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I was wondering on the following and I probably know the answer already: NO.

Is there another number with similar properties as $e$? So that the derivative of $\exp(x)$ is the same as the function itself.

I can guess that it's probably not, because otherwise $e$ wouldn't be that special, but is there any proof of it?

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    $\begingroup$ One of the many definitions of $\exp(x)$ is that it is the unique function such that $f'(x) = f(x)$ and $f(0) = 1$ $\endgroup$
    – user17762
    Aug 17 '11 at 16:45
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    $\begingroup$ To see this, suppose f(x) also has this property. Take the derivative of $e^{-x}f(x)$ to show this function must be constant. $\endgroup$
    – Joe
    Aug 17 '11 at 16:48
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    $\begingroup$ @Timo You keep repeating the word "number" where you mean "function". The derivative of any number (even $e$) is zero. The derivative can be though of as the rate of change with respect to $x$ (or whatever you want to name your variable). If there is no $x$, there is no change. $\endgroup$ Aug 17 '11 at 18:11
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    $\begingroup$ By counterexample, doesn't the function $0(x)$ always fulfill $f(x) = f'(x)$? $\endgroup$ Aug 17 '11 at 22:08
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    $\begingroup$ @Peter - Yes, $0(x)$ always fulfills $f(x)=f'(x)$. However, that is not a counterexample, because $0(x)$ is a function of the form $Ce^x$, as described in the accepted answer. Perhaps the question and title should be edited to propose a proof that $Ce^x$ us the only function for which $f(x) = f'(x)$? $\endgroup$ Aug 18 '11 at 17:52

10 Answers 10

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Of course $C e^x$ has the same property for any $C$ (including $C = 0$). But these are the only ones.

Proposition: Let $f : \mathbb{R} \to \mathbb{R}$ be a differentiable function such that $f(0) = 1$ and $f'(x) = f(x)$. Then it must be the case that $f = e^x$.

Proof. Let $g(x) = f(x) e^{-x}$. Then

$$g'(x) = -f(x) e^{-x} + f'(x) e^{-x} = (f'(x) - f(x)) e^{-x} = 0$$

by assumption, so $g$ is constant. But $g(0) = 1$, so $g(x) = 1$ identically.

N.B. Note that it is also true that $e^{x+c}$ has the same property for any $c$. Thus there exists a function $g(c)$ such that $e^{x+c} = g(c) e^x = e^c g(x)$, and setting $c = 0$, then $x = 0$, we conclude that $g(c) = e^c$, hence $e^{x+c} = e^x e^c$.

This observation generalizes to any differential equation with translation symmetry. Apply it to the differential equation $f''(x) + f(x) = 0$ and you get the angle addition formulas for sine and cosine.

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    $\begingroup$ Whooow, cool trick, thanks for clearing this up :) It took me some time to understand it though xD The question is kinda harder then I expected :P But thanks though :D $e$ is a wonderful number. $\endgroup$ Aug 17 '11 at 20:55
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    $\begingroup$ When I read the question, I thought: one must explain how the Mean Value Theorem enters this argument. But this argument does it quite nicely while keeping that in the background. $\endgroup$ Aug 17 '11 at 22:19
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    $\begingroup$ I like this answer, but I think the proof lacks some detail which would make it clearer, and the statement of the theorem lacks the constant which you mentioned initially. I would add that, since g is constant, and g(x)=f(x)e^(-x) which implies then we have that f(x)=g(x)*e^x, we have that for any constant C, we have some f(x)=Ce^x. Thus, for all possibilities for f we have f=Ce^x. $\endgroup$ Aug 18 '11 at 3:38
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    $\begingroup$ @Qia Do you think that the translational symmetry is more fundamental than the uniqueness theorem? (cf. my answer) To me, it is the latter that is the essence of the matter in general, not the former. $\endgroup$ Aug 19 '11 at 15:02
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    $\begingroup$ This is me again. I've helped you twice to earn gold badge for Great Answer. Congrats! ♥(ˆ⌣ˆԅ) $\endgroup$ Sep 29 '14 at 14:15
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Let $f(x)$ be a differentiable function such that $f'(x)=f(x)$. This implies that the $k$-th derivative, $f^{(k)}(x)$, is also equal to $f(x)$. In particular, $f(x)$ is $C^\infty$ and we can write a Taylor expansion for $f$:

$$T_f(x) = \sum_{k=0}^\infty c_k x^k.$$

Notice that the fact that $f(x)=f^{(k)}(x)$, for all $k\geq 0$, implies that the Taylor series $T_f(x_0)$ converges to $f(x_0)$ for every $x_0\in \mathbb{R}$ (more on this later), so we may write $f(x)=T_f(x)$. Since $f'(x) = \sum_{k=0} (k+1)c_{k+1}x^k = f(x)$, we conclude that $c_{k+1} = c_k/(k+1)$. The value of $c_0 = f(0)$, and therefore, $c_k = f(0)/k!$ for all $k\geq 0$. Hence:

$$f(x) = f(0) \sum_{k=0}^\infty \frac{x^k}{k!} = f(0) e^x,$$

as desired.

