Why is $\frac{dy}{d(ax)}$ = $\frac{1}{a}\frac{dy}{dx}$? I searched on the internet and found the derivation 
$$\frac{dy}{d(ax)}=\frac{1}{a}\frac{dy}{dx}\tag{1}$$
I am confused of why this is true. 
I tried to do the derivation myself, but there was no progress. Here is what I did. 
$$y(x)=2x$$
$$\frac{dy}{dx}=2$$
So I say that 
$$y(2x)=4x$$
According to $(1)$
$$\frac{dy}{d(2x)}= \frac{1}{2}\frac{dy}{dx}=1$$
But how does this even make sense?
 A: Generally, when you have something like
$$\frac{d}{d(g(x))} f(x)$$
I personally find it convenient to make a substitution of $u = g(x)$ and then rewrite $f$ as a function of $u$.

Take note of your error: in your case, you just took the bottom, and plugged in $g(x)$ into $f(x)$ as an input. That's not at all how this sort of situation works. You're essentially checking the rate of change of $f$ with respect to the variable $g(x)$. Typically we just let $g(x) = x$, but that is not always the case, but it can be reduced to that case by manipulations and substitution.

For $f(x) = 2x$, you are correct in that $f'(x) = 2$. However, in finding
$$\frac{d}{d(2x)} 2x$$
we let $u = 2x$. Then $f(x) = 2x \Leftrightarrow f(u) = u$, and thus
$$\frac{d}{d(2x)} 2x = \frac{d}{du} u = 1 = \frac{1}{2} \frac{d}{dx} 2x$$
A: This is simple if you realize that the infinitesimal differential operator $\text{d} f(x) = f'(x) \, \text{d}x$ where $f'(x)$ is the derivative of $f$ with respect to $x$.
In your case $f(x) = ax$ and thus $\text{d} f(x) = a  \,\text{d} x$.
Thus, obtain
$$
\frac{\text{d}y}{\text{d}(ax)}=\frac{\text{d}y}{a\,\text{d}x},
$$
as desired.
A: I know people helped you already, buy maybe this will make it more clear. 
using the chain rule:
$$\frac{dy}{d(ax)}=\frac{dy}{dx}\frac{dx}{d(ax)}=\frac{dy}{dx}\frac{d(\frac{1}{a}(ax))}{d(ax)}=\frac{dy}{dx}\frac{1}{a}=\frac{1}{a}\frac{dy}{dx}$$
A: $$\frac{d}{dx}f(x)=f'(x)$$
To solve for 
$$\frac{d}{d(ax)}f(x)=f'(x)$$
Make a substitution of $u=ax$, we have
$$\frac{d}{du}f\left(\frac{u}{a}\right)=\frac{1}{a}\left(f'\left(\frac{u}{a}\right)\right)=\frac{1}{a}f'(x)=\frac{1}{a}\frac{d}{dx}$$
