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 $\frac{1}{-2\sin(t)}$.

They are the same thing, but one is expressed in terms of $t$ and the other is in terms of $x$.
$$\frac{dt}{dx} = -\frac{1}{\sqrt{4-x^2}} = -\frac{1}{\sqrt{4-(2\cos t)^2}} = -\frac{1}{2\sqrt{1-\cos^2 t}} = -\frac{1}{2 \sin t}$$
where the last step uses the Pythagorean identity: $\sin^2 t + \cos^2 t = 1$.

For parametric equations, the chain rule is often written as
$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$
which makes things a bit more intuitive.
