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I know $d/dx$ means derivative with respect to x (perhaps I am a little unclear on what precisely that means - I'm not quite sure - but I do think I have at least some sense of its meaning). I know (I think!) that... $$\frac{d}{dx} f(x)=f'(x)$$ Here is my question (it's been bugging me incessantly for the last couple weeks): What does $f'(x)$ mean? Does it mean the derivative of $f$ with respect to x? Or, rather, does it mean the derivative of $f$ with respect to x, evaluated at the point x? Or something totally different?

To make my question more clear, let me ask this too: What does $f'(a)$ mean? Does it mean the derivative with respect to $a$? Or the derivative with respect to $x$ evaluated at some point $a$? (so that the "with respect to x" is actually "encoded" in the $f'$ part of the notation!). Etc. Even worse, take $f'(ax)$, which appears in some derivative computation rules. Does that mean derivative with respect to $ax$??

I suspect this confusion may be somewhat related to the maddeningly persistent confusion -- both for me and virtually everyone else -- between a function and its value at a point. Also, teachers tend to use somewhat imprecise notation and language, so I, being someone who likes precision, can sometimes get confused.

While I'm at it, I'll note that this confusion may be related to my additional confusion over language like "the derivative of the sum of two functions," where the two "functions" are, say, $x^2$ and $x^3$. But I thought those were mere polynomials, not functions. One might say $f(x)=x^2$, but one would never say the function itself, $f$, was equal to $x^2$, right? I'm confused. We shouldn't say things like "the function $x^2$," right? :(

Will
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    $f$ is a function, $f'$ is its derivative, $f'(a)$ is the derivative of $f$ evaluated at $a$, i.e. $\lim_{h\to0}(f(a+h)-f(a))/h$. It does not make sense to say "with respect to $x$", because $f$ does not know whether you have called its argument $x$ or $y$ or $a$ or anything else. –  Nov 04 '19 at 06:57
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    Clarification: It does not make sense to say "the derivative of $f$ with respect to $x$". One can say simply "the derivative of $f$", or one can say "the derivative of $f(x)$ with respect to $x$". Then one can make sense of the statement $\frac{\mathrm df(ax)}{\mathrm dx}\ne f'(ax)$, i.e. "the derivative of $f(ax)$ with respect to $x$ is not the same as the derivative of $f$ evaluated at $ax$". –  Nov 04 '19 at 07:03
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    Rahul makes a good point in his second comment. But I would go further and say that "the derivative of $f(x)$ with respect to $x$" is senseless. This question arises, at least partially, because of the misconception that there exists the concept of "function of a variable". If you have a function, @Will, say $f$, it doesn't matter what variable you use. Both $x\mapsto x$ and $t\mapsto t$ mean the same thing. A function is a function, it is not a function of a variable. – Git Gud Nov 04 '19 at 07:18
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    Related questions: 1, 2, 3, 4. – Git Gud Nov 04 '19 at 07:22
  • If you want to understand the history of this big confusion surrounding the word function, you might want to have a look at this, this and this thread. – Michael Bächtold Nov 04 '19 at 07:33
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    An unambiguous alternative I’d like to provide you with: $$\left.\frac{\mathrm df}{\mathrm dx}\right\rvert_{x=t}$$ “The derivative of $f$ with respect to $x$ evaluated at $x=t$. The value associated with this will not have $x$ in it, and the derivative is done around $x$ before ang plugging in and evaluating. Does that make sense? – gen-ℤ ready to perish Nov 04 '19 at 08:15
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    @GitGud: "The derivative of $f(x)$ with respect to $x$" makes perfect sense to me; we are differentiating an expression with respect to a free variable which occurs in it. As Donu Arapura points out in the comments of the excellent MO post Michael Bächtold linked to below, "while professional mathematicians differentiate functions, in basic calculus we differentiate expressions." –  Nov 04 '19 at 09:57
  • @Rahul Interesting comment by Donu there. So their point can apply to my question in that in basic calculus we can "differentiate the expression $x^2$" without worrying too much about the concept of a function as such? We just take the derivative of the expression, essentially? – Will Nov 04 '19 at 15:02

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I'm going to steer clear of high level definitions for functions and instead give you a more intuitive sense for what this notation means. Later on, if you stick with mathematics, you will be exposed to more accurate and rigorous definitions for functions as "mappings" between sets that have special characteristics.

