This question is several years old, but calculus notation can be confusing to newcomers and I thought a layman's non-technical explanation might be useful for some readers.
First, consider the symbol $d \over dx$. The $d$'s in the symbol just refer to differentiation. Notice that there is an $x$ by the $d$ in the "denominator" but there is nothing beside the $d$ in the "numerator." You can read this symbol as "the derivative of ..." When you see this symbol used, there will be a function (for example, $f(x) = x^2 + 3$) either placed next to the symbol, like this: $ \frac {d}{dx} (x^2+3)$, or placed in the "numerator", like this: $ \frac {d(x^2+3)}{dx}$. Either way, this is simply read "the derivative with respect to $x$ of $x^2+3$". The same thing might be expressed as $d f(x) \over dx$, which may be read as "the derivative with respect to $x$ of $f(x)$".
Obviously, you would want to say something about the derivative with respect to $x$ of the function $f(x)$. So you might complete the mathematical statement like so:
$$\frac {d (x^2+3)}{dx} = 2x$$
This statement would be read as "the derivative with respect to $x$ of $x^2+3$ is $2x$". Or, you could say "the derivative with respect to $x$ of $f(x)$ is $2x$".
You will hear people refer to this symbol ($d \over dx$) as an operator which can be interpreted as saying that "we are taking the derivative of this function", or "the derivative of this function is ..." (whatever that function may be).
So, in response to one of Andrew's questions above (contained in a comment) "how do I know where to put the $d \over dx$?", the answer is, you put it wherever you intend to say "the derivative of ... ". For example, to express mathematically that the derivative of $x^3 - 5x$ is $3x^2-5$, you might write:
$$\frac {d}{dx}(x^3-5x) = 3x^2-5$$
You will commonly see the "prime notation" used in the same way. If the function above was defined as $y = x^2 + 3$ instead of $f(x) = x^2 + 3$, then you would commonly see the following alternative notation:
$$y^{'} = 2x$$
The same thing could be expressed as
$$\frac {d (y)}{dx} = 2x$$
In this case, the notation $y^{'}$ means "the derivative of the function $y$". Here the symbol used does not tell you the "with respect to" variable, but in this case it is safe to assume that it is the derivative with respect to $x$. Similarly, the notation $\frac {d (y)}{dx}$ can be spoken as "the derivative with respect to $x$ of the function $y$". So, whenever you want to express mathematically "the derivative of something" then you can say that by using the $d \over dx$ notation (or $d \over dy$ or $d \over d \theta$ or whatever the case may be) along with the function that will be, or has been, differentiated.
Also, referring to Eoin's answer above where Eoin showed
$$ \frac {d}{dx}(y^2+x)= \frac {d}{dx}(1)$$
Andrew asked, "why is the $d \over dx$ before the $(y^2 + x)$ and before the 1?". Eoin was explaining the process of implicit differentiation and how you must differentiate both sides of the equation. He was simply stating mathematically that "the derivative with respect to $x$ of the function $y^2+x$ is equal to the derivative of the constant value 1." At that point Eoin had not actually taken the derivative of either side of the equation. He was simply showing the step indicating that you must differentiate both sides of the equation. Therefore, his notation says that the derivative of the left side of the equation is equal to the derivative of the right side.
Now consider the symbol $dy \over dx$. As noted above (with parentheses), this symbol refers to "the derivative of $y$ with respect to $x$." It is interesting that in calculus you learn to "solve for functions" just as you have solved for variables in the past.
Here's a quiz. Without knowing what the function $f$ is, tell me its derivative with respect to time. Well, it's $df \over dt$. What is the derivative of function $p$ with respect to variable $r$? It's $dp \over dr$. What's the derivative of $\sigma$ with respect to $\omega$? It's $d \sigma \over d \omega$. You don't know what the functions actually are, but you can express them with a symbol and then, when given more information, solve for them.
For example, given the rate at which the radius of a circle is changing (i.e., the derivative of radius with respect to time, $dr \over dt$), you can calculate the rate at which the area is changing (i.e., the derivative of area with respect to time, $dA \over dt$). Similarly, given the rate at which the area is changing, you can calculate the rate at which the radius is changing. Just find a function relating the two things (area and radius of a circle in this case) and then differentiate (in this case, with respect to time):
$$A = \pi r^2$$
$$\frac {d}{dt}(A) = \frac {d}{dt}(\pi r^2)$$
$$\frac {dA}{dt} = 2 \pi r \frac {dr}{dt}$$
(The first line gives the equation to be differentiated. The second line says that I am going to differentiate both sides. The third line actually performs the differentiation.) In this case, you probably don't yet know the rate of change of the area (i.e., $dA \over dt$), or the rate of change of radius (i.e., $dr \over dt$) but that does not stop you from representing those values with a symbol. You also could simplify the notation somewhat if you chose to do so - something like this, perhaps:
$$A = \pi r^2$$
$$A^{'} = \frac {d}{dt}(\pi r^2)$$
$$A^{'} = 2 \pi r r^{'}$$
Both equations say exactly the same thing, namely, the derivative of the area of the circle with respect to time (i.e., its rate of change) is 2 times $\pi$ times the radius times the derivative of the radius with respect to time (i.e., its rate of change). Given $A^{'}$ or $r^{'}$ you can solve for the other.
I hope that helps clarify some of the notation for students new to calculus.