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I am a first year calculus student so I would prefer if answers remained in Layman's terms.

It is common knowledge and seems to me a mantra that I keep hearing over and over again to "not treat differentials/derivatives as fractions".

I am of course, in particular, referring to Leibniz notation.

However, aside from a quick response such as "oh, it's because its not a fraction but rather a type of operator", I never really got a full answer as to why we can't treat it as such. It just kind of sits at the edge of taboo in my mind where it sometimes gets used and sometimes doesn't.

Confusion is further compounded when a lot of things seem to just work out if we treat them just as fractions (e.g. u-substitution / related-rates)


Air is being pumped into a balloon at a rate of $100cm^3/s$. We want the rate of change of radius when the radius is at $25cm$.

$$\text{we are given}\ \frac{dv}{dt}=100cm^3/s$$ $$\text{we want}\ \frac{dr}{dt}\ \text{when}\ r=25cm$$ Thus we will solve this by using the relation $v=\frac{4}{3}\pi r^3$ $$\frac{dv}{dt}=\frac{dv}{dr}\frac{dr}{dt}$$ $$\frac{dv}{dt}\frac{dr}{dv}=\frac{dr}{dt}$$ $$100\frac{1}{4\pi r^2}=\frac{1}{25\pi}$$ So the answer is $\frac{dr}{dt}=\frac{1}{25\pi}$ when $r=25cm$


Note the manipulation of derivatives just as if they were common fractions using algebra.

So my question is: when can I treat differentials/derivatives as fractions and when can I not? Please remember I am a first year student and aim the answer as such as opposed to one that is mathematically rigorous but difficult to comprehend by a beginner

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    $\begingroup$ Related: mathoverflow.net/questions/73492/…, math.stackexchange.com/questions/21199/… $\endgroup$ – Hans Lundmark May 14 '16 at 7:15
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    $\begingroup$ In addition to the link given by @Hans, also look at math.stackexchange.com/questions/linked/21199, there's a good chance that your question was answered there at one point or another. $\endgroup$ – Asaf Karagila May 14 '16 at 8:59
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    $\begingroup$ Sometimes I worry that people have gone too far in abandoning "infinitesimal intuition". It's simple and clear, and you can see how effectively people like Feynman use it when you read physics books. It's not some strange coincidence that it works. There is a saying that "too much rigor teaches rigor mortis". $\endgroup$ – littleO May 14 '16 at 9:11
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    $\begingroup$ In your example Alan, you "treated differentials/derivatives as fractions" despite being told by your instructor (or whoever) not to do so. I find it redundant that they say "do not ..." when you essentially have to do just that when solving many calculus problems by hand. So to reiterate, we know that derivatives are certainly not quotients, but in practice it's entirely safe to "treat" them as such. The notation $\text{d}y/\text{d}x$ is intuitive and handles nicely in equations, but it's also a bit misleading. $\endgroup$ – Corellian May 14 '16 at 18:30
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    $\begingroup$ You should take a look at non-standard analysis $\endgroup$ – bjb568 May 14 '16 at 20:33
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I'll just make two extended comments.

First, if you'd like to treat $dy/dx$ as a fraction, then you need to do two things:

  • (1) Have a clear, precise mathematical definition of what $dy$ and $dx$ are, and
  • (2) Have a way of dividing the quantities $dy$ and $dx$.

There are a few ways of answering (1), but the most common answer among mathematicians -- that is, to the question of "what are $dy$ and $dx$ really?" -- is somewhat technical: $dy$ and $dx$ are "differential forms," which are objects more advanced than a typical calculus course allows.

More problematic, though, is (2): differential forms are not things which can be divided. You might protest that surely every mathematical object you can think of can be added, subtracted, multiplied, and divided, but of course that's not true: you cannot (for example) divide a square by a triangle, or $\sqrt{2}$ by an integral sign $\int$.

Second, every single instance in which expressions like $dy/dx$ are treated like fractions -- like, as you say, $u$-substition and related rates -- are just the chain rule or the linearity of derivatives (i.e., $(f+g)' = f' + g'$ and $(cf)' = cf'$). Every single instance.

So, yes, $dy/dx$ can be treated like a fraction in the sense (and to the extent) that the Chain Rule $dy/dx = (dy/du)(du/dx)$ is a thing that is true, but that's essentially as far as the fraction analogy goes. (In fact, in multivariable calculus, pushing the fraction analogy too far can lead to real issues, but let's not get into this.)

