Consider the function $f(x)$ and let $g(x)=f(cx)$.
By the definition of derivative
$$f'(x)=\frac{df(x)}{dx}=\lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h}\tag{1}$$
and so the definition of $f'(cx)$ is
$$f'(cx)=\frac{df(cx)}{d(cx)}=\lim\limits_{h \to 0} \frac{f(cx+h)-f(cx)}{h}\tag{2}$$
Also
$$g'(x) = \lim\limits_{h \to 0} \frac{g(x+h)-g(x)}{h}$$
$$=\lim\limits_{h \to 0} \frac{f(c(x+h))-f(cx)}{h}$$
$$=\lim\limits_{h \to 0} \frac{f(cx+ch)-f(cx)}{h}$$
$$=c\lim\limits_{h \to 0} \frac{f(cx+ch)-f(cx)}{ch}$$
$$=cf'(cx)$$
Hence
$$g'(x)=cf'(cx)$$
I am not sure if I am seeing an inexistent ambiguity, but $f'(cx)$ seems like ambiguous notation.
As defined above in $(2)$, $f'(cx)$ means the derivative of $f$ relative to $cx$, evaluated at a point we call $cx$, .
Note that this is different than the derivative of $f$ relative to $x$ evaluated at a point $cx$: this derivative is $g'(x)=\frac{df(cx)}{dx}$. In "prime" notation, how do we denote this latter derivative? It would seem to be $f'(cx)$, but I think either this is incorrect, or the definition given in $(2)$ is somehow non-standard.
For example, let $$f(x)=3x^3$$ $$g(x)=f(cx)=3c^3x^3$$
Then
$$f'(x)=\frac{df(x)}{dx}=9x^2$$
$$\left.\frac{df(x)}{dx}\right \vert_{x=cx}=9c^2x^2\ \ (=f'(cx)???)$$
$$g'(x)=\frac{df(cx)}{dx}=c\frac{df(cx)}{d(cx)}=c\cdot f'(cx)=c\cdot 9c^2x^2= 9c^3x^2\tag{3}$$
In this example, $f'(cx)=\frac{df(cx)}{d(cx)}=9c^2x^2$ according to definition I gave in $(2)$.
But perhaps more intuitively, it could also be $f'(cx)= \left.\frac{df(x)}{dx}\right \vert_{x=cx}=9c^2x^2$
Note that both uses of $f'(cx)$ lead to the same result in this example.
Which use of $f'(cx)$ is the "correct" or "standard" one?
Best Answer
This is definitely potentially confusing, but not an ambiguity.
The object written $f$ is a function, which takes an input and gives an output. The object written $f'$ is also a function, which is defined using the function $f$. If $f$ is the function which squares its input (usually written like $f(x)=x^2$), then $f'$ is the function which doubles its input (usually written like $f'(x)=2x$). It's also true that $f'(cx)=2cx$, because $f'(cx)$ does not mean "the derivative of $f(cx)$," it means "the function $f'$ evaluated at an input of $cx$."
Your expression (2) is a correct application of the definition (1), although it would be odd to call it a definition itself. I do not know of a way to write "the derivative of $f(cx)$ as a function of $x$" in prime notation, besides by writing a new function $g(x)$ defined by the rule $g(x)=f(cx)$, from which "the derivative of $f(cx)$ as a function of $x$" is $g'(x)$.