# On proving that $f(x) = f'(x)\iff f(x) = e^x$. (Not aware of a possible duplicate.) [duplicate]

This is just a curious question, but is the following true?

$$f(x) = f'(x)\iff f(x) = e^x.$$

I can prove that $\dfrac{\mathrm d}{\mathrm dx}\left(e^x\right) = e^x$ from using the the formula, $e^x := \operatorname*{\lim}\limits_{n\to 0}(1+n)^{1/n}.$ For those who are not familiar with the proof, it can be found here. Or, you can go here for a similar approach.

However, my question is asking whether or not $f(x) = e^x$ is the only function equal to its own deriv. I suspect it is true, but how can we prove that $e^x$ is the only value equal to its derivative, for any $x$?

I consider it very likely that there exists another question out there, perhaps exactly like this. If so, please comment the link below, and I will go straight to it, delete this post, and give you a muffin. I do not intend on trolling or wasting anyone's time.

## marked as duplicate by vrugtehagel, gimusi, Ethan Bolker, Omnomnomnom, Community♦Mar 21 '18 at 13:20

• No, see $f(x)=0$. In general, $f(x)=Ke^x$ is a solution. – vrugtehagel Mar 21 '18 at 13:14
• There's also $f(x)=2e^x$. I think there are a few more. – Aweygan Mar 21 '18 at 13:14
• We do have $f(x) = Ce^x$ for some constant $C$, though. – Omnomnomnom Mar 21 '18 at 13:14
• I'm sure this is a duplicate. If $f$ is another function that satisfies the differential equation, differentiate $f(x)/e^x$ to show you get a constant. Google google.com/… – Ethan Bolker Mar 21 '18 at 13:15
• Ah yes, I didn't think of that. I kind of left out the case $f(x) = 0$ since that is trivial. Is there a way of proving so? – Feeds Mar 21 '18 at 13:16
If $f=f'$, then let $g=\frac f\exp$. Then$$g'=\frac{\exp\times f'-f\times\exp}{\exp^2}=0.$$Therefore, $g=C$, for some $C\in\mathbb R$. In other words, $f=C\times\exp$.
• I am familiar with the $\text{exp}$ notation, so I understand. I would have upvoted if I did not reach my daily voting limit, hahah. Thank you very much though :) – Feeds Mar 21 '18 at 13:31