Edit: On the OP's request, here are examples of fraction-like manipulations which are not valid: $$\left( \frac{dy}{dx} \right)^2 = \frac{(dy)^2}{(dx)^2} \ \ \text{ or } \ \ 2^{dy/dx} = \sqrt[dx]{2^{dy}}.$$ Because these manipulations are nonsensical, students are often warned not to treat derivatives like fractions.

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    $\begingroup$ @JesseMadnick Over the hyperreals, the first two are legitimate manipulations. The third is legitimate through two applications of the chain rule: $$ \frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}u}{\mathrm{d}v} = \frac{\mathrm{d}y}{\mathrm{d}v} \frac{\mathrm{d}v}{\mathrm{d}x} \frac{\mathrm{d}u}{\mathrm{d}v} = \frac{\mathrm{d}y}{\mathrm{d}v} \frac{\mathrm{d}u}{\mathrm{d}x} $$ $\endgroup$ – u54112 May 14 '16 at 10:45
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    $\begingroup$ This answer is simply wrong. It's written on the assumption that the only way to formalize dy/dx as a ratio is through differential forms, when in fact there are many ways of doing so. One way of doing so is NSA, and examples like $(dy/dx)^2=dy^2/dx^2$ are perfectly OK in NSA. $\endgroup$ – Ben Crowell May 14 '16 at 20:48
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    $\begingroup$ @BenCrowell - your comment is simply wrong. The post does not imply that differential forms are the only formalization of differentials, only the most common (which is true), and that any formalization is too complex for first year calc students, which is even more true of NSA than it is of differential forms. Abstraction is still a foreign concept for them. $\endgroup$ – Paul Sinclair May 14 '16 at 23:27
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    $\begingroup$ @BenCrowell: By not mentioning NSA, I am perhaps lying by omission, granted, but I think the context of $(dy/dx)^2 \neq (dy)^2/(dx)^2$ is clear. If I were to say that the infinite sum $\sum_1^\infty 2^k$ diverges, would you protest that I'm wrong by failing to mention its convergence in the $2$-adics $\mathbb{Q}_{2}$? $\endgroup$ – Jesse Madnick May 15 '16 at 5:41
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    $\begingroup$ @Alan Distinguishing infinitesimals from ordinary real numbers, i.e. the immeasurable from the measurable so to speak. The motivation is that an infinitesimal's magnitude is greater than zero but less than any positive real number. Note: This concept is nothing to sweat over. It tends to invite scary topics like hyperreal numbers or smooth infinitesimal analysis, which are probably more than you bargained for. For the time being, epsilon-delta limits and limits at infinity are very well-equipped to tackle everything in basic calculus. Any decent calculus text/course will cover them. $\endgroup$ – Corellian May 15 '16 at 6:37
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First, $dx$ and $dy$ are in fact differential forms: things that given a point and a vector with this point as origin gives us some value, linear and antisymmetric in the vector argument, continuous / differentiable / smooth in the point argument.

Now, by Newton-Leibniz, any differential form on $\mathbb{R}$ is of the form $dy = f(x)dx$, where $dx$ is a differential form such that $dx(x, h) = h$ (here, $h$ is a one-dimensional vector - you can treat it as a displacement of $x$).

So, we can try to define division like $\frac{dy}{dx} = \frac{f(x)dx}{dx} = f(x)$. While it works for now, it fails in higher dimensions.

Suppose that we are on a plane, having two basis differential forms: $dx_1$ and $dx_2$ ($dx_i$ is just projection on the $i$-th coordinate). Again, any differential form is $dy = f_1(x)dx_1 + f_2(x)dx_2$. Divide by $dx_1$: $\frac{dy}{dx_1} = f_1(x) + f_2(x)\frac{dx_2}{dx_1}$. We could say here that $\frac{dx_2}{dx_1}$ is zero, since the components of a vector are independent, but let's actually do the division. Let $h = (h_1, h_2)$ be the displacement vector, then $\frac{dx_2}{dx_1}(x,h) = \frac{h_2}{h_1}$. Wow, this is surely not equal to zero, but measures some kind of relative displacement in coordinates. The point is that it depends on $h$ now, and the result of the division cannot be just a function of $x$.

What one really wants here is a some kind of dot product, since, for example, dot product with basis vector gives the corresponding coordinate. Here, this "dot product" arises naturally: take a form $dy$, and plug the basis vector in it: $dy(x, e_1) = f_1(x)dx_1(e_1)+f_2(x)dx_2(e_1) = f_1(x)$ (since $dx_1(e_1) = 1$ and $dx_2(e_1) = 0$). Why $e_1$? It is a vector field dual to the form $dx_1$, that's why.