Addendum: About the convergence of the Taylor series. Let us use Taylor's remainder theorem to show that the Taylor series for $f(x)$ centered at $x=0$, denoted by $T_f(x)$, converges to $f(x)$ for all $x\in\mathbb{R}$. Let $T_{f,n}(x)$ be the $n$th Taylor polynomial for $f(x)$, also centered at $x=0$. By Taylor's theorem, we know that $$|R_n(x_0)|\leq |f^{(n+1)}(\xi)|\frac{ |x_0 - 0|^{n+1}}{(n+1)!},$$ where $R_n(x_0)=f(x) - T_{f,n}(x)$ and $\xi$ is a number between $0$ and $x_0$. Let $M=M(x_0)$ be the maximum value of $|f(x)|$ in the interval $I=[-|x_0|,|x_0|]$, which exists because $f$ is differentiable (therefore, continuous) in $I$. Since $f(x)=f^{(n+1)}(x)$, for all $n\geq 0$, we have: $$|R_n(x_0)|\leq |f^{(n+1)}(\xi)|\frac{ |x_0|^{n+1}}{(n+1)!}\leq |f(\xi)|\frac{ |x_0|^{n+1}}{(n+1)!}\leq M \frac{|x_0|^{n+1}}{(n+1)!} \longrightarrow 0 \ \text{ as } \ n\to \infty.$$ The limit goes to $0$ because $M$ is a constant (once $x_0$ is fixed) and $A^n/n! \to 0$ for all $A\geq 0$. Therefore, $T_{f,n}(x_0) \to f(x_0)$ as $n\to \infty$ and, by definition, this means that $T_f(x_0)$ converges to $f(x_0)$.

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    $\begingroup$ You left out one step, which is to show the power series converges to $f(x)$ (which is not true for all $C^{\infty}$ functions). In the case at hand you can use the formulas for the remainder term of a finite Taylor expansion to get convergence. $\endgroup$
    – Zarrax
    Aug 17 '11 at 20:14
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    $\begingroup$ Yes, thanks. I'll come back in a bit and add it to the answer. $\endgroup$ Aug 17 '11 at 23:34
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    $\begingroup$ +1 for submitting this answer while this may not get as many upvotes as Yuan's simple answer. $\endgroup$
    – Isaac
    Aug 19 '11 at 0:12
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Yet another way: By the chain rule, ${\displaystyle {d \over dx} \ln|f(x)| = {f'(x) \over f(x)} = 1}$. Integrating, you get $\ln |f(x)| = x + C$. Taking $e$ to both sides, you obtain $|f(x)| = e^{x + C} = C'e^x$, where $C' > 0$. As a result, $f(x) = C''e^x$, where $C''$ is an arbitrary constant.

If you are worried about $f(x)$ being zero, the above shows $f(x)$ is of the form $C''e^x$ on any interval for which $f(x)$ is nonzero. Since $f(x)$ is continuous, this implies $f(x)$ is always of that form, unless $f(x)$ is identically zero (in which case we can just take $C'' = 0$ anyhow).

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    $\begingroup$ Wow, I like this. It's so simple :D $\endgroup$
    – Beta Decay
    Jan 29 '17 at 11:35
  • $\begingroup$ +1 Very intuitive, this is the way I would have done it. $\endgroup$
    – Toby Mak
    Aug 3 '19 at 4:17
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Hint $\rm\displaystyle\:\ \begin{align} f{\:'}\!\! &=\ \rm a\ f \\ \rm \:\ g'\!\! &=\ \rm a\ g \end{align} \iff \dfrac{f{\:'}}f = \dfrac{g'}g \iff \bigg(\!\!\dfrac{f}g\bigg)' =\ 0\ \iff W(f,g) = 0\:,\ \ W = $ Wronskian

This is a very special case of the uniqueness theorem for linear differential equations, esp. how the Wronskian serves to measure linear independence of solutions. See here for a proof of the less trivial second-order case (that generalizes to n'th order). See also the classical result below on Wronskians and linear dependence from one of my sci.math posts on May 12, 2003.

