How is $\frac{dt}{dx}$ just the reciprocal of $\frac{dx}{dt}$ ?? Came across this problem.  Derivative of a parametric.
$$x=2\cos(t)$$
$$y=2\sin(t)$$
.....Hence,
$$\frac{dx}{dt}=-2\sin(t)$$
$$\frac{dy}{dt}=2\cos(t)$$
As we know, chain rule states:   $$\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}$$
Apparently, this results in  $$\frac{dy}{dx} = 2\cos(t)\frac{1}{-2\sin(t)}$$
That last part seems weird.  How do you get $\frac{dt}{dx} = \frac{1}{-2\sin(t)}$ ?  You just take the reciprocal of $\frac{dx}{dt}$ ??   Seems like a bizarre coincidence.    Inverse is not the same as reciprocal.
If I wanted to get $\frac{dt}{dx}$ explicitly, start with the original:
$$x=2\cos(t)$$
Then put in terms of t, and you'd have:
$$t = \cos^{-1}(\frac{x}{2})$$
Then take $$\frac{dt}{dx} = - \frac{1}{\sqrt{1-\bigl(\frac{x}{2}\bigr)^2}} \cdot \frac{1}{2} = -\frac{1}{\sqrt{4-x^2}}$$
The point is this doesn't look anything like what we used for $\frac{dt}{dx}$ which was $\frac{1}{-2\sin(t)}$.

I don't know if any of these really answer the fundamental question posed by the asker.
Yes, $\displaystyle\frac{dx}{dt}$ and $\displaystyle\frac{dt}{dx}$ indeed have a reciprocal relationship when both are defined. It's no coincidence. In fact, that applies in all complete derivatives, for instance $\displaystyle\frac{dy}{dx}\cdot \frac{dx}{dy} = 1$
This is simply a consequence of chain rule. $\displaystyle\frac{dy}{dx}\cdot \frac{dx}{dy} = \frac{dy}{dy}=1$
It's another of those instances where (in single variable calculus), you can "get away with" thinking that the Leibniz forms behave like fractions. Of course, they're not actually fractions but rather limits, but using this notation, you can manipulate them in much the same way as you would fractions (with some care). There are other questions on this site about the same topic, I believe.
