calculus - What is the logic behind decomposing a derivative operator symbol. In the population growth equation?

$$\frac{dy}{dt} = Ky$$

$$\frac{1}{y}\frac{dy}{dt} = K$$

...then my teacher did something illogical he decomposed the Differential operator $\frac{dy}{dt}$ into $\frac{1}{dt} *dy$ and used this "form" with the laws of arithmetic to "cancel" the $\frac{1}{dt}$ by multiplying both sides by $dt$

$$dt*\frac{dy}{dt}\frac{1}{y} = K* dt$$

$$dy =K *y *dt$$

Why is this possible?
Is it related to the chain rule?

Answer

The proper way to think about this is as follows:

$$\frac{dy}{dt} = Ky \Longleftrightarrow \frac{1}{y}\frac{dy}{dt} = K.$$

Thus

$$\frac{1}{y}\frac{dy}{dt} - K = 0.$$

However

$$\frac{1}{y}\frac{dy}{dt} = \frac{d}{dt}\log|y(t)|.$$

Thus if we integrate our equation with respect to $t$, what we find is that

$$\int \left(\frac{d}{dt}\log|y(t)|-K\right)\,dt = C.$$

Using the fundamental theorem of calculus we get

$$\log|y(t)| - Kt = C$$

which can be easily solved. As you see, we used the relation $\frac{1}{y}\frac{dy}{dt} = \frac{d}{dt}\log|y(t)|$ which relies on chain rule (since $y$ is a function of $t$). The symbolic manipulation your professor did (which many people do) is really just a repackaging of the chain rule. It's not rigorous since the notation $\frac{dy}{dt}$ is not meant to represent a fraction - it is merely notation adapted from $\frac{\Delta y}{\Delta x}$ representing slopes of secant lines. It is convenient notation as such calculations show (you can kind of think of it as a fraction without too many issues in $\text{1D}$).