meaning of the notation f'(-x)

Question

What does $f'(-x)$ essentially mean?

1. $\frac{df(-x)}{dx}$, or

2. $\frac{df(x)}{d(-x)}$, or

3. $\frac{df(x)}{dx}|_{x=-x}$ ?

I am not sure if all the options are different, though! :)

EDIT 1:

Let me explain what I mean by all the options given. In all the cases, $f(x)$ is the original function, say $f(x)=x^3+2x$. Please note that many parts of the explanation are obvious, still I have mentioned them just to avoid any confusion.

1. $\frac{df(-x)}{dx}$ : First we put $-x$ in place of $x$ in $f(x)$ to get $f(-x)=-x^3-2x$. Then we differentiate $f(-x)$ w.r.t $x$ and we get $-3x^2-2$

2. $\frac{df(x)}{d(-x)}$ : This means differentiating $f(x)$ w.r.t. $-x$, so applying the chain rule we get the value as $\frac{df(x)}{dx} \times \frac{dx}{dz} = (3x^2+2) \times (-1) = -3x^2-2$, where $z=-x$ and hence $\frac{dx}{dz}=-1$.

3. $\frac{df(x)}{dx}|_{x=-x}$ : First we differentiate $f(x)$ w.r.t $x$ and in the derivative we put $-x$ in place of $x$ to obtain the result $3x^2+2$.

Another option may be added, though it is not a strong candidate in this case.

4. $\frac{df(-x)}{d(-x)}$ : Which means differentiating $f(-x)$ with respect to $-x$, and we get the result $3x^2+2$.

Now I present a context where $f'(-x)$ is relevant. It's basically the question:

Show that the derivative of an odd function is even (or vice-versa).

The question is discussed here, here and at many other places. In all the answers, to prove the statement, it is shown that $f'(-x)=-f'(x)$. From the question what I interpret is that we have to show $\frac{df(x)}{dx}|_{x=-x} = \frac{df(x)}{dx}$ or in another notation, $f'(x)|_{x=-x} = f'(x)$. While the solutions seem to involve differentiation with respect to $-x$, thus, in my opinion, do not actually prove the statement at hand.

Note: I have changed the title slightly to reflect my question more accurately.

EDIT 2:

David K's vivid explanation cleared my doubt. Here are the conclusions I have come to regarding the notations of concern: (please let me know if there is any flaw)

(i) $f'(g(t))=f'(u)|_{u=g(t)}=\frac{d}{dz}(f(z))|_{z=g(t)}=\lim\limits_{h\to 0} \frac{f(g(t)+h)-f(g(t))}{h}$ [where $h$ is a small change in $g(t)$], hence $f'(x)$ does not necessarily mean that we have to find $f'$ by differentiating $f$ with respect to $x$. It just means that represent $f$ in the form of any 'dummy' variable (say $v$), differentiate it w.r.t. $v$ to get $f'$ and substitute $x$ for $v$ in $f'$ to get $f'(x)$.

(ii) $f'(-x)=\frac{d}{d(-x)}(f(-x))=\frac{d}{dx}(f(x))|_{x=-x}$, hence option 3 and option 4 in my question are actually same and both are correct answer to my original question.

Answer 1

Here's how I would interpret it:

$f$ is a function that takes numbers to numbers. The meaning of $f$ is independent of what you apply it to, as long as you apply it to a number.

$f'$ is the derivative of $f$, that is, $f'$ is a function from numbers to numbers. The meaning of $f'$ also is independent of what you apply it to.

$f'(-x)$ is the function $f'$ applied to $-x$.

Hence $$f'(-x) = \left.\frac{df(u)}{du}\right|_{u=-x}. \tag{1}$$

Update:

Looking at the context in which this notation was found, I am confident that the meaning above is what was intended. The difficulty of applying this definition in the various proofs that the derivative of an odd function is even (or that the derivative of an even function is odd) is not that this definition of the notation contradicts anything in those proofs, but that there are so many sign changes and other notations involved that it is hard to keep account of all of them.

