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I'm reading Kline's Calculus: An intuitive and Physical approach. The author mentions that dy/dx is a confusing term to represent the derivative because dy/dx seems like a quotient when it really isn't and the author also adds that dy/dx should always be taken as a combination and never separately.

My questions:

1) Isn't dy/dx actually a quotient? It is the rate of change of y divided by rate of change of x is it not? For instance when we talk of instantaneous speed we call it ds/dt. Isn't that the quotient of distance covered in an interval of time and the interval of time (when the interval is approaching 0, of course).

2) if dy/dx really isn't a quotient, how are we allowed manipulations such as:

(i) (dy/dx) . (dx/dy) = 1

(ii) [this is seen when we use substitution to integrate]

Integrate $e^(5x + 2)$ $dx$

now we use substitution,

u = $5x + 2$

du/dx = 5

du = 5 $dx$ (How is multiplying by dx on both sides allowed if dy/dx is supposed to be taken as a whole?)

Mikhail Katz
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Siddharth Jossy
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    As you probably know, dy represents an infinitesimal change in y. Thus, du = 5 dx tells me that an infinitesimal change in u is 5 times greater than any infinitesimal change in x. – Kaynex Nov 10 '16 at 13:32
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    But, your question is more so "why are we allowed to algebraically manipulate these symbols?" Multiplying by dx is a mathematical "language gap" if you will. It simply works because the chain rule tells us it should work, and thus we've created a system to reflect that. Choosing to call it dy/dx is a result of the possible manipulations that work. – Kaynex Nov 10 '16 at 13:35
  • I don't have access to the content I want to show, but even if you treat dy/dx as a single object you are still able to prove those results (like $\int \frac{dy}{dx}dx = \int dy$). When taking this approach I always think of dy/dx as not a quotient but behaves like one. If I come across it I can post it as an answer – danwalkerdev Nov 10 '16 at 14:38

2 Answers2

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There are two different notations that look the same:

(1) The differential operator, this is written $\frac{d}{dx}$. This takes a function as input and returns a function. Therefore, if you have $y=f(x)$, then $\frac{dy}{dx}$ means $$ \frac{dy}{dx}=\frac{d}{dx}(y)=\frac{d}{dx}f(x). $$ In other words, the operator $\frac{d}{dx}$ is acting on $f$, turning $f(x)$ into a new function.

(2) Differentials. You may have seen with $u$-substitution, when $y=f(x)$, $dy=f'(x)dx$. This is a relationship between an infinitesimal change in $x$ and an infinitesimal change in $y$. So, when $x$ changes by an incredibly small amount, $y$ changes by a corresponding incredibly small amount. These two small changes are so small that they're infinitesimally close to zero. In order to see what they are, you need to cancel out the infinitesimally small part, so $\frac{dy}{dx}$ is a ratio of two infinitesimally small quantities. Using the formula $dy=f'(x)dx$, we get that $$ \frac{dy}{dx}=\frac{f'(x)dx}{dx}=f'(x). $$

Now, the confusion is that we have that the same notation means two different concepts. Luckily, both concepts evaluate to the same value. Therefore, we can interchange thinking about $\frac{dy}{dx}$ in two different ways (this is why the notation is the same).

Michael Burr
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In a Weierstrassian framework devoid of infinitesimals, we have no choice but claim that dy/dx is a single notation and not a ratio. In Abraham Robinson's framework where we do have infinitesimals, we can also treat it as a ratio. For further discussion see this answer.

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Mikhail Katz
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