I know how to derive, I know how to integrate. I know what to do when I see $\frac{\mathrm{d}}{\mathrm{d}x}$ and such but what does it really mean? I know it means something like derive in terms of $x$, but whats the difference between $\frac{\mathrm{d}y}{\mathrm{d}x}$ and $\frac{\mathrm{d}}{\mathrm{d}x}$?
If someone could give me an explanation in layman's terms that would be very helpful as this has always perplexed me.
Basically, what's that $\mathrm{d}$ mean?
If you have a function $f$ with the independent variable $x$, then $$ \frac{d}{dx} f(x) $$ means the derivative of $f$ with respect to $x$. We also sometimes write this as $f'(x)$. Now if you have a function like $$ f(x) = ax, $$ then the derivative is $$ f'(x) = a. $$ This is clear because in writing $f(x)$ we have indicated that the function $f$ is a function of the variable $x$. If I instead had told you that $$ y = ax $$ and I just asked you to find the derivative, what would you do? You would probably, again, just say that the derivative is $a$. But in this situation it actually isn't clear what is a variable and what is a constant. And therefore we can write $$ \frac{dy}{dx} \quad\text{or}\quad \frac{d}{dx}y $$ to indicate that we are considering $y$ as a function of the variable $x$ and we are considering $a$ as a constant (In the case of multivariable functions we really should be taking about partial derivatives in this case). Now you could also write $$ \frac{dy}{da} $$ and in this case you are saying that $a$ is a variable. So the $\frac{d}{dx}$ notation is very helpful when you have expressions where there are several letters.
So what is $\frac{d}{dx}$? You can consider this as an operator that takes as "input" a (differentiable) function and "outputs" a function.
$\dfrac{dy}{dx} = \dfrac{d(y)}{dx} = \dfrac{d}{dx}(y)$.
$\dfrac{d}{dx}$ is the differential operator. It tells you what operation (differentiation) you are doing and with respect to what variable. $\dfrac{dy}{dx}$ is the actual derivative of a function $y$ with respect to $x$. The operator can be applied to any function, for example $\dfrac{d}{dx}(x^2) = 2x$. If we wrote the same thing with $\dfrac{dy}{dx}$, then we get $\dfrac{dy}{dx}(x^2) = x^2\dfrac{dy}{dx} = x^2y'$. Since $\dfrac{dy}{dx}$ is already differentiating the function $y$, it does not do anything to $x^2$. The former example shows the differential operator being applied to a function. The latter shows the derivative of $y$ being multiplied by a function. The $x^2y'$ expression is often used in differential equations as shorthand to replace the longer to write $\dfrac{dy}{dx}$.
$\dfrac{d}{dx}$ is what analysts would call an operator, meaning that you give it an element of a certain vector space and it gives you another element in that vector space. So then what is a vector space? Well a vector space is any collection of objects that satisfy certain axioms (like you can add two of of these objects together to get another object in the collection, you can multiply an object with a number to get another object in the collection, every object has a negative, etc). Vector spaces are in some way a generalization of the real numbers.
With that out of the way, how does this relate to $\dfrac{d}{dx}$? Well, you can look at $\dfrac{d}{dx}f$ for some function $f$. This is nothing more than the derivative of $f$. Particularly, there is a very nice vector space of functions called the "smooth" functions. These functions are infinitely differentiable and every derivative is continuous (though this is a bit redundant to say). Examples of such functions are polynomials (but these are not all of the smooth functions). Another would be $\arctan(x)$. If you differentiate a smooth function, you get another smooth function! So then if we denote the smooth functions by the symbol $C^{\infty}(\mathbb{R})$, we have that $\dfrac{d}{dx}:C^{\infty}(\mathbb{R})\rightarrow C^{\infty}(\mathbb{R})$.
d/dx means the rate of change of a variable (physical quantity) when 'x' changes. To differentiate is to find the changes. This is the physical significance of differentiation. if $${\delta/\delta x}$$ is used in place of d/dx it means the partial differentiation. In simple words the infinitesimally small rate of changes.
I found this to be incredibly useful (9 years later it reminded me of this old post). It is from an old textbook explaining this. Calculus Made Easy by Silvanus P. Thompson. The entire chapter is available online but I have copied it below (so it is not a "link-only answer"). The book is in the public domain.
The preliminary terror, which chokes off most fifth-form boys from even attempting to learn how to calculate, can be abolished once for all by simply stating what is the meaning–in common-sense terms–of the two principal symbols that are used in calculating.
These dreadful symbols are:
d which merely means "a little bit of."
Thus dx means a little bit of x; or du means a little bit of u. Ordinary mathematicians think it more polite to say "an element of," instead of "a little bit of." Just as you please. But you will find that these little bits (or elements) may be considered to be indefinitely small.
∫ which is merely a long S, and may be called (if you like) "the sum of."
Thus ∫dx means the sum of all the little bits of x; or ∫dt means the sum of all the little bits of t. Ordinary mathematicians call this symbol "the integral of." Now any fool can see that if x is considered as made up of a lot of little bits, each of which is called dx, if you add them all up together you get the sum of all the dx's, (which is the same thing as the whole of x). The word "integral" simply means "the whole." If you think of the duration of time for one hour, you may (if you like) think of it as cut up into 3600 little bits called seconds. The whole of the 3600 little bits added up together make one hour.
When you see an expression that begins with this terrifying symbol, you will henceforth know that it is put there merely to give you instructions that you are now to perform the operation (if you can) of totalling up all the little bits that are indicated by the symbols that follow.
That's all.
