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

$$f'(cx)=\frac{df(cx)}{d(cx)}=\lim\limits_{h \to 0} \frac{f(cx+h)-f(cx)}{h}\tag{2}$$


$$g'(x) = \lim\limits_{h \to 0} \frac{g(x+h)-g(x)}{h}$$ $$=\lim\limits_{h \to 0} \frac{f(c(x+h))-f(cx)}{h}$$ $$=\lim\limits_{h \to 0} \frac{f(cx+ch)-f(cx)}{h}$$ $$=c\lim\limits_{h \to 0} \frac{f(cx+ch)-f(cx)}{ch}$$ $$=cf'(cx)$$



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 $$f(x)=3x^3$$ $$g(x)=f(cx)=3c^3x^3$$


$$f'(x)=\frac{df(x)}{dx}=9x^2$$ $$\left.\frac{df(x)}{dx}\right \vert_{x=cx}=9c^2x^2\ \ (=f'(cx)???)$$ $$g'(x)=\frac{df(cx)}{dx}=c\frac{df(cx)}{d(cx)}=c\cdot f'(cx)=c\cdot 9c^2x^2= 9c^3x^2\tag{3}$$

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?

  • 3
    $\begingroup$ $f'(cx)$ means $f'$ evaluated at $cx$. It is not equal to $g'(x)$. $\endgroup$ Apr 3, 2022 at 19:48
  • $\begingroup$ As I derived in my question, $g'(x)=cf'(cx)$. However, the definition of $f'(cx)$ is given in $(2)$. Is the latter definition correct/standard? $\endgroup$
    – evianpring
    Apr 3, 2022 at 19:50

3 Answers 3


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 $g'(x)$.

  • $\begingroup$ How do you conciliate your explanation with my definition $(2)$? Is my definition non-standard? According to that definition, and using your example, we'd have: $f(x)=x^2$, $f(cx)=(cx)^2$, $f'(cx)=2cx$. $\endgroup$
    – evianpring
    Apr 3, 2022 at 19:49
  • $\begingroup$ @evianpring I've updated my answer. The expression in (2) follows logically from the definition of the function $f'$ from (1). $\endgroup$ Apr 3, 2022 at 19:55
  • $\begingroup$ Is $\left . \frac{df(x)}{dx} \right \vert_{x=cx}=\frac{df(cx)}{d(cx)}$ a true statement for all x? The left side means "the derivative function $f'$ evaluated at a point $cx$, ie the value of the slope of $f$ at $cx$", the right side means "the derivative of the function $f$ at a point $cx$, ie the slope of of $f$ at $cx$". Writing it out like this, it would seem this is true. $\endgroup$
    – evianpring
    Apr 3, 2022 at 20:00
  • 1
    $\begingroup$ Actually I think the root of the confusion is the expression $g'(x)=\frac{df(cx)}{dx}$. We can't actually write $f'(cx)$ here, because as you said, there is no prime notation for "the derivative of $f(cx)$ relative to $x$". $\endgroup$
    – evianpring
    Apr 3, 2022 at 20:09
  • 1
    $\begingroup$ @evianpring Yes, $\frac{df(x)}{dx}\bigg|_{x=cx}=\frac{df(cx)}{d(cx)}$. $\endgroup$ Apr 3, 2022 at 20:16

I just want to emphasize one thing. You don't take the derivative of a function "relative to something". The derivative of a function $f:\mathbb R \to \mathbb R$ is always taken "with respect to" one thing and one thing only: its entry.

If you want to see how the quantity $f(cx)$ varies when $x$ varies, then you should think of it as defining a new function $g$ by $g(x)=f(cx)$ (because you want to vary $x$) and then differentiating this new function $g$.

My main point is that you have one operation that acts on functions and spits out functions: the derivative. It takes $f$ and gives you $f'$ back. It takes $g$ and gives you $g'$ back. After that, you can evaluate these functions wherever you want. The relation that you correctly found

$$g'(x) = cf'(cx)$$

reads like this: when you evaluate the function $g'$ at the point $x$, you obtain the same number as if you were to evaluate the function $f'$ at the point $cx$ and multiply the result by $c$.

That's it. No "with respect to"s involved.


This is just a supplementary part to the nice answer of @KarlSchildkraut. I think it is helpful to consider the complete definition of functions and look at their relationship somewhat more detailed.

The setting more detailed:

We start with a differentiable function \begin{align*} &f:\mathbb{R}\to\mathbb{R}\\ &x\mapsto f(x) \end{align*}

When we now consider \begin{align*} g(x)=f(cx) \end{align*} we have a function $h$ \begin{align*} &h:\mathbb{R}\to\mathbb{R}\\ &x\mapsto h(x)=cx\\ \end{align*} and a function $g$ \begin{align*} &g:\mathbb{R}\to\mathbb{R}\\ &x\mapsto g(x)=\left(f\circ h\right)(x)\\ \end{align*} which is the composition of $f$ with $h$. It follows according to the definition of $h$ and $g$ : \begin{align*} g(x)=\left(f\circ h\right)(x)=f(h(x))=f(cx) \end{align*}

Differentiation: Two different situations

As @KarlSchildkraut said there is no ambiguity, but we have different situations instead. On the one hand we have

\begin{align*} \color{blue}{f^{\prime}(cx)=f^{\prime}(u)\big|_{u=cx}} \end{align*} which is the differentiated function $f$ evaluated at a point $u=cx$. On the other hand we have \begin{align*} \color{blue}{g^{\prime}(x)}&\color{blue}{=\left(f\circ h\right)^{\prime}(x)}=\left(f(h(x)\right)^{\prime}=\left(f(cx)\right)^{\prime}\\ &=f^{\prime}(h(x))h^{\prime}(x)\\ &\,\,\color{blue}{=f^{\prime}(cx)c} \end{align*} which is the differentiation of a composition of functions. So, we have two different situations.

  • $\begingroup$ And just to make it crystal clear, after the last step we could write $=f'(u) \vert_{u=cx}c$, correct? $\endgroup$
    – evianpring
    Apr 7, 2022 at 22:04
  • $\begingroup$ @evianpring: Yes, this is correct. $\endgroup$ Apr 7, 2022 at 22:10

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