So, although it looks like a fraction, it's actually more a dot product.

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    $\begingroup$ IMO, $dy / dx$ is still ratio (and in your final example, $dy / dx_1$ is undefined since they aren't multiples of one another... unless $x_1$ and $x_2$ are dependent in which case $dy / dx_1$ is the function you would get by treating it as a ratio), whereas it is $\partial y / \partial x_1$ is more like a vector product. $\endgroup$ – user14972 May 14 '16 at 12:12
  • $\begingroup$ As a physicist I believe that $\mathrm{d}y/\mathrm{d}x$ is a ratio because it has the dimensions of that ratio. As for generalising to two independent variables we have $\mathrm{d}f = \frac{\partial f}{\partial x_1} \mathrm{d}x_1 + \frac{\partial f}{\partial x_2} \mathrm{d}x_2$, which indeed reduces to the one dimensional case when one of the variables is held constant. $\endgroup$ – u54112 May 14 '16 at 12:54
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    $\begingroup$ This answer is wrong for the same reason that Jesse Madnick's answer is wrong. dx and dy are in fact differential forms Nope, this is simply untrue. There are multiple ways of formalizing these symbols, and differential forms are just one of these ways. $\endgroup$ – Ben Crowell May 14 '16 at 20:50
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Suppose $\Delta x$ is a tiny (but finite and nonzero) real number and $\Delta f$ is the amount that a function $f$ changes when its input changes from $x$ to $x + \Delta x$. Then, it's not true that $\Delta f = f'(x) \Delta x$ (with exact equality), but it is true that $\Delta f \approx f'(x) \Delta x$. You are free to manipulate $\Delta x$ and $\Delta f$ however you like, just as you would with any real numbers, so long as you rememember that the equations you derive are only approximately true. You can hope that "in the limit" you will obtain exactly true equations (as long as you are careful).

For example, suppose that $f(x) = g(h(x))$. Then \begin{align} f(x + \Delta x) &= g(h(x+\Delta x)) \\ &\approx g(h(x) + h'(x) \Delta x) \\ &\approx g(h(x)) + g'(h(x)) h'(x) \Delta x, \end{align} which tells us that \begin{equation} \frac{f(x+\Delta x) - f(x)}{\Delta x} \approx g'(h(x)) h'(x). \end{equation} And it certainly seems plausible that if we take the limit as $\Delta x$ approaches $0$ we will obtain exact equality: \begin{equation} f'(x) = g'(h(x)) h'(x). \end{equation}

These kinds of arguments, introducing tiny changes in $x$ and making linear approximations using the derivative, are the essential intuition behind calculus.

Often, arguments like this can be made into rigorous proofs just by keeping track of the errors in the approximations and bounding them somehow.

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    $\begingroup$ Khinchin (Eight Lectures on Mathematical Analysis, section 32) does exactly this, and then defines $df = f'(x) \Delta x$; that is, $df$ is a function of $x$ and $\Delta x$. Then he shows rigorously that $\alpha = df - \Delta f$ is not merely bounded but is infinitesimal when compared with $\Delta f$, and that $df$ is the only linear function of $x$ and $\Delta x$ that is so small. This characterization can be used to define $df$ without $f'$ if desired. Using this, it's not merely plausible but actually provable that $\frac{df}{dx} = \frac{dg}{dh}\frac{dh}{dx}$ as you stated. $\endgroup$ – MJD Jun 16 '17 at 17:10
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The historical truth of the matter is, when Calculus was being invented, mathematicians were not already using limits or $\epsilon$ - $\delta$ arguments or anything nearly rigorous. All of their arguments were built on reasoning algebraically using infinitesimals - quantities assumed to be infinitely small yet still non-zero. When Leibnitz introduced the ratio $dy/dx$ he intended it to be an actual ratio of infinitesimals. When a change of variables in an integral produced an expression such as $dy = 2x$ $dx$, it literally meant that the infinitesimal $dy$ was thicker than $dx$ by a factor of $2x$ (and this scaling was necessary to get the integral - as a sum of infinitely many infinitesimals - to work out right).

"But aren't infinitesimals logical nonsense?", you ask. In a word, yes - and a whole lot of not-at-all-stupid people repeatedly pointed this out, even at the time. The general response was, essentially: "Well it all works, so stop bugging us about it." But after a couple hundred years of that, mathematicians finally had to fix the problems that infinitesimals were causing; it is at this point that $\epsilon$ - $\delta$ arguments and limits were invented.