Theorem $\ \ $ Suppose $\rm\:f_1,\ldots,f_n\:$ are $\rm\:n-1\:$ times differentiable on interval $\rm\:I\subset \mathbb R\:$ and suppose they have Wronskian $\rm\: W(f_1,\ldots,f_n)\:$ vanishing at all points in $\rm\:I\:.\:$ Then $\rm\:f_1,\ldots,f_n\:$ are linearly dependent on some subinterval of $\rm\:I\:.$

Proof $\ $ We employ the following easily proved Wronskian identity:

$\rm\qquad\ W(g\ f_1,\ldots,\:g\ f_n)\ =\ g^n\ W(f_1,\ldots,f_n)\:.\ $ This immediately implies

$\rm\qquad\quad\ \ \ W(f_1,\ldots,\: f_n)\ =\ f_1^{\:n}\ W((f_2/f_1)',\ldots,\:(f_n/f_1)'\:)\quad $ if $\rm\:\ f_1 \ne 0 $

Proceed by induction on $\rm\:n\:.\:$ The Theorem is clearly true if $\rm\:n = 1\:.\:$ Suppose that $\rm\: n > 1\:$ and $\rm\:W(f_1,\ldots,f_n) = 0\:$ for all $\rm\:x\in I.\:$ If $\rm\:f_1 = 0\:$ throughout $\rm\:I\:$ then $\rm\: f_1,\ldots,f_n\:$ are dependent on $\rm\:I.\:$ Else $\rm\:f_1\:$ is nonzero at some point of $\rm\:I\:$ so also throughout some subinterval $\rm\:J \subset I\:,\:$ since $\rm\:f_1\:$ is continuous (being differentiable by hypothesis). By above $\rm\:W((f_2/f_1)',\ldots,(f_n/f_1)'\:)\: =\: 0\:$ throughout $\rm\:J,\:$ so by induction there exists a subinterval $\rm\:K \subset J\:$ where the arguments of the Wronskian are linearly dependent, i.e.

on $\rm\ K:\quad\ \ \ c_2\ (f_2/f_1)' +\:\cdots\:+ c_n\ (f_n/f_1)'\: =\ 0,\ \ $ all $\rm\:c_i'\:=\ 0\:,\ $ some $\rm\:c_j\ne 0 $

$\rm\qquad\qquad\: \Rightarrow\:\ \ ((c_2\ f_2 +\:\cdots\: + c_n\ f_n)/f_1)'\: =\ 0\ \ $ via $({\phantom m})'\:$ linear

$\rm\qquad\qquad\: \Rightarrow\quad\ \ c_2\ f_2 +\:\cdots\: + c_n\ f_n\ =\ c_1 f_1\ \ $ for some $\rm\:c_1,\ c_1'\: =\: 0 $

Therefore $\rm\ f_1,\ldots,f_n\:$ are linearly dependent on $\rm\:K \subset I\:.\qquad$ QED

This theorem has as immediate corollaries the well-known results that the vanishing of the Wronskian on an interval $\rm\: I\:$ is a necessary and sufficient condition for linear dependence of

$\rm\quad (1)\ $ functions analytic on $\rm\: I\:$
$\rm\quad (2)\ $ functions satisfying a monic homogeneous linear differential equation
$\rm\quad\phantom{(2)}\ $ whose coefficients are continuous throughout $\rm\: I\:.\:$

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    $\begingroup$ Careful not to divide by zero. $\endgroup$
    – Mark
    Aug 17 '11 at 18:47
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    $\begingroup$ @Mark My hints often ignore degenerate cases to focus on the essence of the matter. $\endgroup$ Aug 17 '11 at 19:07
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    $\begingroup$ Since first-year calculus students usually haven't heard of the Wronskian, why not just say $(f/g)' = 0$ and therefore $f/g$ is constant? $\endgroup$ Aug 17 '11 at 22:22
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    $\begingroup$ @Mic I presumed the "is constant" inference was clear. The point of mentioning the Wronskian is to emphasize that this is a special case of more general results (for those who may know such). On Wronskians see also my sci.math post here. $\endgroup$ Aug 17 '11 at 22:31
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    $\begingroup$ @MrPink The links have been updated, thanks. $\endgroup$ Sep 12 '20 at 19:39
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Here is a different take on the question. There is a whole spectrum of different discrete "calculi" which converge to the continuous case, each of which has it's special "$e$".

