Interpretation of parametric differentiation 
"How does parametric differentiation make sense both mathematically and graphically?"

I mean the method is simple to understand since one only has to differentiate both equations and divide (generally), but what does it actually mean to differentiate 2 questions and then divide them to find the differential of one with respect to another?
I've worked out the idea for the mathematical aspect and realized that $da/db$ is just a ratio so it's just the idea that $a:b = 2:3$ and $b:c = 6:11$ and if one is required to find $a:c$, he'd divide both. I think same thing holds for differentiation. But then what's the purpose of the notion familiarly stated with derivatives "derivative of one with respect to another?"
But I was struggling with the graphical interpretation, I got ideas that the graph would become a curved surface in 3D and the tangent will become a plane or that it might be related to partial derivatives and such but I think this is beyond me (class 12 student in India). A detailed explanation is welcome though. 
Here's one from a snippet of a book,
“ x and y are given as functions of a single variable e.g. x =f(t) and y=g(t) are 2
functions of a single variable. In such a case x and y are called parametric functions or parametric equations and t is called the parameter. To find dy/dx in case of parametric functions, we first obtain
relationship between x and y by eliminating the parameter t and then we differentiate it with respect to x. But, it is not always convenient to eliminate the parameter. Therefore dy/dx can also be by the following formula, 
$$
\frac{d y}{d x}=\frac{d y / d t}{d x / d t}
$$
I have also found this alongside with it,
“To prove it, let Δx and Δy be the changes in x and y respectively corresponding to small change is Δt in t. Then,
$$
\frac{\Delta y}{\Delta x}=\frac{\Delta y / \Delta t}{\Delta x / \Delta t} \Rightarrow \frac{d y}{d x}=\lim _{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}=\frac{\lim _{\Delta t \rightarrow 0} \frac{\Delta y}{\Delta t}}{\lim _{\Delta t \rightarrow 0} \frac{\Delta x}{\Delta t}}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}
$$
So I guess, this takes care of the mathematical aspect (though I'm still not convinced with it) but what about the graphical interpretation?
 A: The "proper" way to do the things you did above. Consider the function $u(x)$. Now lets assume that $x$ is actually again a Function or something that depends on $parameter$. So in your case lets say $x = x(v)$. Now when we want to calculate the derivative of $u$ we with respect to $v$, where we need the chain rule:
$$
\frac{d}{dv} u(x(v_0)) = \frac{du}{dx}(x(v_0)) \cdot \frac{dx}{dv}(v_0).
$$
Now, let's not evaluate the derivative, then the above expression becomes
$$
\frac{du}{dv} = \frac{du}{dx} \frac{dx}{dv}.
$$
Observe, that this last expression is a function, and so are $\frac{du}{dx}$ and $\frac{dx}{dv}$ are both functions! 
Now, in the beginning of the development of calculus (e.g. the 17 century). A lot of things were not known, especially about limits and convergence and that stuff. So, when people, namely Newton and Leibniz, developed this idea of taking a derivative, they used a lot of intuition and good will. Leibniz very much argued in a similar spirit as you, say, hey look, lets denote by $du$ a very very small difference of $u$ and by $dx$ lets denote a very very small difference of $dx$, then even so these difference will super close to zero (but never really) zero, then the fraction $\frac{du}{dx}$ actually could be something meaningful and let's called it the derivative. 
Today, we do not argue like this any more (at least if you are a mathematician). But Leibniz invented by this an amazing symbolism that you see in play above. Because starting from $\frac{du}{dv}$ simply by multiplying by $1 = \frac{dx}{dx}$ you actually get the chain rule effortless (as hell!). Now lets keep this idea, that $\frac{du}{dx}$ is actually a fraction, then you can also get your expression
$$
\frac{du}{dx}\frac{dx}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}}
$$
and with a little bit of magic this completely unjustified manipulation of symbols actually gives the right answer (in the right circumstance!).
Maybe so much for now. Is it helpful? Not sure if I actually addressed  the question properly. 
A: What you're doing is applying the chain rule. You have (following your example), $$u=f(v),$$ where $v=g(x).$ Then of course $$u'_x=f'_vv'_x.$$
Well, in particular, for your example, since $v=\sqrt{\cos x},$ it follows that $\sin x=\sqrt{1-\cos^2x}=\sqrt{1-v^4},$ so that $$u=\log \sqrt{1-v^4},$$ explicitly.
