Can We Cancel the Derivative of dt in These Equations?

[Another]

problem in this book : classical mechanics goldstein

Why can we cancel the derivative of dt from these equations?

e.g.
$\frac{d(x)}{dt} + \frac{b sin\theta}{2} \frac{d(\theta)}{dt} = asin\theta \frac{d(\phi)}{dt}$
$x +\frac{b \theta sin\theta}{2} = a \phi sin\theta$

because I think
$\frac{d(x)}{dt} + \frac{b sin\theta}{2} \frac{d(\theta)}{dt} = asin\theta \frac{d(\phi)}{dt}$
$\frac{d}{dt}(x - (b/2) cos\theta) = asin\theta \frac{d(\phi)}{dt}$ due to $sin\theta$ dependent on t. we can't cancel dt

Or it is just only divider. So We can cancel

 

[fresh_42]

The short answer is: because it works.
The long answer is: because of the connection between tangent bundles and differential forms.

The long answer is not really trivial and I'm not sure I can explain it without making mistakes.

Let's consider an easy case: $\dot{x}=\dot{y}.$ If we treat the expressions $\dfrac{dx}{dt}$ as ordinary quotients, then $dx=dy$ is obvious. We could either imagine that the infinitesimals are tiny distances, which is rigorously seen wrong, or we could invoke L'Hôpital's rule, which is a bit better, since $\frac{dx}{dt}$ is an abbreviation of a limit process.

Let's see what $dx=dy$ tells us. Integration yields $\int dx = \int dy$ and thus $x=y+C$, which is what we would expect from $\dot{x}=\dot{y}.$