It seems like part of your confusion stems from a lack of understanding as to what a function is. It's helpful to think of a function as some operation that's being defined, and we typically give that operation a name like $f$ or $g$. The notation $f(x) = x^2$ simply means "there exists an operation, called $f$, performed on the variable $x$, and that operation is performed by taking the square of $x$." Another example is $g(x)=2x^3+4$. This is simply notation that means "there exists an operation, called $g$, performed on the variable $x$, and that operation is $2x^3+4$." You are correct in saying $x^2$ and $2x^3+4$ are polynomials. In the notation just used, they are more generally being referred to as operations with specific names, $f$ and $g$, respectively. Any given mathematical expression is not either a function or a polynomial; it can be both.

Now, as for derivatives... the notation $f'(x) = x +4$ simply means "there exists an operation, called $f'$, performed on the variable $x$, and that operation is $x+4$." Now, the addition of the prime symbol $'$ after the $f$ is the convention that is used to let you know that this particular function, or "operation," is actually the derivative of another function, that function being $f$. So in short we say $f'(x)$ is "the derivative of $f$ with respect to $x$."

You also seem to be confused by the phrase "with respect to $x$." Remember that $x$ is simply a variable. It represents some arbitrary number in the domain, but it is NOT any particular number in the domain. So $f'(x)$ meaning "the derivative of $f$ with respect to $x$" is another way of saying "the derivative of $f$ taken at some arbitrary number in the domain of $f$." It is difficult to tell you what $f'(a)$ means without any context, but this is most likely the author or your teacher using the letter $a$ to represent a particular number in the domain of $f$ as opposed to any arbitrary number in the domain of $f$. In other words, $f'(a)$ is "the derivative of $f$ where $x=a$." Some people may still use the words "the derivative of $f$ with respect to $a$," but what they mean is "the derivative of $f$ where $x=a$." As you mentioned already, you could also interpret this as "the derivative of $f$ at the point where $x=a$." All of these are different ways of saying the same essential thing.

If you see $f'(ax)$, this means that the input to the function $f$ is some arbitrary number $x$ multiplied by some particular number $a$.

The notation $\frac{d}{dx}f(x)$ is just another way expressing $f'(x)$. They represent and mean the exact same thing; the reason they are both used is part tradition and part utility. Calculus was developed independently by Newton and Leibniz long ago, and they each developed their own notation that remains in use today, albeit with some changes along the way. Sometimes $f'(x)$ is preferred because it is more compact; at other times $\frac{d}{dx}$ is preferred because it clearly demonstrates that a derivative is the ratio of two infinitesimal quantities, which in some proofs or applications is quite useful.

