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It's been over 15 years since I last did anything with calculus at school, and I've forgotten most things, so please be gentle with your answers. :)

One thing that has always bugged me when getting into more advanced stuff was that the derivative simbols are suddenly manipulated in a way that doesn't make sense to me. For example, let's take the simple equation

$$y=12x^2+24x+45$$

An alternative notation for the same thing would be:

$$f(x)=12x^2+24x+45$$

Here we merely substituted $y$ for $f(x)$. It means the same thing.

If memory serves me right, when we take the first derivative of this function, we have three ways we can write it:

$$y'=24x+24$$ $$f'(x)=24x+24$$ $$\frac{dy}{dx}=24x+24$$

Now, it's this third notation that I'm talking about here. To my understanding the $\frac{dy}{dx}$ doesn't really mean anything by itself. It's just another way to specify that we're talking about a derivative of a function, just like the $'$ in the other two notations.

But then sometimes I come across an equation like this:

$$m\frac{\text{d}^2\mathbf{r}}{\text{d}t^2} = \frac{kqq'}{|\mathbf{r}|^2},$$

And here suddenly the $d$ is being treated like a variable, not to mention the $t$ and... And if memory serves me right, I've seen ever more extreme examples where the fraction $\frac{dy}{dx}$ itself is taken apart and each of the components used as a variable. I think this was popular back in university when I was talking course on differential equations. Barely passed that, half of it didn't make sense, and this was a large part of the reason why. I don't remember why I never asked my teacher about it back then.

So... what does it mean when we start to break the $\frac{dy}{dx}$ apart?

Vilx-
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  • So if you take the power rule for differentiation using $\frac{df(x)}{dx}=\frac{f(x+h)-f(x)}{h}$ you get that the derivative of $x^2$ is equal to $2x$. Since $\frac{d}{dx}(af(x)+bg(x))$ is $af'(x)+bg'(x)$ also (by the same rule), the derivative of $12x^2+24x+45$ actually becomes $12(2x)+24(1)$, or $24x+24$, or $24(x+1)$, or anything else of the like that you want it to be! – Pineapple Fish May 27 '21 at 15:26
  • @PineappleFish - Whoops! Thank you, fixed it! Just goes to show how long I haven't done anything with these things. ^^) – Vilx- May 27 '21 at 15:28

4 Answers4

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Rather then consider the $\operatorname{d}$ alone, one should view the expression $\operatorname{d}\!y/\operatorname{d}\!x$ as the following: $$ \frac{\operatorname{d}\!y}{\operatorname{d}\!x} = \frac{\operatorname{d}\!}{\operatorname{d}\!x} y, $$ that is $ \operatorname{d}/\operatorname{d}\!x$ is the differential operator that applies to the function $y$. In view of this, we have the second derivative as $$ \frac{\operatorname{d}\!}{\operatorname{d}\!x} \frac{\operatorname{d}\!}{\operatorname{d}\!x}y = \frac{\operatorname{d}^2}{\operatorname{d}\!x^2}y = \frac{\operatorname{d}^2\!y}{\operatorname{d}\!x^2}. $$

Ken Hung
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  • The problem with this notation-logic is that the "d" in the denominator should also be squared, or equivalently it could be (dx)^2. Unless of course we assume that "dx" is just a single symbol, but this is kind of ugly. – Juan Perez Jun 23 '21 at 13:50
  • Well, if you consider writing the expression as $ d^2/(dx)^2 $, then the expression is not pretty either. So I guess originally people write in the way how we write now partly because of aesthetic reason. – Ken Hung Jun 25 '21 at 02:51
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This just means the 2nd derivative of $r$ with respect to $t$

$$\frac{\text{d}^2\mathbf{r}}{\text{d}t^2}$$

So it's again just a notation. It has always looked confusing to me too. But that's what it is, just a notation. The $d$ is not treated like a variable. Even though you see $d^2$ there you can't e.g. take square root of it (and get $|d|$ as a result), or anything like that.

