How do we show that the function which is its own derivative is exponential?

Hint. If $f$ is such a function then calculating the derivative of $f(x)/e^x$ provides useful information about $f$.

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 $f(x)=f(0)e^{x}$.

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.