$f'(ax)$ vs. $[f(ax)]'$ - what's the difference?
Andrew Henderson 
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.)
$\endgroup$ 33 Answers
$\begingroup$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
$\endgroup$ 5 $\begingroup$Consider $$f(x)=x^2$$
We have $$f'(x) = 2x $$ and $$f(5x) = 25x^2$$
Note that $$f'(5x) = 2(5x)=10x$$ while
$$(f(5x))' = (25x^2)'= 50x$$
$\endgroup$ $\begingroup$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 $ax$.
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