See also:

https://en.wikipedia.org/wiki/Second_derivative

peter.petrov
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  • OK, I think there were worse examples too. But I'll have to find one first. – Vilx- May 27 '21 at 07:23
  • Ahh, found something in Wikipedia: $P(y)dy+Q(x)dx=0$ – Vilx- May 27 '21 at 07:27
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    This is kind of slightly different, even though related. It's called differential. https://en.wikipedia.org/wiki/Differential_of_a_function Note also this quote from Wikipedia: "The precise meaning of the variables dy and dx depends on the context of the application and the required level of mathematical rigor." – peter.petrov May 27 '21 at 07:28
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    I would suggest you get a good book on calculus, and go through the basic definitions in the right order. Otherwise I think the meanings will remain kind of fuzzy to you. – peter.petrov May 27 '21 at 07:30
  • Ahh, wait, so in differential equations it means something else? Beautiful. OK, at least I know where to start if I want to go deeper. Thanks! :) – Vilx- May 27 '21 at 07:36
  • I mean that when you don't have the whole thing $\frac{dy}{dx}$ but you have $dy$ and $dx$ separately, you should probably read them as differentials. – peter.petrov May 27 '21 at 07:41
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Regarding the splitting of the $dy/dx$ thingie ...

Suppose we have $$ y = 2x + 5 $$ Then, as you say, we can write $$ y' = 2 $$ or $$ \frac{dy}{dx} = 2 $$ Here, the $dy/dx$ just means the same as $y'$ -- it's the derivative of $y$ with respect to $x$. It's a single indivisible symbol, not really a fraction. It tells us that the rate of change of $y$ with respect to $x$ is equal to $2$. In other words, $y$ is changing twice as fast as $x$ is. So far, so good.

But then some people would feel at liberty to rearrange it, as if it were a fraction, and they'd write $$ dy = 2\,dx $$ In my view, this is a bit sloppy, but you can make sense out of it. It says that if you change $x$ by a small amount $dx$, then $y$ will change by an amount $dy$ that's twice as large. That's certainly consistent with the earlier statement that $y$ is changing twice as fast as $x$.

I personally think it's confusing to use $dy$ and $dx$ to denote small changes in this way. I'd much prefer to use $\delta$ instead of $d$ for this purpose, so I would write $$ \delta y = 2 \, \delta x $$ This makes exactly the same statement about the sizes of small changes, but it avoids the confusion with the $dx$ and $dy$ used earlier.

bubba
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  • In truth, today on Wikipedia it was the first time I saw $\delta$ being used. Since it looks so similar to $d$ (and I expect even more so in handwriting), maybe I just haven't noticed it before? – Vilx- May 27 '21 at 09:30
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    You can't complain about the sloppiness of $dy=2dx$ and then go on to write $\delta y = 2 , \delta x$! You have to write $\delta y \approx 2 , \delta x$ or similar. Which raises the question: what does the $\approx$ symbol mean? And you are kind of back where you started. – TonyK May 27 '21 at 09:34
  • @TonyK Excellent observation, and I was going to say the same. – K.defaoite May 27 '21 at 15:34
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    @RossPresser You're thinking of $\partial$, not $\delta$. Two different symbols. – bubba May 28 '21 at 01:24
  • @bubba Thank you, I was wrong. Deleted comment. – Ross Presser May 28 '21 at 15:54
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There are a variety of opinions on this. The thing to keep in mind is that d is not a variable, but an operator. Therefore, $dy$ is a shorthand for $d(y)$, and means "the differential of $y$". When you add in the superscript numbers, in the case of differentials, it means we are applying the operator that number of times. So, $d^2y = d(d(y))$.

You really can split the fraction $\frac{dy}{dx}$. Infinitesimals are valid mathematics, and differentials represent infinitesimal values.

You can't split the fraction $\frac{d^2y}{dx^2}$, but that isn't the fault of the differentials, but of the notation. If you want a notation for second derivatives where you can split the fraction, it is:

$$ y'' = \frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$$

This comes from just applying the quotient rule to the first derivative (i.e., actually treating the first derivative as a quotient). You can get higher-order derivatives by continuing the process.

johnnyb
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