The key idea in those proofs is generally that idea that you can write $f(-x)$ as $f(g(x))$ where you have defined the function $g$ by the equation $g(x) = -x$. The proofs also use the chain rule, for which a general formula is

$$ \left.\left(\frac{d}{du} (f(g(u)))\right) \right|_{u=x} = \left.\left(\frac{d}{du}(f(u))\right)\right|_{u=g(x)} \cdot \left.\left( \frac{d}{dv} (g(v)) \right)\right|_{v=x}. \tag{2}$$

Usually this is written in more compact notation, for example, $$ \frac{d}{dx} f(g(x)) = f'(g(x)) g'(x), $$ but in Equation $(2)$ I chose to spell out the various parts of the formula in a much more detailed way, with extra parentheses added to try to avoid any possibility of applying the functions and operators in the wrong order.

For either the "$f'$ is even" or "$f'$ is odd" theorems, we are given a rule to find $f(-x)$ in terms of $f(x)$, and we need to show how to express $f'(-x)$ in terms of $f'(x)$. Importantly, when we say that $f'$ is odd (for example), we are talking about a function named $f'$ (which happens to be related to another function named $f$), which we then evaluate on some input number in parentheses, as in Equation $(1)$.

Example: let $f(x) = x^3 + 2x$, that is, $f$ is the function $f : x \mapsto x^3 + 2x$, or equivalently (since we can use any variable name in that definition), $f : t \mapsto t^3 + 2t$. Then $f'(x) = 3x^2 + 2$, that is, $f'$ is the function $f' : t \mapsto 3t^2 + 2$. Therefore $f'(-x) = 3(-x)^2 + 2 = 3x^2 + 2.$ This implies $f'(-x) = f'(x)$, that is, $f'$ is an even function.

The general proof that the derivative of an odd function is even can be explicated as follows. Let $f$ be an odd function, that is, $f(-x) = -f(x)$ for all $x$. Define a function $g$ such that $g(x) = -x$, so that $\frac{d}{dx}g(x) = -1$ and $$f(x) = -f(-x) = -f(g(x)). \tag{3}$$ Then $$\frac{d}{dx}(f(x)) = \frac{d}{dx}(-f(g(x))),$$ and we can simplify both sides of this equation to get $$f'(x) = -\frac{d}{dx}(f(g(x))). \tag{4}$$ But the chain rule says that

\begin{align} \frac{d}{dx} (f(g(x))) = \left.\left(\frac{d}{du} (f(g(u)))\right) \right|_{u=x} &=\left(\left.\frac{d}{du}(f(u))\right|_{u=g(x)}\right) \cdot \left.\left( \frac{d}{dv} (g(v)) \right)\right|_{v=x} \\ &=\left(\left.\frac{d}{du}(f(u))\right|_{u=-x}\right) \cdot \left.\left( -1 \right)\right|_{v=x} \\ &= f'(-x) \cdot (-1) \\ &= -f'(-x). \tag{5} \end{align}

Plug this in at the far right-hand end of Equation $(4)$, and we find that $$f'(x) = -(-f'(-x)), $$ that is, $f'(x) = f'(-x)$, showing that $f'$ is an even function.

If you follow the (more compactly written) proofs in this answer or in the "official" solution (according to this question) carefully, you should find that they agree at every step with the facts above. It's unfortunately easy to lose track of a sign change somewhere along the way; note that there is one "sign change" in Equation $(3)$ due to $f$ being odd, and another "sign change" in Equation $(5)$ due to the chain rule and the fact that $g'(x) = -1$.

Notice what happens with $\frac{d}{dx} (f(-x))$ when $f$ is odd. Since $f(-x) = -f(x)$, we find that $$\frac{d}{dx} (f(-x)) = \frac{d}{dx} (-f(x)) = -\frac{d}{dx} (f(x)) = -f'(x),$$ which does not look at all like what we want to prove; we need $f'(-x) = f'(x)$, not $-f'(x)$, to show that $f'$ is an even function.

The expression $\frac{d}{d(-x)} (f(x))$ is just as bad, because if we set $t = -x$ we have $$\frac{d}{d(-x)} (f(x)) = \frac{d}{dt} (f(-t))$$ which is just $\frac{d}{dx} (f(-x))$ using the variable name $t$ instead of $x$.