Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is there a way to prove that the derivative of $e^x$ is $e^x$ without using chain rule? If so, what is it? Thanks.

share|cite|improve this question
do you know about Taylor series? – crf Sep 20 '12 at 1:49
What is your definition of $e^x$? – lhf Sep 20 '12 at 2:00
In my Analysis class, we defined $e^x$ as the solution of $f'(x) = f(x)$ with $f(0) = 1$. So um, that works. – Ben Millwood Sep 20 '12 at 2:16
How do you do this with the chain rule? – Chris Eagle Sep 20 '12 at 2:20
@ChrisEagle let $y=e^x$ then $\ln(y)=x$ hence $\frac{1}{y}y'=1$ thus $y'=y$ aka $\frac{d}{dx}e^x=e^x$ – James S. Cook Sep 20 '12 at 2:48

Define $e$ implicitly by $\lim_{h \rightarrow 0} \frac{e^h-1}{h}=1$. Calculate, $$ \frac{d}{dx} e^x = \lim_{h \rightarrow 0} \frac{e^{x+h}-e^x}{h} = e^{x}\lim_{h \rightarrow 0} \frac{e^h-1}{h} = e^x.$$ This definition assumes that properties of exponential functions are somehow known.

In contrast, the definition that defines the $\ln(x) = \int_{1}^{x} \frac{dt}{t}$ allows you to derive properties of the natural log. Then the exponential function is introduced as the inverse function and its properties can be induced from those already proven for the natural log.

Logically the definition of the natural log as primary has advantages. But, pedagogically if you wish to discuss the exponential function before integral calculus then some sort of chicanery is required.

share|cite|improve this answer
But isn't the hardest part left? Namely demonstrating that $\lim_{h\to 0}\frac{e^h-1}{h}=1$... – Bart Patzer Sep 20 '15 at 11:12
@BartPatzer well, if you study $y=2^x$ and $y=3^x$ it is clear these have tangent lines at $x=0$ and it is clear that the slopes are $0.69..$ and $1.097..$ hence as there is a continuous selection of bases for $y=a^x$ between $a=2$ and $a=3$ it is reasonable to suppose such a base $a$ exists as to produce slope $1$ at $x=0$. The implicit definition of $e$ I state here is merely the formulation of the slope of the tangent for such a base. .. – James S. Cook Sep 20 '15 at 18:57
The demonstration that every value between $\ln(2)=0.69..$ and $\ln(3) =1.097...$ is attained as the slope of one and only one base $a$ is a delicate question beyond the scope of calculus 1. But, this is no worse than the supposition that the formula $f(x)=a^x$ is well-posed for $x$ a real value. So, it's a reasonable definition. Of course, the power series definition for $e$ is more logically economical. – James S. Cook Sep 20 '15 at 18:57

To follow up on "what is your definition of $e^x$", if your definition is as the solution to the differential equation $y'=y$ such that $y(0)=1$, then you have nothing to prove!

share|cite|improve this answer

Hmmm.... Well, how precisely have you defined $e^x$? Depending on the answer, the approach will vary. If you've defined $$e^x=\sum_{k=0}^\infty\frac{x^k}{k!},$$ then it will follow fairly readily that $e^x$ is its own derivative, using Taylor series properties.

If on the other hand you've defined $$e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n,$$ then you may have a slightly harder way to go. I think using the difference of $n$th powers formula may help.

share|cite|improve this answer
I think you mean to use an exponent of $n$, not $1/n$, in the second definition. – Owen Biesel Sep 20 '12 at 3:31
Thanks, Owen! Fixed. – Cameron Buie Sep 20 '12 at 3:55
You don't need to interchange limits, see my answer. – celtschk Sep 20 '12 at 10:54
I've seen it, celtschk, and I'm not altogether convinced, yet. Let me know when you've addressed the issues I pointed out. – Cameron Buie Sep 20 '12 at 16:26
Note that my corrected version still needs no interchange of limits. – celtschk Sep 20 '12 at 18:50

