the solutions are of the form $Cg(x)$ for some solution $g(x)$. Let $exp(x)$ be defined as a solution to this differential equation satisfying the initial condition. Then clearly $exp'(x) = exp(x)$. Then you can easily see that $exp'(x)$ is a differentiable function, so by induction $exp(x)$ is an infinitely differentiable function with Taylor expansion $$ \exp(x) = \sum_{k=0}^\infty \frac{x^k}{k!}. $$ I mentioned it to show the importance of which definition one decides to choose ; it can change the whole structure of an argument.

Hope that helps,

  • $\begingroup$ I'm not sure if you can just define $e^x = 1 + \sum_{n=1}^{\infty} \frac{x^n}{n!}$. After all, $e$ is a real number, not some operator, so you need to define first what $e$ is, not $e^x$. IMO, the way to go would be to prove that $1 + \sum_{n=1}^{\infty} \frac{1}{n!}$ and $\lim_{n \rightarrow \infty} (1 + \frac 1n)^n$ converge to the same limit, define the limit as $e$, and then prove that $e^x = \lim_{n \rightarrow \infty} (1 + \frac 1n)^nx = \lim_{n \rightarrow \infty} (1 + \frac xn)^n = 1 + \sum_{n=1}^{\infty} \frac{x^n}{n!}$. $\endgroup$ Sep 4, 2012 at 14:20
  • $\begingroup$ @RijulSaini: What's really going on is an abuse of notation. An alternate notation that may address your objection is the following: define $\exp(x)$ in one of the two ways given above, and then define (or show, if you have already defined $e$) $e = \exp(1)$. Then prove identities about $\exp$, and use them to show that $\exp(x) = e^x$ whenever the latter was already defined (i.e., when $x$ is rational). $\endgroup$ Sep 4, 2012 at 18:06
  • $\begingroup$ @Rijul Saini : There is nothing wrong in defining something with a very suggestive notation. I could define $\mathrm{logh(x) = e^{e^x}$ and it would be morally incorrect because the subword "log" in my chosen notation suggests something that's completely not coherent with the log function, but it mathematically holds. In the case of $e^x$ what I just did is just a choice of notation (that is very pertinent). $\endgroup$ Sep 5, 2012 at 16:58
  • $\begingroup$ You seem to disagree with the fact that $e^x$ is a notation that suggests an exponential function such as $2^x$, so I would need to define $e$ before ; note that $e^x$ is probably the first exponential function one would define over $\mathbb R$ or $\mathbb C$, because the other ones (such as $2^x$) would be extended from the rationals to the reals using $e^x$. So I don't think there's something morally incorrect to define $\mathrm{exp}(x)$ first and $e$ afterwards. $\endgroup$ Sep 5, 2012 at 16:59
  • $\begingroup$ Sure, I agree there's nothing incorrect in defining $\exp(x)$ and $e=\exp(1)$. But wouldn't you need to prove $\exp(x) = e^x$ in the $e$ raised to the power $x$ sense to actually ensure finding the derivative of $\exp(x)$ and $e^x$ amounts to the same thing? $\endgroup$ Sep 5, 2012 at 17:08

Suppose you have an exponential function, like $f(x)=2^x$.

The derivative is $$ f'(x) = \lim_{h\to0} \frac{f(x+h)-f(x)}{h} = \lim_{h\to0}\frac{2^{x+h}-2^x}{h} = \lim_{h\to0}\left(2^x\cdot\frac{2^h-1}{h} \right). $$ So far, just algebra. Now watch this: $$ = 2^x\lim_{h\to0}\frac{2^h-1}{h}. $$ This can be done because $2^x$ is "constant" and "constant" means "not depending on $h$".

But this is equal to $(2^x\cdot\text{constant})$. But in this case "constant" means "not depending on $x$". "Constant" always means "not depending on something", but "something" varies with the context.

What's the "constant"? In the case above, it's not hard to show that the constant is somewhere between $1/2$ and $1$. It we'd started with $4^x$ instead, then it would be fairly easy to show that the "constant" would be more than $1$. For a base somewhere between $2$ and $4$, the "constant" is $1$. That base is $e$.

If you want to talk about how it is knonw that $2$ is too small and $4$ is too big, to be the base of the "natural" exponential function (i.e., the one for which the "constant" is $1$), I can post further on this.

Later edit: OK, how do we know that $2$ is too small and $4$ is too big, to serve in the role of the base of the "natural" exponential function? Look at the graph of $y=2^x$. It gets steeper as you go from left to right. As $x$ goes from $0$ to $1$, $y$ goes from $1$ to $2$. So the average slope between $x=1$ and $x=2$ is $$\dfrac{\text{rise}}{\text{run}} = \frac{2-1}{1-0} = 1.$$ Since it gets steeper going from left to right, the slope at the left end of this interval, i.e. at $x=0$, must be less than that. Thus we have $\dfrac{d}{dx}2^x = (2^x\cdot\text{constant})$ and the "constant" is less than $1$. (Thinking about the interval from $x=-1$ to $x=0$ in the same way shows that the "constant" is more than $1/2$.)

