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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!

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    $\begingroup$ probably as unique as every other real number. $\endgroup$ – Jorge Fernández Hidalgo Jul 22 '16 at 21:09
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    $\begingroup$ Yes $e^x$ is the only function that is its own derivative. $\endgroup$ – Gregory Grant Jul 22 '16 at 21:09
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    $\begingroup$ @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 $Ce^x$. $\endgroup$ – Erick Wong Jul 22 '16 at 21:11
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    $\begingroup$ @GregoryGrant, well except for $ae^x$ :) $\endgroup$ – Pax Kivimae Jul 22 '16 at 21:16
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    $\begingroup$ @user1717828 That's included in $ce^x$. $\endgroup$ – Gregory Grant Jul 23 '16 at 0:40
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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.

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  • $\begingroup$ The same idea as in Jason's answer but without differential equations. $\endgroup$ – Jyrki Lahtonen Jul 22 '16 at 21:27
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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 $A$

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    $\begingroup$ 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. $\endgroup$ – Simply Beautiful Art Jul 22 '16 at 21:16
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    $\begingroup$ @SimpleArt I dont think so, since its not a question of existence of $e^x$, but rather uniqueness. $\endgroup$ – Pax Kivimae Jul 22 '16 at 21:17
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    $\begingroup$ @SimpleArt This is a slightly different fact. $\endgroup$ – JasonM Jul 22 '16 at 21:18
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    $\begingroup$ 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. $\endgroup$ – Erick Wong Jul 22 '16 at 21:21
  • $\begingroup$ @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. $\endgroup$ – JasonM Jul 22 '16 at 21:24
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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.

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    $\begingroup$ This shows that $e^x$ is a solution, not that it is a unique solution. $\endgroup$ – Aditya Jul 23 '16 at 2:50
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    $\begingroup$ @Aditya Not really meant to be an answer in the first place, more of an interesting fact. $\endgroup$ – Simply Beautiful Art Jul 23 '16 at 13:37
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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 $$ f(x) = y_0 \cdot \exp(x). $$

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