1

Related to but distinct from a question I asked here earlier today:

What's the difference between $f′(ax)$ and $[f(ax)]′$ ? That is, why aren't they the same thing?

I know they can't be the same because I know $[f(ax)]′ = a *f′(ax)$. But I still don't quite get why they're different, at a fundamental, theoretical level.

Perhaps my understanding would be helped by letting, say, $y=g(x)=f(ax)$ and $y=h(x)=f'(ax)$.

Part of the problem, I suspect, is that transformations have always confused me a bit (e.g., I remember always asking my teacher whether horizontal transformations created a new function or rather merely altered the inputs to the same function... Still a bit confused about that.)

Will
  • 820
  • 1
    $f(a\cdot)$ changes much quicker than $f(\cdot)$. Changing $x$ to $x+\epsilon$ in the linear map $x \mapsto ax$ results in a change of $a\epsilon$, whereas changing $x$ to $x+\epsilon$ in $x \mapsto x$ results in a change of $\epsilon$. $[f(ax)]'(x_0)$ is the rate of change of $x \mapsto f(ax)$ at the point $x_0$, whereas $f'(ax_0)$ refers to the rate of change of $x \mapsto f(x)$ at the point $ax_0$. They are completely different things. – mathworker21 Sep 25 '19 at 02:17
  • 1
    $f'(ax)$ is the derivative of $f$ evaluated at $ax$, while $[f(ax)]' = a f'(ax)$ – rims Sep 25 '19 at 02:21
  • 1
    One issue is that the notation is not helpful to understanding. The notation $f'$ is well defined. The notation $(f(x))'$ is well understood, but misleading since really it means $\phi'$ where $\phi(x) = f(\alpha x)$. Even $(x \mapsto f(\alpha x))'$ is better (for me). – copper.hat Sep 25 '19 at 02:25

3 Answers3

2

Consider $$f(x)=x^2$$

We have $$f'(x) = 2x $$ and $$f(5x) = 25x^2$$

Note that $$f'(5x) = 2(5x)=10x$$ while

$$(f(5x))' = (25x^2)'= 50x$$

Mohammad Riazi-Kermani
  • 68,728
2

Perhaps it's hard to tell because some details haven't been written out. What are we differnetiating by in $[f(ax)]'$? What does the prime outside mean? If we discard the prime notation, we see that $[f(ax)]'$ means the derivative of $f(ax)$ with respect to $x$. However, for $f'(ax)$ means the derivative of $f(ax)$ with respect to $ax$.

cxx
  • 524
  • Wait, what does it mean for a derivative to be "of $f(ax)$," and "with respect to $ax$"? I'm confused! – Will Sep 25 '19 at 21:39
2

Start with some function $f$ (no need to name the variable). Then the derivative of $f$ is $f'$. If $a$ and $x$ are numbers then so is $ax$ and $f'(ax)$ is just the value of $f'$ at input $ax$.

I think the real problem is that the second expression is confusing. In your question you seem to suggest that you know that.

To make sense of $[f(ax)]'$, first define a new function $g$ by declaring that for every $x$, $$ g(x) = f(ax). $$ Then $g'$ is the derivative of $g$. The chain rule tells you that $$ g'(x) = af'(ax). $$

For help with what scaling the axis means geometrically, see Mechanics of Horizontal Stretching and Shrinking

Ethan Bolker
  • 95,224
  • 7
  • 108
  • 199
  • You write "if a and x are numbers." But isn't x a variable? Why can we treat it as an ordinary number? – Will Sep 25 '19 at 02:39
  • ... So then isn't $f'(ax)$ also a function, assuming you choose a value for a, which you could call, say, $h(x)$ and graph as, e.g. for a being 2, $y=h(x)=f'(2x)$? – Will Sep 25 '19 at 02:42
  • 1
    @Will You should try to learn to think about functions independent of "variables". In your example, $f$ is a function. THen $f'$ is another function, related to $f$: it's the function that describes the slope of the graph of $f$. When you write $f(\text{something})$ you mean the result of evaluating $f$ at something. So $f'(ax)$ is the slope of $f$ at the point $ax$. It does not mean "differentiate the function whose value at $x$ is $f(ax)$", – Ethan Bolker Sep 25 '19 at 18:15
  • Incisive point - I'll need to mull on that a bit. So, to clarify, is $[f'(ax)]'$ "the derivative with respect to $x$ of the function whose value at $x$ is $f(ax)$?" Whereas $f'(ax)$ simply means the derivative of $f$ with respect to $x$, evaluated at the point $ax$? – Will Sep 25 '19 at 21:37
  • 1
    @Will I would never write $[f(ax)]'$ (or $[f'(ax)]' $ as in your comment) because it is confusing. Better to let $g(x) = f(ax)$ and then use the chain rule to write $g'(x) = af'(ax)$. – Ethan Bolker Sep 25 '19 at 21:45