Pick some $t>0$. Consider the equation $$f(x)=\frac{f(x+t)-f(x)}{t}$$ It is not hard to show by induction that there is a function $C_t:[0,t)\to \mathbb{R}$ so that $$f(x)=C_t(\{\frac{x}{t}\})(1+t)^{\lfloor\frac{x}{t}\rfloor}$$ where $\{\cdot\}$ and $\lfloor\cdot\rfloor$ denote fractional and integer part, respectively. If I take Qiaochu's answer for comparison, then $C_t$ plays the role of the constant $C$ and $(1+t)^{\lfloor\frac{x}{t}\rfloor}$ the role of $e^x$. Therefore for such a discrete calculus the right value of "$e$" is $(1+t)^{1/t}$. Now it is clear that as $t\to 0$ the equation becomes $f(x)=f'(x)$, and $(1+t)^{1/t}\to e$.

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  • $\begingroup$ I'd like some clarification after "it is not hard to show..." Is $C_t$ a special notation? $\endgroup$
    – Pedro Tamaroff
    Mar 19 '12 at 18:51
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    $\begingroup$ @Peter: it's just a function Gjergji defined for convenience. $\endgroup$ Mar 24 '12 at 14:48
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    $\begingroup$ @J.M. The definition is kind of a leap. I'm interested in that leap. $\endgroup$
    – Pedro Tamaroff
    Mar 24 '12 at 16:27
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    $\begingroup$ Very cool... falemenderit. $\endgroup$
    – obataku
    Apr 29 '13 at 16:45
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The solutions of $f(x) = f'(x)$ are exactly $f(x) = f(0) e^x$. But you can also write it as $b a^x$, if you want a different basis. Then $f'(x) = b \log(a) a^x$, and so if you want $f'=f$ you need $\log(a)=1$ and $a=e$ (except for the trivial case $b=0$).

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The proof they use at High School, so not as deep or instructive, but it doesn't require as much knowledge.

$$\begin{eqnarray*} \frac{dy}{dx} &=& y\\ \frac{dx}{dy} &=& \frac 1 y\\ x &=& \log |y| + C\\ y &=& A\exp(x) \end{eqnarray*} $$

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    $\begingroup$ Why can you switch from $\frac{dy}{dx}=y$ to $\frac{dx}{dy}=\frac1y$? Last I recall they are not teaching non-standard analysis in high school. $\endgroup$
    – Asaf Karagila
    Jul 2 '12 at 22:03
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    $\begingroup$ In the British high school system, the ability to switch between dy/dx and dx/dy is a compulsory part of the math curriculum (in particular it's in "Core 3", a module taken in the final year i.e. when students are 17-18, and the 3rd of 4 compulsory pure mathematics modules). It's not a piece of non-standard analysis. However, this is a High School Course, and no formal real analysis is taugh that justifies the result. $\endgroup$ Jul 3 '12 at 12:32
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    $\begingroup$ British students are not taught to treat dy/dx as a fraction! They are explicitly warned against doing so! And it has not been treated as a fraction here! NB neither standard nor non-standard analysis is taught at High School - since most students who'll use calculus will be doing physics or engineering undergrad, not math. However since this is a question often asked by High Schoolers, I wanted to give an explanation using only High School methods. In Core 3, British students are expected to solve questions like $dy/dx=3y\sqrt x$ so $dy/dx=y$ should be an "easy" question for them. $\endgroup$ Jul 3 '12 at 19:27
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    $\begingroup$ @AsafKaragila: Dear Asaf, The formula that Just uses is standard, and has many justifications that have nothing to do with non-standard analysis. It presupposes that the relationship $y = y(x)$ is invertible, so that we can write $x$ as a function of $y$, but then again the formula only makes sense when $dy/dx$ is non-zero, in which case we know (inverse function theorem, if you like) that indeed $y = y(x)$ is so invertible. As with Bill's answer, this one has complications if we allow for the possibility a priori that $y(x)$ vanishes at some values of $x$. (One advantage of ... $\endgroup$
    – Matt E
    Jul 3 '12 at 22:24
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    $\begingroup$ @AsafKaragila Or, you can use the chain rule to rigorously justify JustPassingThru's proof (en.wikipedia.org/wiki/…). $\endgroup$
    – Poder Rac
    Aug 13 '20 at 15:19
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Let $x \in C^1$ on the whole line be a solution to $\dot{x}(t) = x(t)$, $x(0) = 1$. Using the Taylor expansion with remainder, show that necessarily $x(t) = e^t$.

