3

I still have some trouble grasping the true mathematical meaning of $dx$. I have the intuitive understanding that it is an "infinitely tiny amount of $x$" but I don't know in what cases I can use it.

Here is an example :

is $\lim_{x\to 0} \frac{2x + 32x^2}{e^x}$ equivalent to $\frac{2.dx + 32.dx^2}{e^{dx}}$ ?

Thanks a lot

s89ne
  • 125
  • $dx$ is the difference between 2 values of $x$, your case is obviously unfit for such use – Makina Dec 16 '18 at 15:58
  • Then is $\lim_{\Delta x\to 0} \frac{2\Delta x + 32(\Delta x)^2}{e^{\Delta x}}$ equivalent to $\frac{2.dx + 32.dx^2}{e^{dx}}$ ? – s89ne Dec 16 '18 at 16:00
  • No, because the limit you gave either doesn’t exist or evaluates to a real number. What you wrote is an expression that’s neither. This feels similar to how people treat $\frac{df}{dx}$. Luckily notation and theory work out to where you can treat it like a fraction in same cases without running into problems, but it’s still not correct. – Will Fisher Dec 16 '18 at 16:01
  • 2
    Depending on the context, $\textrm{d}x$ on its own has different meanings. It can be a measure or a differential form (off the top of my head). I know no context where it would stand for an infinitely tiny amount of $x$. The usual notation for a very small amount is just $\varepsilon$, but it is usually not infinitesimal. But even if $\textrm{d}x$ was an actual infinitesimal, the two expressions would not be the same (not before applying standard part to the right hand side). – tomasz Dec 16 '18 at 16:02
  • 1
    What $dx$ precisely is seems confusing. Is there a rigorous mathematical definition of it, or should it always be understood intuitively depending on the context? – s89ne Dec 16 '18 at 16:11
  • In reality, I stumbled upon this problem while trying to write down what a derivative is. The rigorous convincing form I found is for say $f(x) = x^2$ : $$f'(x)= \lim_{\Delta x\to 0} \frac{\Delta (f(x))}{\Delta x} = 2x$$ and I was pondering why the notation $\frac{d(f(x))}{dx}$ was equivalent to the one I found – s89ne Dec 16 '18 at 16:17
  • Possible duplicate of Is $\frac{\textrm{d}y}{\textrm{d}x}$ not a ratio? –  Dec 16 '18 at 17:16
  • @s89ne Since $f'$ is defined as the limit of a fraction, it's convenient to have a notation for it that looks like a fraction. If $g=f'$ then the frequent statement $df=gdx$ is an abbreviation for the corollary $\Delta f=g\Delta x+ o(\Delta x)$ (see https://en.wikipedia.org/wiki/Big_O_notation#Little-o_notation). As Yves Daoust's answer already noted, we can define a linear map $df$ that makes the $d$ comparison statement equivalent to the $\Delta$ comparison statement. – J.G. Dec 16 '18 at 17:18
  • I couldn't find the actual linear approximation equation in the answer to your question that @J.G. makes reference to, but that part is nicely explained here. – Antoni Parellada Dec 16 '18 at 20:58
  • @AntoniParellada I was referring to its first inline equation. – J.G. Dec 16 '18 at 21:08

2 Answers2

2

Let's start with your $f'(x)= \lim\limits_{\Delta x\to 0} \frac{\Delta (f(x))}{\Delta x} = 2x$ when $f(x)=x^2$

Sometimes the derivative $f'(x)$ is written $\frac{df}{dx}$ or $\frac{d}{dx} f(x)$. All three of these are simply notation for $\lim\limits_{y \to x} \frac{f(y)-f(x)}{y-x}$ or $\lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h}$ and some people describe this as the limit of the ratio of the change of $f(x)$ to the change in $x$ leading to your $\lim\limits_{\Delta x\to 0} \frac{\Delta (f(x))}{\Delta x}$ or $\lim\limits_{\Delta x\to 0} \frac{\Delta f}{\Delta x}$: the notation $\frac{df}{dx}$ is suggestive of this limit but the $dx$ does mean anything on its own here

Since the derivative operation can be applied to different functions, further notation lets $\frac{d}{dx}$ represent a derivative operator. It can be applied more than once: for example a double derivative would involve a double operation and is sometimes written $f''(x)= \frac{d^2}{dx^2}f(x)=\frac{d^2f}{dx^2}$; with $f(x)=x^2$ you would get $\frac{d^2f}{dx^2} = 2$. Again this is notation, and the $dx$ is part of the notation without having a stand-alone meaning; in particular it is not being squared

You will also see $dx$ as part of the notation for integrals: for example $\int_6^9 x^2\, dx = 171$. But while this might be suggestive of a sum (the long s in the integral sign) of lots of pieces for $f(x)$ with widths represented by $dx$, it is again notation and the $dx$ does not have an independent meaning here, just being part of a wider expression for the integral operation

Some people write expressions such as $d(fg)=f\, dg + g\, df$ for the product rule. Typically this is shorthand for $\frac{d}{dx}(fg)=f \frac{d}{dx}g + g \frac{d}{dx}f$ and is another example of the use of notation

There is nothing to prevent people from giving a particular meaning to a bare $dx$ so long as they define it in a meaningful way. Then again it would still be notation, with that definition

Henry
  • 157,058
0

The differential being an infinitely tiny amount is a misconception. In modern setting, the differential is a linear map, specifically the linear part of the variation of a function when you vary the variable. In other words, when a function is differentiable there is a constant $c$ such that

$$f(x+dx)=f(x)+c\cdot dx+o(dx)$$ ($o(dx)$ is a sublinear term as a function of $dx$) and by definition $$df(x)=c\cdot dx.$$

The constant $c$ is well know to be the derivative of $f$, and $dx$ can be an arbitrarily large increment.

Actually, differentials have little to do with limits.

On the picture, the function is in blue. Its differential (at $x=1$) in magenta and the remainder in black.

enter image description here