RyRy the Fly Guy
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    Just a small nitpick: When you say "there exists an operation, called $f$, performed on the variable $x$..." it might give the wrong impression that $f$ "cares" about the variable it operates on. So one might believe that the operation $g$, which performs squaring on the variable $y$ is different from your $f$. More generally the idea of a variable, as used in calculus, was never formalized in modern mathematics. And it might be interesting to note that neither Newton nor Leibniz used the notation $f(x)$ or the modern idea of function. – Michael Bächtold Nov 04 '19 at 08:07
  • @RyRy the Fly Guy Thanks! So, to clarify: The derivative of a function is itself a function, and the derivative function $f'$ is defined without regard to the choice of variable? (That is, you could use any variable?) – Will Nov 04 '19 at 14:29
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    Yes, the derivative of a function is itself a function. And, yes, the choice of variable is arbitrary. In other words, in the function $f(x)$, we represent our variable with the letter $x$, but we could just as easily represent our variable with the letter $t$ or $v$ and call it $f(t)$ or $f(v)$. Sometimes, however, mathematicians use letters to represent particular numbers in the domain and not just any number in the domain. Such would be the case with $f(a)$ described above. Whenever mathematicians do this, they usually say this explicitly so there is no confusion. – RyRy the Fly Guy Nov 04 '19 at 14:49
  • @RyRytheFlyGuy And is $y'(x)$ the same as just $y'$? Sometimes I see both. (I know this is a bit different from my original question but it's related to notation.) Does this have something to do with the (arguably) abuse of notation $y=y(x)$? – Will Nov 04 '19 at 15:00
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    yes, they are the same. When you see $y=y(x)$, it means "there exists a variable $y$ whose value is equal to the result of an operation, also called $y$, performed on the variable $x$." And $y'=y'(x)$ has the same analogous meaning. This repetitive use of the letter $y$ is sometimes useful when you are dealing with multiple functions that determine the outputs of multiple variables and you want to keep track of which goes with which. – RyRy the Fly Guy Nov 04 '19 at 15:18
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    For example, if I'm given $y=f(x)$ and $z=g(x)$, I have to remember "the operation $f$ determines the variable $y$, and the operation $g$ determines the variable $z$." But if instead I write $y=y(x)$ and $z=z(x)$, now all I have to remember is "the operation $y$ determines the variable $y$, and the operation $z$ determines the variable $z$." Remembering that "$y$ goes with $y$" and "$z$ goes with $z$" is just easier. I suspect you are taking calculus I and they are exposing you to this kind of notation so that you are familiar with it. – RyRy the Fly Guy Nov 04 '19 at 15:22
  • @RyRytheFlyGuy One more thing I just thought of: If we have $d/du[u^2]$ I am informed that equals $2u$. Is that an example of how changing the variable name doesn't change the underlying function? Still trying to grapple with how you can take derivatives with respect to $u$, particularly for purposes of the chain rule $dy/dx = dy/du * du/dx$. – Will Nov 05 '19 at 06:43
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    @Will Yes, that is yet another example of how changing the variable does not change the underlying function. I think you should submit the chain rule question as another post. It's a bit much to reply to in comments. In short, when you see $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$, they are showing you how to find the derivative of a composite function $(y \circ u)(x)$. – RyRy the Fly Guy Nov 05 '19 at 13:55
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When declaring a function we need to specify 3 things: a domain, a codomain and a mapping rule. In particular, when $f$ takes only one variable as input, we declare it as

$$f\colon A\to B, \quad f(x)=\text{some expression in }x.$$

The symbol $f(x)$ represents an element of $B$, but not the function itself. The symbol $f$ is the function. The symbol $x$ is irrelevant here, we could have used $\diamond$ if we wanted to.

We can obviously make functions that take functions as input and produce functions as output. Such functions are sometimes called operators; the derivative is an example of an operator, called the differential operator, and both $d/dx$ and $'$ is what we use to denote it. The fact that $x$ appears here is again irrelevant, we could have used $\frac{d}{d\diamond}$ if it was clear from the context that we use $\diamond$ to represent the indeterminate.

As $d/dx$ is a funciton, it has a domain (of which e.g. real valued differentiable functions in one variable are elements of) and a codomain. So if $f$ is in its domain, then $\frac{d}{dx}f$ is the corresponding element in the codomain, i.e. the output is a function in the codomain, and not a function in the codomain evaluated at some point. Since $'$ represents the same operator, $f'$ represents the same element in the codomain as $\frac{d}{dx}f$.

So what is $\frac{d}{dx}f(x)$? It is $\left[\frac{d}{dx}f\right](x)$, i.e. the function $\frac{d}{dx}f$ evaluated at $x$. Similarly, $f'(x)$ is the function $f'$ evaluated at $x$.

Then, what does $f'(a)$ mean? The same thing as $f'(x)$, except we replace $x$ by $a$. But be careful: $f(a)'$ doesn't mean the same thing.

About you other concern: first, polynomials are functions. Communicating mathematics can be very different to doing mathematics. When we say that a function is $x^2$, it is implicitly understood that we are talking about the square function, i.e. the real-valued (also, possibly complex-valued) function $f$ in one real (or complex) variable such that $x\mapsto x^2$. But after a while it gets really tedious to keep repeating all of that, when we all know that "square function" or "the function $x^2$ is to be understood as the same thing.

Randy Marsh
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