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.
Thank you in advance.
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, $f=C\times\exp$.