quick question on an example of the derivative as a linear map. After reading many answers on the subject I feel like I am close to finally understanding why the derivative is a linear map. I think that if someone helps me understand the following example I might "get it"
So I have $f(x)= e^x$ then $f(x)' = e^x$ is not linear, but instead the function defined by $y \rightarrow e^x y$ is linear, but what is $y$? is it introduced for the sole purpose to have a linear map? Say that I fix x, then for different values of $y$ what am I computing? 
So I should not call $f(x)'$ the derivative any more if the derivative is a linear map, then what should I call $f(x)'$?
 A: You are confusing the derivative of a function with the differential operator. 
The derivative of a function is general is not linear, as you show: $f'(x) $ can be equal to $x^2$, $x^{29}$, $e^x$, etc, all non - linear function.
What is linear is the differential operator: this is NOT a function, is an operator that goes from a space of functions to another space of functions, very different^[1] than something like $x \mapsto e^x$ which is a function and maps numbers to other numbers.
So what it means in general for an operator (or a function) to be linear? It means that 
$$f(ax + by) = af(x) + bf(y)$$
where $f$ is the operator, $a,b$ are numbers and $x,y$ elements in the domain of the opeator. 
So if for a function $f:\mathbb R \mapsto \mathbb R$ being linear means that 
$$f(ax + by) = af(x) + bf(y)$$ with $a,b,x,y \in \mathbb R$ (and this then implies that $f(x) = ax + b$ for some $a,b$) for an operator  $L$ it means that 
$$L(af + bg) = aL(f) + bL(g)$$ where $L$ is our operator, $a,b \in \mathbb R$ and $f,g$ are functions in the domain of the operator. 
Now if $L = \frac d{dx}$ is the differential operator that associates to a function its derivative, that is $L(f) = f'(x)$, then we say $\frac d{dx}$ is linear because 
$$L(af + bg) = (af + bg)' = af' + bg' = aL(f) + bL(g)$$
Hope it's more clear now.
[1] Okay, it's not very different. Technically they are the same thing, just their domain are different (reals vs space of functions). It's good to keep in mind their distinctions though :)
A: Think of the derivative (more properly the differential) as a rule that assigns a linear map to each point of the domain. This linear map approximates the change in the function’s value for a small displacement from that point.  
In your example, you have the rule $f'(x)=e^x$. This rule assigns the map “multiply by $e^{x_0}$” to the point $x_0$. For any given value of $x_0$, this is multiplication by a constant, which is obviously linear. This linear map can then be used to approximate $f(x_0+h)$ by $f(x_0)+e^{x_0}h$. (I used $h$ instead of $y$ to avoid confusing a displacement from a point in the domain with a value in the function’s range.)
A: I think you may be misunderstanding what is being called linear.
It isn't the function $f$ that is linear or not, it is the action of taking a derivative -- the differential operator $\frac d{dx}$ is linear in the sense that
$$\frac d{dx}(a\cdot f + b\cdot g) = a\cdot \frac d{dx}(f) + b\cdot \frac d{dx}(g)$$
where $a$ and $b$ are constants and $f$ and $g$ are differentiable functions. Linearity here means that the operator "distributes" across a sum of terms, and the multiplicative constants "pull through".
A: The linearity of a system, map, function, etc. is tested by the superposition principle. For example, consider a function $f(x)$. It is linear if
$f(\alpha x_1 + \beta x_2) = \alpha f(x_1) + \beta f(x_2)$.
In this case, the derivative is also linear operator because
$D[\alpha y_1(x) + \beta y_2(x)] = \alpha D[y_1(x)] + \beta D[y_2(x)]$.