When using the definition $$\mathrm e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n$$ you can proceed as follows: $$\begin{aligned} \frac{\mathrm d}{\mathrm dx}\mathrm e^x &= \lim_{h\to 0}\frac{\mathrm e^{x+h}-\mathrm e^x}{h}\\ &= \lim_{h\to 0}\frac{\lim\limits_{n\to\infty}\left(1+\frac{x+h}n\right)^n - \lim\limits_{n\to\infty}\left(1+\frac xn\right)^n}{h}\\ &= \lim_{h\to 0}\lim_{n\to\infty}\frac{\left(1+\frac{x+h}n\right)^n - \left(1+\frac xn\right)^n}{h} \end{aligned}$$ Now $$\left(1+\frac{x+h}{n}\right)^n = \sum_{k=0}^n{n\choose k}\left(\frac{h}{n}\right)^k\left(1+\frac{x}{n}\right)^{n-k}$$ and therefore $$\begin{aligned} \frac{\mathrm d}{\mathrm dx}\mathrm e^x &= \lim_{h\to 0}\lim_{n\to\infty}\sum_{k=1}^n{n\choose k}\frac{h^{k-1}}{n^k}\left(1+\frac{x}{n}\right)^{n-k}\\ &= \lim_{h\to 0}\left(\lim_{n\to\infty}{n\choose 1}\frac{1}{n}\left(1+\frac{x}{n}\right)^{n-1}+h\lim_{n\to\infty}\sum_{k=2}^n{n\choose k}\frac{h^{k-2}}{n^k}\left(1+\frac{x}{n}\right)^{n-k}\right)\\ &= \lim_{n\to\infty}{n\choose 1}\frac{1}{n}\left(1+\frac{x}{n}\right)^{n-1}+\lim_{h\to 0}h\lim_{n\to\infty}\sum_{k=2}^n{n\choose k}\frac{h^{k-2}}{n^k}\left(1+\frac{x}{n}\right)^{n-k}\\ &= \lim_{n\to\infty}\left(1+\frac{x}{n}\right)^{n-1}\\ &= \lim_{n\to\infty}\frac{\left(1+\frac{x}{n}\right)^{n}}{1+\frac{x}{n}}\\ \end{aligned}$$ Since the limit for numerator and denominator exists independently, we can calculate them separately. The numerator is just the definition of $\mathrm e^x$, and the limit of the denominator is $1$, so we arrive at $$\frac{\mathrm d}{\mathrm dx}\mathrm e^x = \mathrm e^x$$

share|cite|improve this answer
Not following the first line after "and therefore" (in the middle). According to the line before it, shouldn't we have the fraction $\cfrac{h^{k-1}}{n^k}$, rather than $\cfrac{h^{k-1}}k$? Also, observe that you passed the $\lim\limits_{n\to\infty}$ right through the $\sum_{k=1}^n$ on the line after that. That's not good.... – Cameron Buie Sep 20 '12 at 16:24
@CameronBuie: Oops, you're right. Also, I now notice that I've forgotten the binomial coefficients in the sum. I'll try to fix it. – celtschk Sep 20 '12 at 16:38
@CameronBuie: Now better? – celtschk Sep 20 '12 at 16:49
(+1): Indeed it is. – Cameron Buie Sep 20 '12 at 18:45
From line 3 to line 4 in the last part you not really valid; you'd need to show that the inner limit exists, or at least that the expression in it stays bounded... – Max May 3 '13 at 3:21

If the definition of $e^x$ is "the differentiable solution to $f(x+y) = f(x)f(y)$ with $f'(0) =1$, this way works:

Putting $y = 0$, $f(x) = f(x)f(0)$ for all $x$, so $f(0) = 1$.

$(f(x+h)-f(x))/h = (f(x)f(h)-f(x))/h = f(x)(f(h)-1)/h = f(x)(f(h)-f(0))/h $. Taking the limit as $h \to 0$, $f'(x) = f'(0)f(x)$.

We now can use the differential equation approach.

Note: If this seems familiar, I have used this answer previously in a similar context.

share|cite|improve this answer
how do you work backwards from that to show that $f^{\prime}(0)=1$? I'm currently trying to figure that out here:… – Jessy Cat Feb 23 at 16:36
Because $f'(0) = \lim_{h \to 0} (f(h)-f(0))/h$. – marty cohen Feb 23 at 18:51
but if I am trying to prove that $f^{\prime}(0) = 1$, how do I know that $\lim_{h \to 0} (f(h)-f(0))/h = 1$?? I can't use the thing I'm trying to prove in order to prove it. – Jessy Cat Feb 23 at 19:12

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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