Now look at $y=4^x$, and use the interval from $x=-1/2$ to $x=0$, and do the same thing, and you see that when you write $\dfrac{d}{dx}4^x = (4^x\cdot\text{constant})$, then that "constant" is more than $1$.

This should suggest that $4$ is too big, and $2$ is too small.

You can show that $3$ is too big by using the interval from $x=-1/6$ to $x=0$ and doing the same thing. But the arithmetic is messy.

  • $\begingroup$ This is a very easy-to-understand way of explaining it - thanks! $\endgroup$
    – voithos
    Sep 4, 2012 at 3:19
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    $\begingroup$ So basically you are defining $e$ to be the value of $a$ that makes $a^x$ be the derivative of $a^x$. It's not clear to me that this answers the question. $\endgroup$ Sep 4, 2012 at 3:20
  • $\begingroup$ @Michael : Agreeing with Gerry here ; the pertinent part of your answer lies in "I can post further on this", because the fact that $2$ is too small and $4$ is too big is essentially the part that says that the limit of $\frac{a^h - 1}h$ exists. This looks to me like the question, not the answer. $\endgroup$ Sep 4, 2012 at 3:23
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    $\begingroup$ I have seen this approach in at least one calculus book, probably more than one. The idea is to take the high school exponentiation rules for granted, and to define $e$ as described in the answer by Michael Hardy. There are, to put it mildly, some unsatisfactory elements to taking this approach, but it may very well be precisely the approach the OP's course is taking. $\endgroup$ Sep 4, 2012 at 3:28
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    $\begingroup$ If integral calculus is required to define the logarithm then it makes it difficult to introduce the the exponential function early in the first calculus course. Andre is correct in that the reason for defining $e$ implicitly by $\lim_{h \rightarrow 0} \frac{e^h-1}{h}=1$ is to build on assumed laws of exponents from highschool. Of course, you could take a more geometric view and phrase it in terms of tangents, or use the goal of the OP $\frac{d}{dx}e^x=e^x$ as the definition. I totally understand the elegance of the integral definition for $\ln(x)$, but it relegates exponential $\endgroup$ Sep 4, 2012 at 4:46

Apostol's approach in his Calculus text is to define the natural logarithm by $$\log x=\int_1^xt^{-1}\,dt$$ from which you immediately get that the derivative of $\log x$ is $1/x$. Then Apostol defines the exponential function as the functional inverse of the logarithm. So if $y=e^x$, then $x=\log y$; now differentiate with respect to $x$, using the chain rule, to get $$1={1\over y}{dy\over dx}$$ So, ${dy\over dx}=y=e^x$, as desired.

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    $\begingroup$ Also, that definition of log allows an easy proof that log(x) + log(y) = log(xy). $\endgroup$ Sep 4, 2012 at 5:49
  • $\begingroup$ About the last line..the solution to the diff.eq is $Ce^{x}$ $\endgroup$
    – Belgi
    Sep 4, 2012 at 8:25
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    $\begingroup$ @Bel, no differential equation is being solved here - I'm just taking the displayed equation, multiplying both sides by $y$, and then noting that earlier $y$ was defined to be $e^x$. $\endgroup$ Sep 4, 2012 at 9:28
  • $\begingroup$ This is how most calculus books (I have seen) do it. $\endgroup$
    – GeoffDS
    Sep 4, 2012 at 14:53
  • $\begingroup$ Isn't the integral definition that you give the definition for the natural logarithm $\ln x$ and not $\log x$? If I recall correctly $\ln(e^y) = y$ while $\log(10^y) = y$. $\endgroup$ Sep 4, 2012 at 17:45

If you take the series definition of $e^x$ then you have

$$\begin{eqnarray*} e^x &=& 1 + x + \frac{x^2}{2!} + \ldots \\ \implies \frac{e^x - 1}{x} &=&1 + \frac{x}{2!} + \frac{x^2 }{3!} + \ldots \\ \implies \lim_{x \rightarrow 0} \frac{e^x - 1}{x} &=& \lim_{x\rightarrow 0} 1 + \frac{x}{2!} + \frac{x^2 }{3!} + \ldots \\ &=& 1.\end{eqnarray*}$$

