I was not sure how to word this question, but I'm wondering how to evaluate $$\frac{dy}{d(ax)}$$
where $y=f(x)$ and $a$ is a constant.
I was thinking I could write $y=f(ax/a)$ and rename $u=ax$, so $y=f(u/a)$. Then $$ \frac{dy}{d(ax)}=\frac{dy}{du}=\frac{1}{a}f'(u/a)=\frac{1}{a}f'(x) $$
This suggests a simple "hack" by evaluating the differential $d(ax)=adx$, so
$$ \frac{dy}{d(ax)}=\frac{dy}{adx}=\frac{1}{a}f'(x) $$
I have a few questions:
Is there a name for this type of problem, where the you're taking the derivative with respect to a scaled or different independent variable? Is it just a form of variable subsitution?
Under what conditions can you simply evaluate the differential in the "denominator" of the derivative?
EDIT
Writing this problem as a composition of functions.
Let $u(x)=ax$ and $h(x)=x/a$. Then, we can write
\begin{align*} \frac{dy}{d(ax)}&=(f\circ h)'(u)\\ &=f'(h(u))h'(u)\\ &=f'(x)\frac{1}{a} \end{align*}
I thought that perhaps I could use $(f\circ h\circ u)$, but it just gives me the original function.
\begin{align*} \frac{dy}{d(ax)}&\neq(f\circ (h \circ u))'(x)\\ &=f'(h(u(x))h'(u(x))u'(x)\\ &=f'(x)\frac{1}{a}a\\ &=f'(x) \end{align*}
Let $y=f(x)$ and $h(x) = ax$ and $g = f \circ h$. Then $$ g'(x) = f'(h(x))h'(x) = af'(ax). $$
To see that geometrically, suppose (for example) that $a=2$. Then the graph of $g$ on the interval between $0$ and $1$ is just the graph of $f$ on the interval between $0$ and $2$, compressed horizontally into half its width. So it is twice as steep.
You can rewrite that as $$ f'(ax) = \frac{1}{a}g'(x) $$ if that is what you mean by "$dy/dax$".
See Mechanics of Horizontal Stretching and Shrinking