Intuition behind why the derivative of $\ln(cx)$ always equals $1/x$, for any $c>0$? I understand how to prove that algebraically using implicit differentiation:
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However, when I hope to gain an understanding through the graphs, I had a hard time wrapping my head around why the transformed lnx function still has the same slope at every point of x as the parent function.
Graphs of the 3 functions that I talked about
 A: It's because of the identity $\ln(ab)=\ln a+\ln b$, which holds for all $a,b>0$.
A: Here's a more general way of asking the question. Let $c\neq0$.
In order for $f(x)$ and $g(x):=f(cx)$ to have the same slope for all $x$, we need $f'(x)\equiv g'(x)\equiv c\cdot f'(cx)$. (The $\equiv$ sign means that these equations need to hold for all values of $x$: in order words, these functions must be equal, not just coincide at a particular point $x$.)
Which functions $h$ have the property that $h(x)\equiv c\cdot h(cx)$? Or, in other words, $h(cx)\equiv c^{-1}\cdot h(x)$?
Geometrically, this last identity says that compressing the graph of $h$ horizontally by a factor of $c$ is the same as stretching the graph of $h$ vertically by a factor of $c^{-1}$. 
Algebraically, you can see that any function of the form $ax^{-1}$ solves this functional equation. In fact, any sum of such functions will work, too: we can have $h(x)=\sum a_ix^{-1}$.
Note that you can see clearly that no polynomial will work, though: substituting $cx$ for $x$ in $\sum a_ix^i$ for $i\geq0$ will not allow you to factor $c^{-1}$ out of each term.
