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?
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.
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.