I'll start with an example.
In physics, $x(t)$ represents the $x$-position of a particle, and $v(t)$ its ($x$-)velocity. To determine the total displacement of a particle on the interval $[a, b]$, we can use the formula
$$\Delta x = \int_a^b{v(t)~dt}$$
To me, this makes sense, because $v = \frac {dx}{dt}\\$, so the above equation is equivalent to:
$$\Delta x = \int_a^b{v(t)~dt} = \int_a^b{\frac {dx}{dt}~dt} = \int_a^b{dx} = x(b) - x(a) = \Delta x$$
However, I've been told that you can't just "cancel" the $dt$ differential because it's not "proper."
Another example uses parametric arc length:
$$\ell = \int_a^b{\sqrt{\left ( \frac {dx}{dt} \right )^2 + \left ( \frac {dy}{dt} \right )^2} dt}$$
Now take a standard function, $y = f(x)$. We can define $x = t$, and then we have $y(t) = f(t)$, $x(t) = t$, and $\frac{dx}{dt} = 1$. Then, $\frac {dy}{dt} = \frac {{dy}~/~{dx}}{{dx}~/~{dt}} = \frac {{dy}~/~{dx}} 1 = \frac {dy}{dx}$, and our formula simplifies to
$$\ell = \int_a^b{\sqrt{1 + \left ( \frac {dy}{dx} \right )^2} dt}$$
This is indeed the correct formula for arc length of a function (also derivable with Pythagorean theorem), but it relies on being able to cancel the $dx$s in $\frac {{dy}~/~{dx}}{{dx}~/~{dt}}$.
So, my question is, when, if ever, can you cancel differentials?