We have that $\dot{x} = x$ implies $x^{(n)} = x^{(n-1)}$ for all $n \ge 1$, and by induction on $n$, we have that $x(t)$ is $C^\infty$ with $x^{(n)} = x$ for all $n$. Thus, if $x(0) = 1$ and $\dot{x} = x$, Taylor's Theorem gives$$x(t) = \left( \sum_{k=0}^{N-1} {{t^k}\over{k!}}\right) + {{x^{(N)}(t_1)}\over{N!}}t^N,$$for $t_1$ between $0$ and $t$. But $x^{(N)} = x$, so if$$M = \max_{|t_1| \le |t|} |x(t)|,$$which we know exist by compactness of $[-|t|, |t|]$, then$$\left| x(t) - \sum_{k=0}^{N-1} {{t^k}\over{k!}}\right| < {{Mt^N}\over{N!}}.$$The right-hand side heads to $0$ as $N \to \infty$, so the series for $e^t$ converges to $x(t)$.

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Ok, so all of the above answers either assume that the derivative of $e^x$ is itself, the derivative of $\log(x)$ is $1/x$, or some other theorem regarding Taylor series. Here is an argument which is, in some sense, from first principles.

Suppose you want to find a function which is it's own derivative. Suppose we have a real valued function $f(x)$ which is suitably well behaved such that every higher order "antiderivative" of $f$ to exist.

Let us denote the antiderivative of $f_0=f$ to be $f_1$, the antiderivative of $f_1$ to be $f_2$ and so on.

Suppose we consider the sum

\begin{equation} E(x) = \sum_{i=0}^\infty f_i(x). \end{equation}

Suppose we assume that the series is absolutely and uniformly convergent so that it can be differentiated term by term (this will be justified later on).

We end up with the equation

$$ \frac{d}{dx} E(x) = E(x) - f(x). $$

Since $E$ is our guess answer for the question, we take our $f$ to be identically zero. This gives us that $f_1(x) = c_1$, $f_2(x) = c_1x+c_2$ and so on. But note here that the constants $c_1,c_2,\ldots$ should be chosen such that the series defining $E$ should converge absolutely and uniformly. Equivalently we can choose $c_i$ such that $C=\sum c_i < \infty$. Therefore we end up with the expression,

$$ E(x) = C + C\sum_{i=1}^\infty \frac{x^i}{i!}. $$

This seems to answer the question but there is a slight caveat here.

Suppose $E$ is a function which is its own derivative, why should it be the sum of some $f_i$'s as given above?

Well, this question is answered if we can characterize the starting function $f$ in terms of $E$, which can be seen from the equation $f(x) = E(x) - E'(x)$. In fact this forces the choice of $f$ to be zero! Now, the claim is proved.

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  • $\begingroup$ Unfortunately I do not know how to number equations inside answers. If that were possible, then the answer could be written in a bit more accessible way. $\endgroup$ Jan 25 at 16:14
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Note that $e$ is defined by the following Limit: $e=\lim_{n \rightarrow \infty}(1+ \frac{1}{n})^n$. Then: $e^x=\lim_{n \rightarrow \infty}(1+ \frac{1}{n})^{nx}$. Applying the Definition of the derivative $f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$ one obtains: $(e^x)'=\lim_{h \rightarrow 0} \frac{ \lim_{n \rightarrow \infty}((1+ \frac{1}{n})^{n(x+h)}-(1+ \frac{1}{n})^{nx})}{h} = \lim_{h \rightarrow 0}( \lim_{n \rightarrow \infty}(1+\frac{1}{n})^{nx} \lim_{n \rightarrow \infty}(\frac{(1+\frac{1}{n})^{nh}-1}{h}))$

$= e^x \lim_{h \rightarrow 0} \lim_{n \rightarrow \infty}(\frac{(1+\frac{1}{n})^{nh}-1}{h})$.

Now one can replace $h$ by $n$ by the relation $h= \frac{C}{n}$ with a finite constant $C$, because if $n \rightarrow \infty$ then $h$ tends to Zero. Hence:

$\lim_{h \rightarrow 0} \lim_{n \rightarrow \infty}(\frac{(1+\frac{1}{n})^{nh}-1}{h}) = \lim_{h \rightarrow 0} (\frac{(1+\frac{h}{C})^{C}-1}{h}) = \lim_{h \rightarrow 0} (\frac{(1+C \frac{h}{C} + \frac{C(C-1)}{2}(\frac{h}{C})^2+O(h^3)-1}{h}) = \lim_{h \rightarrow 0} (1 + \frac{C(C-1)}{2}\frac{h}{C^2}+O(h^2)) = 1$

Therefore $(Ce^x)'=C(e^x)'=Ce^x$.

q.e.d.

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    $\begingroup$ Late to the party but all this proves is that $\left(Ce^x\right)'=Ce^x$, not that $f(x)=f'(x)\iff f(x)=Ce^x$. $\endgroup$
    – Jam
    Sep 7 '16 at 0:53

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