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To my knowledge, dy/dx is equal to the limit of (f(x+h) - f(x)) / h as h approaches zero. That is, dy is equal to the difference in the y value (f(x+h) - f(x)) and dx is equal to the difference in the x value (h) and dy/dx is equal to the rate of change of the y function as the x function increases.

As well as this, in the chain rule you multiply dy/du * du/dx in order to cancel out the du's like you would with normal fractions.

Despite this, all of my teachers insist that dy/dx is not a ratio, it is just a symbol for the gradient or for the derivative of y with respect to x that can't be split up and manipulated like a normal fraction can be.

On a side note, the symbol d/dx (y) to mean "differentiate y with respect to x" confuses me too, as i thought that dy and dx were atomic symbols, but this notation splits up d and y even further.

Adam
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1 Answers1

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Because $\frac{\mathrm{d}y}{\mathrm{d}x}$ is not a ratio of numbers, variables, number-valued functions, or any other familiar sort of thing.

Instead, it's a ratio of a different type of object called a "differential form", but this object is usually not (explicitly) introduced in the introductory setting.

However, the result of the ratio is a familiar sort of thing, and so you can be taught how to work with it without ever discussing the new kind of object. Doing so means that you have to treat $\frac{\mathrm{d}y}{\mathrm{d}x}$ as a whole.

You learn algebraic identities that mimic some of the properties of ratios — it is those algebraic identities you are using when you do arithmetic with these things.

Incidentally, your idea of multiplying the ratios so you can cancel out a common factor doesn't work with differential forms either: you can't multiply $\mathrm{d}y$ with $\mathrm{d}u$ in any familiar sort of way. Proving that $\frac{\mathrm{d}y}{\mathrm{d}u} \frac{\mathrm{d}u}{\mathrm{d}x} = \frac{\mathrm{d}y}{\mathrm{d}x}$ must be done in a totally different way, which basically boils down arguing that this equation must hold if it holds when you multiply both sides by $\mathrm{d}x$, and using a different arithmetic rule to simplify that product, namely $\frac{\mathrm{d}s}{\mathrm{d}t} \mathrm{d}t = \mathrm{d}u$.