# How unique is $e$?

Is the property of a function being its own derivative unique to $e^x$, or are there other functions with this property? My working for $e$ is that for any $y=a^x$, $ln(y)=x\ln a$, so $\frac{dy}{dx}=\ln(a)a^x$, which equals $a^x$ if and only if $a=e$.

Considering equations of different forms, for example $y=mx+c$ we get $\frac{dy}{dx}=m$ and $mx+c=m$ only when $m=0$ and $c=0$, so there is not solution other than $y=0$. For $y=x^a$, $\frac{dy}{dx}=ax^{a-1}$, which I think equals $x^a$ only when $a=x$ and therefore no solutions for a constant a exist other than the trivial $y=0$.

Is this property unique to equations of the form $y=a^x$, or do there exist other cases where it is true? I think this is possibly a question that could be answered through differential equations, although I am unfortunately not familiar with them yet!

• probably as unique as every other real number. Commented Jul 22, 2016 at 21:09
• Yes $e^x$ is the only function that is its own derivative. Commented Jul 22, 2016 at 21:09
• @acernine Use the product rule on $ye^{-x}$ and you'll see that if $y'=y$ then the derivative of $ye^{-x}$ is $0$, making it constant, so $y$ must take the form $Ce^x$. Commented Jul 22, 2016 at 21:11
• @GregoryGrant, well except for $ae^x$ :)
– Pax
Commented Jul 22, 2016 at 21:16
• @user1717828 That's included in $ce^x$. Commented Jul 23, 2016 at 0:40

Assume that $f(x)$ is a function such that $f'(x)=f(x)$ for all $x\in\Bbb{R}$. Consider the quotient $g(x)=f(x)/e^x$. We can differentiate $$g'(x)=\frac{f'(x)e^x-f(x)D e^x}{(e^x)^2}=\frac{f(x)e^x-f(x)e^x}{(e^x)^2}=0.$$ By the mean value theorem it follows that $g(x)$ is a constant. QED.

• The same idea as in Jason's answer but without differential equations. Commented Jul 22, 2016 at 21:27

Consider the equation $y'=y$. Our goal is to solve for the function $y=f(x)$. Roughly speaking $$\frac{dy}{dx}=y \implies \frac{dy}{y}=dx \implies \int\frac{dy}{y}=\int dx \implies\ln(y)=x+C \implies y=e^{x+C}=Ae^x$$

for some constant $A$

• You have to figure out what $\frac d{dx}\ln(y)$ means first, to do that. Which usually requires knowing $\frac d{dx}e^x=e^x$. Big circle of going nowhere. Commented Jul 22, 2016 at 21:16
• @SimpleArt I dont think so, since its not a question of existence of $e^x$, but rather uniqueness.
– Pax
Commented Jul 22, 2016 at 21:17
• @SimpleArt This is a slightly different fact. Commented Jul 22, 2016 at 21:18
• This is standard, but the OP confesses unfamiliarity with differential equations and there is a bit of untidiness here in dealing with the $y=0$ and $y<0$ cases. Commented Jul 22, 2016 at 21:21
• @ErickWong Ah, unfortunately I only skimmed the question so I didn't see that last part. Yes, I agree the $y \leq 0$ cases are untidy, but if one wants a general understanding, I think this argument suffices, even if one is unfamiliar with differential equations. I'll add some comments to explain. Commented Jul 22, 2016 at 21:24

This may not be an answer you are looking for, but its a nice one to consider.

Consider $y=\cos(ix)-i\sin(ix)$.

You may find that:

$$\frac{dy}{dx}=-i\sin(ix)-i^2\cos(ix)=\cos(ix)-i\sin(ix)$$

Thus, $y'=y$ is satisfied. Since $y(0)=1$, $y'(0)=1$, $\dots$, then by Taylor's theorem, we have $e^x=\cos(ix)-i\sin(ix)$, or more commonly known as

$$e^{ix}=\cos(x)+i\sin(x)$$

Which is Euler's formula for complex exponents.

• This shows that $e^x$ is a solution, not that it is a unique solution. Commented Jul 23, 2016 at 2:50
• @Aditya Not really meant to be an answer in the first place, more of an interesting fact. Commented Jul 23, 2016 at 13:37

The equation $$\frac{\mathrm{d}}{\mathrm{d}x} f(x) = f(x)$$ is a linear (thus Lipschitz continuous), first-order ordinary differential equation on $\mathbb{R}$. By the Picard-Lindelöf theorem, such an equation has a unique solution for any initial condition of the form $$f(0) = y_0$$ with $y_0 \in \mathbb{R}$. In particular, for the condition $$f(0) = 1$$ the unique solution is $f = \exp$, so given that condition, $e \equiv \exp(1) = f(1)$ is unique.

For the general initial condition, you get, because the ODE is linear, that the solution is always $$f(x) = y_0 \cdot \exp(x).$$