What is a constant? The word "constant" is used in such expressions as "The derivative of a constant is $0$."  What does it mean?
(I will post my own answer here, but I'm sure lots of others can have fun with their own answers.)
PS: I'd have thought this was obvious, but maybe not, judging by some comments: The phrase "The derivative of a constant is $0$" is Only an example, not the topic of the question.
 A: "Constant" always means "not depending on something" but what the "something" is depends on the context.  When one says that
$$
\text{If $c$ is a constant, then }\frac{d}{dx} (cf(x)) = c\frac{d}{dx} f(x),
$$
then "constant" means not depending on $x$, i.e. $c$ does not change as $x$ changes.
Example:
If one wants to prove that $\dfrac{d}{dx} 2^x = (2^x\cdot\text{constant})$, one may write
\begin{align}
\frac{d}{dx} 2^x & = \lim_{h\to0}\frac{2^{x+h}-2^x}{h} = \lim_{h\to0}\left(2^x\cdot \frac{2^h-1}{h}\right) \\[12pt]
& = 2^x\cdot\lim_{h\to0}\frac{2^h-1}{h} \text{ since $2^x$ is a constant} \\[12pt]
& = (2^x\cdot\text{constant}) \text{ since the limit is a constant}
\end{align}


*

*In the first "since", the word "constant" means not depending on $h$.  The factor $2^x$ does not change as $h$ goes to $0$.

*In the second "since", the word "constant" means not depending on $x$.  This constant remains the same as $x$ changes.


A moral: A leisurely account of what something is held not to depend on, sometimes as long as a whole sentence or maybe even two, can be worth calling the audience's attention to.  It may what enables someone to understand something.  One who hears only that something is a "constant" may miss the essential point.
Example: Suppose one want to prove that every function holomoprhic at a point $z_0$ in $\mathbb C$, i.e. complex-differentiable in some open neighborhood of $z_0$, is expressible as a convergent power series near $z_0$ ("Holomorphic functions are analytic."). One may start with something like
$$
f(z) = \frac{1}{2\pi i}\int_\gamma \frac{f(w)}{w-z}\, dw
$$
and after some algebra and analysis one gets it to
$$
\sum_{n=0}^\infty (z-z_0)^n \underbrace{\frac{1}{2\pi i}\int_\gamma \frac{f(w)}{(w-z_0)^{n+1}} \, dw}
$$
The crucial thing to know about the expression over the $\underbrace{\text{underbrace}}$ is that the variable $z$ does not appear in it, i.e. it is "constant" as a function of $z$.  It is not constant as a function of $n$, but that is not what matters here.
A: A constant is one of these:


*

*a literal object such as the scalar $3$ or the vector $\left[\begin{array}{ccc}0 & 0 & 0\end{array}\right]$.

*a symbol denoting just one object, rather than a variable. For instance $\pi$.

*a function (possibly multi-variable) which maps all domain values to the same object, essentially ignoring its arguments and producing a constant as its value

*a formula consisting only of operations on constants

*a multi-variable function in which some arguments are constrained to be constants, in some context, and whose value does not vary with the remaining arguments. (Sometimes constrained arguments are called parameters, and there is a semicolon notation for writing them).

*a numeric approximation to some unchanging quantity found in nature, such as Planck's constant.

*any one of the solutions to an equation, or the set itself. The constant $2$ can also be expressed as "the solution to $2x - 4 = 0$". (Solutions to equations are not always constants, of course).

A: A constant is 0-ary function into some codomain.  
It is often better to talk about a "constant in X" to specify which codomain X is being discussed, as that often determines it's properties with respect to the rest of an expression.  A constant is often represented by the element of the codomain that the 0-ary function evaluates to.
A: Hint $\ $ TFAE for any operator $\rm\:f \to f'$ satisfying the Product Rule $\rm\:(fg)' = f'g + fg'$
$\rm(1)\quad (cf)'\! =\, cf'\ $ for all $\rm\:f$
$\rm(2)\quad c'\! = 0$
Proof $\ \ (1\Rightarrow 2)\ \ $ Putting $\rm\:f=1\:$ in $\,(1)\,$ gives $\rm\: c' = c 1' = 0\ $ by the Lemma below.
$\rm(2\Rightarrow 1)\ \  $ By the Product Rule $\rm\: (cf)'\! = c'f + cf' = cf'\ $ by $\rm\:c' = 0$.
Lemma $\rm\ \ 1' = 0.\ \ $ Proof $\ $ By the Product Rule$\rm\ \  (1 = 1\cdot 1)' \Rightarrow 1' = 1'+1'\Rightarrow\ 1' = 0.$
