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 $\frac{1}{-2\sin(t)}$.
 A: 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.
A: 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.
A: Your last step is wrong. It should be
$$\frac{dt}{dx}=-\frac12\cdot\frac{1}{\sqrt{1-\frac{x^2}{4}}} $$ $$\frac{dt}{dx}=-\frac12\cdot\frac{1}{\sqrt{1-\cos^2 t}} $$
A much better approach would be to differentiate    $\space x=2\cos t $ with respect to $x $
$$1=-2\sin t \frac{dt}{dx} $$
A: For parametric eq. we write chain rule as, 
$$
\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}
$$
If you are interested in why? or How? then i will tell you,
$$
\frac{dy}{dx}=\lim_{h\to 0}\frac{y(x+h)-y(x)}{x+h-x}
$$
Divide numerator and denominator by $t+p-t$ where $p\to0$,
$$
\lim_{h\to 0}\frac{\frac{y(x+h)-y(x)}{t+p-t}}{\frac{x+h-x}{t+p-t}}=\lim_{h\to 0}\frac{\lim_{p\to 0}\frac{y(x+h)-y(x)}{t+p-t}}{\lim_{p \to 0}\frac{x+h-x}{t+p-t}}
$$
[ Now try to understand this:
Consider $t$ changes to $t+p$ , due to this change $x$ changes to $x+h$ $(=x(t+p))$.
So we can say that $y(x+h)=y(x(t+p))=y(t+p)$ becuase all changes are relative to each other.  ]
$$
\lim_{h\to 0}\frac{\lim_{p\to 0}\frac{y(t+p)-y(t)}{t+p-t}}{\lim_{p \to 0}\frac{x(t+p)-x(t)}{t+p-t}}=\frac{\lim_{p\to 0}\frac{y(t+p)-y(t)}{t+p-t}}{\lim_{p \to 0}\frac{x(t+p)-x(t)}{t+p-t}}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}
$$
[$\lim_{h \to 0} $ is redundant here]
Note: These are completely my thoughts, comments are welcomed
A: It’s no coincidence that you can take the reciprocal of $dx/dt$ to find $dt/dx$. You can see this by differentiating both sides of $f^{-1}\circ f=\operatorname{id}$ and applying the chain rule. If $b=f(a)$, we get $(f^{-1})'(b)f'(a)=1$. (The Inverse Function Theorem guarantees that $f^{-1}$ and its derivative exist for the functions we’re looking at here.)
This generalizes to functions $f:\mathbb R^n\to\mathbb R^n$, but you can’t simply take the reciprocal of a partial derivative to find the corresponding partial of the inverse function. Instead, you have to invert the entire Jacobian matrix to find the inverse partial derivatives.  
P.S. This also holds for $f:\mathbb R^n\to\mathbb R^m$: if $\operatorname{d}f$ is continuous and is invertible at $\mathbf v$, then $f^{-1}$ exists in a neighborhood of $f(\mathbf v)$ and is also continuously differentiable with $\operatorname{d}(f^{-1})_{f(\mathbf v)}=(\operatorname{d}f_{\mathbf v})^{-1}$, i.e., differentiation and inversion “commute.” In this most general case, however, there’s not a recipe for inverting the differential.
