How do we show that the function which is its own derivative is exponential? In my calculus class, to show that $\frac{d}{dx}e^x=e^x$ we did something like this:
$$\lim_{h \to 0} \frac{a^{x+h} - a^x}{h} = a^x \lim_{h \to 0} \frac{a^h-1} h,$$
and then we defined $e$ to be the base $a$ that makes the limit
$$\lim_{h \to 0} \frac{a^h-1}{h}$$
equal to $1$. Now, my question is this: how can we know beforehand that the function which is its own derivative is an exponential function? In other words, if we have 
$$ \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = f(x),$$
how do we know $f(x)$ is of the form $ka^x$ (ignoring the trivial case $f(x) = 0$)? How can we show that there aren't functions of other forms that give the same result?
 A: OK, so we are given the differential equation
$$f'(x)=f(x) \tag{1}$$
Of course this implies that $f$ is differentiable.
Let's first make the observation that whenever $f(x)$ is a solution to this differential equation, then for arbitrary $s,c\in\mathbb R$,
$$g(x) = s\,f(x-c) \tag{2}$$
is also a solution.
Let's further have a look at the general form the solutions may have. If at some $x$, $f(x)>0$, then due to continuity we have $f(x)>0$ in some neighbourhood. Since $f'(x)=f(x)$ this also implies that $f(x)$ is monotonously growing in that neighbourhood. Clearly if it is positive and growing, it cannot reach $0$ for any larger value of $x$. Analogously, if it is anywhere negative, it will remain negative for all larger $x$.
Thus we have by now restricted the possible solutions to functions of one of the following types:


*

*Either the function is everywhere strictly positive and monotonously growing.

*Or the function is zero up for all $x\le x_0$, and strictly positive for all $x>x_0$ (at $x_0$ if has to be $0$ due to continuity, as a limit from the left must reproduce the function value).

*Or the function is zero everywhere.

*Or the function is the negative of one of the first two types.
Of course the constant function $f(x)=0$ is one solution of $(1)$. If there are other solutions, there must be at least one $x$ so that $f(x)\ne 0$. Because of $(2)$ we can thus assume wlog that $f(0)=1$. Also, for convenience, define for the first case $x_0=-\infty$.
Now assume $f(x)\ne 0$. Then $f(x+1)/f(x)$ is well defined in a neighbourhood, and we have
\begin{align}
\frac{\mathrm d}{\mathrm dx}\,\frac{f(x+1)}{f(x)}
&= \frac{f(x)f'(x+1) - f(x+1)f'(x)}{f(x)^2}\\
&= \frac{f(x)f(x+1) - f(x+1)f(x)}{f(x)^2} = 0\tag{3}
\end{align}
In other words, $f(x+1)/f(x)$ is a constant. This holds for all $x>x_0$.
So since we have assumed that $f(0)=1$, we have $f(1)=f(0+1)=af(0)=a$, $f(2)=a(1+1)=af(1)=a^2$. Also, assuming $x_0<-1$, $af(-1)=f(0)$, so $f(-1)=a^{-1}$ etc. Or in short, for any $n\in\mathbb Z$ with $n>x_0$, $f(n)=a^n$.
But then, in $(3)$, instead of $f(x+1)$ we could have used $f(x+1/k)$ with the same result, giving $f(n/k)=b^n$. For $n=k$, we thus get $a = f(1) = f(k/k) = b^k$, that is, $b=a^{1/k}$. So in summary, we have for all $q\in\mathbb Q$ with $q>x_0$ that $f(q)=a^q$.
But since $f$ is differentiable, it is in particular continuous, and thus determined by its values at rational points. And since $a^x$ is continuous, too, we get $f(x)=a^x$ for all $x\in\mathbb R$ with $x>x_0$.
Continuity of $a^x$ then also implies $\lim_{x\to x_0+0}f(x) = \lim_{x\to x_0+0}a^x = a^{x_0}$. However since by assumption $f(x)$ is continuous, $\lim_{x\to x_0+0}f(x) = f(x_0)$. But we previously have seen, also by continuity, that $f(x_0)=0$. Since the equation $0=a^{x_0}$ has no solution in $\mathbb R$, this means that the second case cannot apply, and the function therefore is strictly positive everywhere, and thus is $a^x$ everywhere.
In other words, the function is an exponential function.
A: First, notice that

