Why is the symbol for derivatives $\frac{d}{dx}$? Is this notation for derivatives just a symbol, or is there actually a reason behind it? I understand the meaning behind $\frac{dy}{dx}$ since that is the formula for calculating the slope. But can $d$ be separated from $y$ as in $\frac{d}{dx}y$?
 A: $\frac{\mathrm d}{\mathrm dx}$ is an operator, so $\frac{\mathrm dy}{\mathrm dx}$ precisely means $\frac{\mathrm d}{\mathrm dx}(y)$.
We take $\frac{\mathrm d}{\mathrm dx}$ as the symbol for the differential operator, because it's then intuitive to remember things like the chain rule. It's just notation.
A: It's just a (purely symbolic) generalization from $\frac{dy}{dx}$.
If the dependent variable does not have a name but is just an expression, we would have to write something like
$$ \frac{d(x^2+2x+1)}{dx} $$
or whatever. In particular for higher derivatives this would mean that the function could end up deep in a nest of (formal) fraction bars, so it makes for easier notation to pretend the $y$ in $\frac{dy}{dx}$ is just a factor that we can move out of the fraction.
As a bonus, this then gives us a conveniently memorable notation for the operator that takes an expression to its derivative.
A: Suppose you have a formula for $y$ in terms of $x$, like
$$
y = x^3 + 2x
$$
Then you can, at some point $x = a$, compute the slope:
$$
\frac{dy}{dx} = 3a^2 + 2
$$
And you realize (after some work) that this really works for any $a$, and so maybe you write it as a function
$$
\frac{dy}{dx}(a) = 3a^2 + 2
$$
to be read "the slope at $a$ is $3a^2 + 2$, for any real number $a$." But once you've done that, you might as well write
$$
\frac{dy}{dx}(x) = 3x^2 + 2.
$$
And what you've done is to start with a function $y$ of the variable $x$, and produce a new function of $x$, called the derivative. That new function usually doesn't get written as I've done, but instead is written 
$$
\frac{dy}{dx} = 3x^2 + 2.
$$
It's natural to think of "going from $y$ to this new function" as an operation on $y$, sort of a function that gets applied to a function to produce a new one. 
What should we call this new operation? Well, we could give it a fancy name, $\frac{d}{dx}$, and then its application to $y$ would be something like 
$$
\frac{d(y)}{dx},
$$
i.e., the "d by dx operation, applied to $y$". And that conveniently (if you forget that in this case the parens mean "apply a function" rather than "multiply", morphs into 
$$
\frac{dy}{dx},
$$
The appeal of this notational "pun" is so great that it's been ensconced in mathematics, and has managed to both confuse generations of calculus students and to help many mathematicians rapidly perform computations that would be far messier without it. 
A: This concept will make sense much later if you study differential geometry and or differential topology. You can think of $d/dx$ as a $\mathbb{R}$-values operator on continuous function which obeys the Leibniz rule i.e the product rule. You define the operation by,
$$ \frac{d}{dx}\Bigr|_{x=a} (f) = \frac{d f}{dx}\Bigr|_{x=a}$$
The other property I mentioned (Leibniz) is expressed as,
$$ \frac{d}{dx}\Bigr|_{x=a} (f \cdot g)= f(a) \ \frac{d}{dx}\Bigr|_{x=a}(g) +  g(a) \ \frac{d}{dx}\Bigr|_{x=a}$$
Later on in differential geometry or differential topology, you'll see that we can think of operators with these properties as tangent vectors i.e in the case $f: U \subset \mathbb{R} \to \mathbb{R}$, we would say $d/dx$ spans the tangent space for $\mathbb{R}$. This seems nonsensical because you have too few dimensions here, but the statement means $v= \lambda(v) \ d/dx$ where $v \in \mathbb{R}$.