  • $\begingroup$ Do you mean $1$ on the last line? $\endgroup$
    – Sasha
    Sep 4, 2012 at 3:05
  • $\begingroup$ Sorry I made a typo above. $\endgroup$
    – user38268
    Sep 4, 2012 at 3:06
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    $\begingroup$ How do you obtain Taylor series for $e^x$? Don't you need to have its derivative first? $\endgroup$
    – Arashium
    Feb 6, 2015 at 18:55

$$e^x := \lim_{n \to \infty} \left(1+\frac{x}{n} \right)^n \implies \\ \frac{d}{dx} e^x = \lim_{n \to \infty} n \cdot \frac{1}{n} \cdot\left(1+\frac{x}{n} \right)^{n-1}=\lim_{n \to \infty} \left(1+\frac{x}{n} \right)^{n-1}=e^x$$


Note that even though the expression $\lim_{h\to 0}{e^h-1 \over h}$ is indeterminate, you can rewrite it as $$\lim_{h\to 0}{e^h-1\over h} = \lim_{h\to 0}{f(h)-f(0)\over h} = f'(0)$$ where $f(x)=e^x$, and where we have used the definition of $f'(0)$. The same can be done with any other base; if $g(x)=a^x$, then the same calculation as you have given shows that $$g'(x)=a^x\lim_{h\to 0}{a^h-1\over h}=a^x\lim_{h\to 0}{g(h)-g(0)\over h}=a^x\cdot g'(0)$$ So it is the slope at the origin which appears here. The formula for $g'(x)$ is simplest when the slope at the origin is equal to $1$, and this is one of the many possible ways to define $e$.

  • $\begingroup$ But this assumes without proof that the exponential(s) are differentiable... $\endgroup$
    – N. S.
    Jul 17, 2013 at 16:06

just use the series definition of $e^x$, put in $h$ and subtract 1 and divide by $h$ to get the desired result.

using the series definition of $e^x$ the only question that needs to be answered is that the derivative of each term summed up is same as summing up and then differentiating.


Well, $\ln e^x = x$

So $\frac{d}{dx} \ln e^x = \frac{d}{dx} x \implies \frac{1}{e^x} \frac{d}{dx} e^x = 1$ by chain rule.

Then multiply both sides by $e^x$ and you get $\frac{d}{dx} e^x = e^x$

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    $\begingroup$ You assume knowing the derivative of the logarithmic function, which is another problem. $\endgroup$ Sep 4, 2012 at 3:11
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    $\begingroup$ That's fine if you know the derivative of the logarithm function, but perhaps that's circular reasoning. $\endgroup$ Sep 4, 2012 at 3:12
  • $\begingroup$ Yeah . . . . you're assuming lots of stuff is already established, but how do you establish it without already knowing in advance that $(d/dx)e^x=e^x$? $\endgroup$ Sep 4, 2012 at 3:12

Let's define $b^x$ as

$$ b^x = a_0 + a_1x +a_2 \frac{x^2}{2!} + \ldots =\sum_{k=0}^{\infty} a_k\frac{x^k}{k!}$$

for $x=0$, $ b^0 =1 $ Thus $a_0=1$

$$ b^{x} = 1 + a_1x +a_2 \frac{x^2}{2!} + \ldots = 1 + \sum_{k=1}^{\infty} a_k\frac{x^k}{k!} \tag 1$$

$$\frac{d}{dx} b^x = \lim_{h \to 0} \frac{b^{x+h}-b^x}{h}=b^x \lim_{h \to 0} \frac{b^{h}-1}{h}=b^x \lim_{h \to 0} \frac{(1 + a_1h +a_2 \frac{h^2}{2!} + \ldots)-1}{h}$$ $$\frac{d}{dx} b^x=b^x \lim_{h \to 0} (a_1 +a_2 \frac{h}{2!} + \ldots)=a_1b^x $$

Let's select $a_1=1$ then we define $b=e$ and then we need to find all $a_k$ values and $e$.

After selecting $a_1=1$ and $b=e$, we have :

$$\frac{d}{dx} e^x=e^x \tag 2$$

According to this defination, $a_k=1$ for $k \geq 2$


for $a_1=1$ and $b=e$ , And using relation (1)

$$ e^x = 1 + x +a_2 \frac{x^2}{2!} + \ldots =1 + x+ \sum_{k=2}^{\infty} a_k\frac{x^k}{k!}$$

$$ \frac{d}{dx} e^x = 1 + a_2 x +a_3 \frac{x^2}{2!} + \ldots =$$

According to the result (2), $$ \frac{d}{dx} e^x =e^x= 1 + x +a_2 \frac{x^2}{2!} + \ldots= 1 + a_2 x +a_3 \frac{x^2}{2!} + \ldots $$

