Question regarding the definition of a functional I know that a functional is defined as a map
$$f : M \mapsto \mathbb{R} \tag{1}$$
or
$$f : M \mapsto \mathbb{C} \tag{2}$$
(or similar). Therefore, expressions like the following are functionals:
$$ F[y] = \int_{a}^{b} \mathrm{d}x \sqrt{1-y'^{2}(x)} \tag{3}$$
$$ G[y, y'] = \int_{x_{1}}^{x_{2}} \mathrm{d}x \frac{y^{3}(x) + y'(x) +x^{2}}{\cos(y(x))} \tag{4}$$
$$ H[y] = \exp(y^{2}(x) \delta(x-x_{0})) \tag{5}$$
because they depend on the function ($y(x)$) and its derivatives only, but not on the independent variable. Thus by plugging an implicit map $y: z \mapsto f(z)$, one gets a real number (that is how I understand it, please correct me if I am wrong).
My question regards expressions which also involve the independent variable, like
$$A[y(x)] = y(x) \sin(y(x)) \tag{6}$$
$$B[x(t)] = x(t) \int_{0}^{1} \mathrm{d}w \frac{e^{-w}}{w + t} x(w) \tag{7}$$
Can these be regarded as functionals? Are they some kind of generalised functionals?
The reason for my question is the fact that I am learning about functional derivatives. And while I have seen many examples of how to find the functional derivatives of $F, G, H$, I am not clear on how (or whether it is possible) to find the functional derivative of something like $A$ or $B$.
 A: 
"expressions like the following are functionals: [...] because they depend on the function (()) and its derivatives only, but not on the independent variable."

Thousands of people would disagree with that definition of a "functional" since the Feynman-Vernon influence functional does depend on the independent variable $t$ [see Eq. 9 of this 2012 paper or Eq. 8 of this 1995 paper]. These are just two of the many papers over a period of 6 decades that use this functional (an older paper from 1963, which has 3000+ citations on Google Scholar is this one).
In your comment, you said:

"Just to clarify,  and  are not functionals, but operators."

But no one said that an operator cannot be a functional.

"while I have seen many examples of how to find the functional derivatives of ,,, I am not clear on how (or whether it is possible) to find the functional derivative of something like  or ."

Your $A$ is a composite function. It can also be written with round brackets: $A(y(x))$ to indicate that it's an ordinary function $A(x)$ in which the argument is an ordinary function $y(x)$. Knowing $x$ is enough to give you $A(y(x))$, whereas for $B[y(x)]$ we certainly need the square brackets to indicate that we need to know all of $y(x)$ (not just $y$ at a specific value of $x$) to obtain $B$.
Your $B$ is precisely the same as the following operator in Annie Cuyt's PhD thesis, except you changed $s$ to $w$ and the numerator was changed from $t$ to $\exp(w)$. It was differentiated in her thesis as you can see below:

She used the Frechet derivative, where $L$ is the derivative (see one page earlier in the same thesis):

A: The idea that a map from functions to anything else could "depend on the independent variable" is quite vague and ambiguous, but I suspect I know what is intended.
If we are being careful, in modern times, a function $f$ can be defined in various ways, and, at the end of the definition, there is no "independent variable" (by any name). Yes, $f(x)$ may be expressed by a formula in terms of $x$, but the scope of this auxiliary "$x$" is limited to that definition.
For some sort of operator/functional $T$ on functions $f$, to say that $Tf$ "depends on $x$" is not literally sensible, though I can guess what is intended... Ok, the $f$'s are functions on some space, for example, the real line, and the outputs $Tf$ are also functions on the real line (as opposed to just being real or complex numbers). It's not that they "depend on the independent variable". That is a too naive description of functions, etc.
Yes, at a very introductory level, it is useful to think about "variables" and "independent variables", and such, but it is even more important to eventually discard this viewpoint, since it amounts to "painting oneself into a corner". After all, it is not true that "$x$" is any sort of sacred name for the input to a function, etc. :)
