# Why isn't it mathematically rigorous to treat dx's and dy's as variables? [duplicate]

This question already has an answer here:

If I do something like:

$$\frac{dy}{dx} = D$$

$$dy = D \times dx$$

People would often say that it is not rigorous to do so. But if we start from the definition of the derivative:

$$\lim_{h \to 0}{\frac{f(x + h) - f(x)}{h}} = D$$

And by using the properties of limits we can say:

$$\frac{\lim_{h \to 0}{f(x + h) - f(x)}}{\lim_{h \to 0}{h}} = D$$

And then finally:

$$\lim_{h \to 0}(f(x + h) - f(x)) = D \times (\lim_{h \to 0} h)$$

Isn't this the same? Or am I missing something?

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• Your claim after "And by using the properties of limits we can say:" is false. – smcc Jul 15 '16 at 14:06
• $dx$ and $dy$ are not variables. They're an operation like addition, subtraction, integration, differentiation, etc. – KingDuken Jul 15 '16 at 14:06
• @KingDuken Even if I define dx and dy to be the denominator and numerator of the fraction (of the definition of the derivative)? – Anonymous Jul 15 '16 at 14:09
• @smcc How so? I do not understand. – Anonymous Jul 15 '16 at 14:12
• Anyway, there actually are nonstandard systems out there that let you manipulate infinitesimals rigorously. The most common one is "hyperreal analysis". You can get a gentle introduction to this in a calculus book by Keisler. But it doesn't look very much like you might naively expect. – Ian Jul 15 '16 at 14:19

I can spot the following mistake in your attempt:

1. And by using the properties of limits we can say:

$$\frac{\lim_{h \to 0}{f(x + h) - f(x)}}{\lim_{h \to 0}{h}} = D$$

You cannot actually do that, as smcc has said. You must note that $$\lim_{x\to 0} \frac{f(x)}{g(x)}=\frac{\lim_\limits{x\to 0} f(x)}{\lim_\limits{x\to 0} g(x)} \,\,\,\,\,\,\,\,\,\,\mathrm {iff \lim_\limits{x\to 0} g(x) \not = 0}$$ So what you have said is not right.

Now coming to the actual question, if $dx$ and $dy$ can be treated as variables, most of the mathematicians treat $\frac{d}{dx}$ as a mathematical operator (like $+,-,*,/$) which acts on the variable $y$. So that way, you can clearly understand what is the variable and what is not.

However, if you are strict enough to observe from the "limit" viewpoint, then observe that $\frac{dy}{dx}$ is nothing but $\lim_\limits{\Delta x\to 0}\frac{\Delta y}{\Delta x}$. Now $\frac{\Delta y}{\Delta x}$ is a fraction with $\Delta y$ and $\Delta x$ in the numerator and denominator. So you can view them as variables now.

Looks a bit weird, I know, but it entirely depends on how you want to support your argument and from which point of view you want to make your claim.

• $D$ is not for the euler's notation. It is just a number. The value of the derivative. – Anonymous Jul 15 '16 at 15:00
• Using $D$ as the letter for the numerical value of the derivative is a bit contrary to certain standard notation (e.g. the notation used in Evans), but it is clear in this context. – Ian Jul 15 '16 at 15:01
• @Anonymous Okay, I have got it. Going to edit my answer. But tell me, is the derivative equal to D for all $x$ or just some $x=x_0$? – SchrodingersCat Jul 15 '16 at 15:05
• @SchrodingersCat for all x's. A general one. – Anonymous Jul 15 '16 at 15:08