Now We need to equal all cooeffients of $x^k$,

we find $a_{2}=1$ and $a_{k+1}=a_{k}$ for $k \geq 2$

If we solve that relation, we get: $a_{k}=1 $ for $k \geq 2$

Thus $ e^x$ can be written as power series as shown below

$$ e^x= 1 + x + \frac{x^2}{2!} + \ldots= \sum_{k=0}^{\infty} \frac{x^k}{k!} $$

and to find $e$: put $x=1$, $e^1=1 + 1 + \frac{1^2}{2!} + \ldots= \sum_{k=0}^{\infty} \frac{1}{k!}$

Note: If we select $a_1=m$ and follow the same way as shown in the proof, we will get $$ b^{x}= 1 + mx + \frac{(mx)^2}{2!} + \ldots= \sum_{k=0}^{\infty} \frac{(mx)^k}{k!} =e^{mx}=(e^m)^x$$ $$ b^{x}=(e^m)^x$$ Thus $ b=e^m$


$$\frac{d}{dx} b^x=a_1 b^x =m b^x=\ln(b) b^x $$

$$\frac{d}{dx} (b^x)=\ln(b) b^x $$

  • $\begingroup$ I don't get it. How can you conclude $\frac{d}{dx} b^x=b^x \lim_{h \to 0} (a_1 +a_2 \frac{h}{2!} + \ldots)=a_1b^x$? (You could have $a_n = (n+1)!$ for $n \ge 2$) $\endgroup$ Sep 4, 2012 at 14:13
  • $\begingroup$ @RijulSaini : I got lim value for $h-->0$ and result is $a_1b^x$ Please put (h=0) in limit $a_1+a_2.0/2!+a_30/3!+....=a_1$ $\endgroup$
    – Mathlover
    Sep 4, 2012 at 14:48
  • $\begingroup$ Well, like I said before, you can't conclude that if $a_2/2!+a_3/3! + \cdots$ is not bounded. Take for example, $a_n = (n+1)!$ $\endgroup$ Sep 4, 2012 at 14:52

Using the series definition just differentiate the series. You can regard all terms of the sum in the series as functions and thus use the sum rule ("1" consists of a constant function). In other words, where $D$ indicates the derivative, $D[f(x)+g(x)]=D(f(x))+D(g(x))$. So $$D(1+x+\frac{x^2}{2!}+\ldots)=D(1)+D(x)+D\left(\frac{x^2}{2!}\right)+\ldots=0+1+x+\frac{x^2}{2!}+\ldots$$

  • $\begingroup$ -1: That derivatives commute with infinite sums needs to be verified. In fact, it does not hold unrestrictedly: see e.g. p. 251 of math.uga.edu/~pete/2400full.pdf for an example involving sequences of functions, and note that every sequence of functions is also a sum of the successive differences of the original sequence of functions. $\endgroup$ Sep 8, 2012 at 5:05
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    $\begingroup$ The interchange of the sum and the derivative is valid in this case by considerations involving uniform convergence: see e.g. Theorem 285 of the notes linked to above. But it takes some real work -- beyond the non-honors freshman calculus level -- to establish this. $\endgroup$ Sep 8, 2012 at 5:08

We know that
$\frac{d}{dx} \ln(e^{x}) = 1$ (1)
since $e^{x}$ and $\ln(x)$ are inverse functions. We also know
$\frac{d}{dx} e^{x} = \frac{1}{e^{x}}$ by (1). That also means $\frac{d}{dx} e^{x} {\frac{e^{x}}{e^{x}}} = e^{x}$ by the chain rule. Source: Khan Academy

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    $\begingroup$ I think you meant $$\dfrac{d}{dx}\ln(e^x) = \dfrac{1}{e^x}\cdot \dfrac{d}{dx}e^x,$$ because clearly $$\dfrac{d}{dx}e^x \neq \dfrac{1}{e^x}.$$ $\endgroup$
    – Bman72
    Apr 26, 2019 at 12:48

l'Hôpital's rule applies to indeterminate form of the limit you have in $$ \frac{d}{dx} e^x = e^x \lim_{h \to 0} \frac{e^h-1}{h} $$ So you get: $$ \lim_{h \to 0} \frac{e^h-1}{h} = \lim_{h \to 0} \frac{\frac{d}{dh}(e^h-1)}{\frac{dh}{dh}}=\lim_{h \to 0}\frac{e^h}{1}=1 $$

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    $\begingroup$ Welcome to StackExchange. Please read the question carefully. Clearly, application of l'Hôpital's rule is circular, as you're using the derivative of $e^x$ in the proof of the derivative of $e^x$. $\endgroup$ Sep 4, 2012 at 14:09
  • $\begingroup$ Absolutely agree with Rijul. Try to do the above assuming no knowledge of the derivative of $e^x$. You can't do it. $\endgroup$
    – GeoffDS
    Sep 4, 2012 at 14:10
  • $\begingroup$ Sorry, didn't really thought through. $\endgroup$ Sep 4, 2012 at 14:33

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