Can we interpret $e^x$ as an unique fixpoint of the derivative $\frac{d}{dx}$ and what would it mean? I'm not sure if there is some similar question, but I didn't find something on this topic as I wanted, so please spare me in case I overlooked something. 
Back in school I've learned that if you try to find a function that solves $\frac{d}{dx}a^x = a^x$ and try different values for a that you will get closer to 2.71..., which is later on defined as $e$ the Euler's number.
Now I've also learned that we can reinterpret the derivative of a function as a linear transformation whose graph is (after an appropriate translation) the best linear approximation to the graph of the original function, with the Jacobian matrix as this linear transformation - as of Wikipedia, but I'm sure we also thought about it like that in my real analysis class back then.
Back to my initial question in this context this would kind of mean that $e^x$ is an unique fixpoint of the "map" (I'm not sure if this terminology is correct for the derivative, I think it's an operator) $\frac{d}{dx}$. Even more it would mean that the best linear approximation of $e^x$ is the function itself. 
When it comes to operators my knowledge is bit rudimentary, so I don't know if there is a necessary condition for an operator to have a fixpoint. I only know some fixpoint theorems, so I would welcome some suggestions on this.
So in general I have my troubles understanding what this interpretation means for $e^x$and $e$ itself. Does this mean we cannot approximate $e^x$ with a linear approximation - could we do it nonlinear then and how? Also why is it $e$ - is it like $\pi$ just a coincidence for a constant? How can we view the derivative in this context - maybe also geometric or algebraic?
Thank you if you read until here, I hope you can help me understanding this a bit better.
 A: In order to speak about a “fixed point” of a linear map, we need a map from some vector space $V$ to itself.
For this case, the space can be $V=C^{\infty}(\mathbb{R})$, the functions having derivatives of any order on the real line.
The map $D\colon V\to V$ that associates to each function its derivative is linear. A fixed point is thus a function $f$ which is equal to its derivative:
$$
f(x)=f'(x) \qquad\text{for every $x\in\mathbb{R}$}
$$
There are several ways for finding the solutions. A simple one is to assume $f$ is a solution and to consider
$$
h(x)=e^{-x}f(x)
$$
Then $h'(x)=-e^{-x}f(x)+e^{-x}f'(x)=-e^{-x}+e^{-x}f(x)=0$. Therefore $h$ is constant, having zero derivative over $\mathbb{R}$. So, for every $x$,
$$
c=e^{-x}f(x)
$$
where $c=h(0)$, and therefore
$$
f(x)=ce^x
$$
Conversely, any such function satisfies the requirement, so we have found all solutions to the fixed point problem to be of the form
$$
c\exp
$$
where $\exp$ is the standard exponential function defined by $\exp(x)=e^x$ and $c$ is any scalar (real number).
We can notice that the set of fixed points forms a line in $V$, that is, a subspace of dimension one.
In other terminology, $1$ is an eigenvalue of $D$, with dimension one eigenspace.
A: Let $C_{1}(1)$ denotes the set of all one variable functions which their first derivative exists. Now the operation $\frac{d}{dx}:C_1(1)\rightarrow C_0(1)$ can be defined easily by the set of relations for derivation, where $C_0(1)$ is the set of all one variable functions. From that we can understand that this operation is linear as you mentioned. To understand your question it is convenient to see your new space (set of all one variable functions that has first derivative). Now the fixed point of $\frac{d}{dx}$ in this space has meaning and is $e^x(=\exp(x))$. You only need to re-think on the space where you are working on.