*

*In order to show that $\dfrac d {dx} a^x = ( a^x\cdot\text{constant}),$ you don't need to show that only exponential functions satisfy the equation $f'(x) = \big(f(x)\cdot\text{constant}\big).$

*You already see in the discrete case that exponential functions grow at a rate proportional to their current size: $$ \frac{\Delta 2^x}{\Delta x} = \frac{2^{x+1} - 2^x}{1} = 2^x.$$
To show that no other functions satisfy $f'(x) = f(x),$ can be done by supposing $f$ is such a function and considering
$$
g(x) = \frac{f(x)}{e^x}.
$$
Then
$$
g'(x) = \frac{e^x f'(x) - f(x) e^x }{e^{2x}} = \frac{f'(x) - f(x)}{e^x} = \frac{f(x)-f(x)}{e^x} = 0.
$$
If $g'(x) = 0$ for all values of $x$, then $g$ is constant, so $g(x) = \Big( f(x)\cdot \text{constant}\Big).$
Note that the mean value theorem was tacitly used above: It is used in the proof that functions whose derivatives are everywhere $0$ are constant.
A: So essentially we are looking for the solution to the differential equation $\frac{dy}{dx}=y$ where $y=f(x)$. This can be solved by using separation of variables $\frac{dy}{y}=dx$ and then we integrate both sides and solve for $y$ and get $ke^x$ now when you solve a differential equation there is something about a uniqueness of the solution and that all solutions are contained by changing the constant $k$. So therefore you know the only functions where $y=y'$ are in the form $ke^x$
A: If $f'=f$, take $g(x)=f(x)e^{-x}$. Then $g'(x)=f'(x)e^{-x}-f(x)e^{-x}=0$ and so $g$ is constant. The constant is $g(0)=f(0)$. Therefore, $f'=f$ implies $f(x)=f(0)e^{x}$.
A: Hint. If $f$ is such a function then calculating the derivative of $f(x)/e^x$ provides useful information about $f$.
A: Consider the more general equation:
$$y=a^x \tag{1}$$
We apply logarithmic rules:
\begin{align}
\ln y &= x\ln a\\
\frac{\mathrm d}{\mathrm dx}\,\ln y &= \frac{\mathrm d}{\mathrm dx}\,x \ln a\\
\frac{\mathrm d}{\mathrm dx}\,\ln y &= \ln a \frac{\mathrm d}{\mathrm dx}\,x\\
\frac{\mathrm d}{\mathrm dx}\,\ln y &= \ln a \tag{2}\\
\end{align}
We apply the chain rule to the LHS of (2)
\begin{align}
\frac{\mathrm d \ln y}{\mathrm dy}\,\frac{\mathrm dy}{\mathrm dx} &= \ln a\\
\frac{1}{y}\,\frac{\mathrm dy}{\mathrm dx} &= \ln a\\
\frac{\mathrm dy}{\mathrm dx} &= y \ln a \\
\frac{\mathrm d}{\mathrm dx}\, a^x &= a^x \ln a \tag{3}
\end{align}
Thus, (3) is the general result for the derivative w.r.t. $x$ of a function where $a$ is raised to $x$.
We ask ourselves, for what value of $a$ is our constant $\ln a$ equal to 1 and the answer is, of course, $e$
\begin{align}
\frac{\mathrm d}{\mathrm dx}\, a^x &= a^x \ln a \\
\frac{\mathrm d}{\mathrm dx}\, e^x &= e^x \ln e \\
\frac{\mathrm d}{\mathrm dx}\, e^x &= e^x
\end{align}
A: \mathrm
We want an expression which when differentiated yields itself.
The first possible case is when $f(x)=0.$
Second, let's solve the initial value problem $f(0)=1$ and $f(x)=f'(x)$.
Now taking $f(x)=1,$ and then integrating the previous function and adding the result to obtain a better solution to the initial value problem, we get,
$f(x)=1 + x$, and then further integrating the second element, and repeating we get,
$f(x)=1 + x + (x^2)/2.$
Repeating the process will eventually yield the taylor series expansion for the exponential function as $f(x)=\sum_{i=0}^n (x^n)/k!$
And then we define the nth root of this expression as being the constant e.
Also, by differentiating the inverse of the function $f(x)$, we arrive at $(f^-1)'(x)=1/x$, which is the natural logarithm.
The essence of the exponential function with base e is that it is its own derivative and as such can be taken as its definition.
PS: We can also define e to be the positive upper limit of integration in the integral definition of the natural logarithm, and deduce that $e^x$ is its own derivative.
