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Why is this algebra calculus trick legal?

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

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

...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?

Community
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Conor
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    This is indeed an illegal move. However, it highlights why we write derivatives to look like fractions: even though such manipulations are not correct, they very often get the correct answer in the end. – Arthur Jun 04 '16 at 14:39

2 Answers2

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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 $\text{1D}$).

Cameron Williams
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This kind of manipulation can be made rigorous, but even without rigor the two equations $$ \frac{dy}{dt} = Ky $$ and $$ dy = K y \, dt $$ can be seen to be consistent.

The first says the rate of change of $y$ (with respect to $t$) is proportional to the value of $y$. The second says that a small change $dt$ in $t$ causes the small change $Ky\, dt$ in $y$.

(I think there's an error in your version of the second equation. If you fix it I'll edit this answer.)

Ethan Bolker
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  • I know people like to make the differential argument, but it's a repacking of limits without the usual rigor of limits and has the same internal inconsistencies as what OP's professor did in my opinion. I don't agree that this is a rigorous approach (differentials are again borrowed ideas from $\Delta x$), but I am probably a severe minority in this. I won't downvote since it's a reasonable answer to most people; I'm just expressing my view. – Cameron Williams Jun 04 '16 at 14:49
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    @CameronWilliams I agree (and said) that it's not rigorous. But is definitely a useful way to think about calculus when you're trying to do real problems/applications. In this case both equations are good ways to understand exponential growth. If Newton, Leibniz and then Euler had to argue with the rigor of some freshman calculus courses today they might not have been able to invent and then exploit calculus. – Ethan Bolker Jun 04 '16 at 14:54
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    This is very true. I might have misinterpreted what you were referring to when you said "[it] can be made rigorous." By "this kind of manipulation", I thought you were referring to your own argument, but I see now that you meant OP's professor's argument. A lot of people like to appeal to differentials from differential geometry as a justification for this manipulation, however I feel like that is not anywhere close to being what calculus students think of as $dx$. There it's almost by definition that the manipulations work out in that case.. – Cameron Williams Jun 04 '16 at 14:57