So, if infintesimals are out, why still $dy/dx$? Aside from mere historical inertia, the fact is that treating these expressions as fractions is no accident - Leibnitz had thought hard about notation, and chose this form so that simple intutive algebraic manipulation of formulae would tend to yield correct results in analysis. In short, the notation is a great intuition builder (Newton's notation, not so much - which is why it is more rarely used).

For the modern analyst: Think of these expressions as 'bookkeeping terms', whose proper manipulation helps maintain internal relations among formulae which are required for your arguments to adhere to the underlying theorems that justify them (ultimately tied to relationships among the $\epsilon$s and $\delta$s in their proofs).

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For better or worse, introductory calculus is only expressed in terms of functions and real-valued expressions.

On the one hand, this has the advantage that you don't need to learn about any "new" sorts of objects, since functions and real-valued variables are presumably something already familiar to you.

Differentials $\mathrm{d}x$ are a new kind of object you haven't learned about yet — by the above philosophy, your class wants to avoid dealing with them.

Fortunately, the whole expression $\frac{\mathrm{d}y}{\mathrm{d}x}$, when it makes sense, has the advantage that it is an ordinary real-valued expression; thus, you can avoid having to deal with new sorts of objects if you are always careful to work with such expressions as a whole.

However, this does not cripple you, because you learn or can prove theorems like

$$ \frac{\mathrm{d}z}{\mathrm{d}w} \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}z}{\mathrm{d}x} \frac{\mathrm{d}y}{\mathrm{d}w} \qquad \text{and} \qquad \frac{\frac{\mathrm{d}z}{\mathrm{d}x}}{\frac{\mathrm{d}y}{\mathrm{d}x}} = \frac{\mathrm{d}z}{\mathrm{d}y} \qquad \text{and} \quad \frac{\mathrm{d}x}{\mathrm{d}x} = 1 $$

which basically encompass everything you would want to with these expressions, even if you did know about the sort of object that $\mathrm{d}x$ and did want to treat Leibniz notation as a ratio.


Looking towards the future, the interpretation as a ratio becomes less useful.

Differentials still make sense, but often $\mathrm{d}x$ and $\mathrm{d}y$ aren't multiples of one another, so $\frac{\mathrm{d}y}{\mathrm{d}x}$ wouldn't even make sense.

Worse, the similar notation for partial derivatives; e.g. $\frac{\partial z}{\partial x}$ has its own quirks and problems that make it actively misleading to think of it as a ratio of ${\partial z}$ to ${\partial x}$; a particular theorem of note is that

$$ \frac{\partial y}{\partial x} \frac{\partial z}{\partial y} \frac{\partial x}{\partial z} = -1$$

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    $\begingroup$ In fact the problem is that the partial derivative point you made is that not all the relevant information is being specified. When you label these differentials with what is held fixed, you see there is no contradiction; for example, the first $\partial y$ is really $(\partial y)_z$ while the second is $(\partial y)_x$, and there is no reason to believe these should be the same (even "up to a higher order infinitesimal", whatever that means). $\endgroup$ – Ian May 14 '16 at 12:19
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    $\begingroup$ @Ian What if you used $(\partial y)_w$ instead (and similar for the other differentials)? $\endgroup$ – PyRulez Nov 30 '17 at 21:12
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$dy\over dx$ is by definition a limit of a function that maps $x$ to $y$.

it is a symbol, a way of writing that is agreed upon. it could as well be a little doggy sign but that would be unhelpful.

that certain "fraction-like" way of expressing a certain limit comes to help us, humans, use a proven mathematical law. that law states that the derivative of a composite function $g \circ f$ equals to the derivative of $g$ times the derivative of $f$. that is not trivial.

the fact that many of us get confused as to why we can treat it like it's a fraction comes to show how efficient this notation really is.

let's math it out:

let $f:x\rightarrow y$ be differentiale at any $x$.

let $g:y\rightarrow z$ be differentiale at any $y$.

${dz\over dx}:=\lim_{\Delta x\to0}{\Delta z\over \Delta x}=\lim_{\Delta x\to0}{\Delta z\over \Delta y}{\Delta y\over \Delta x} $

now since we know $f$ and $g$ are differentiable we know that their limits exist, which means we can do the following:

$\lim_{\Delta y\to0}{\Delta z\over \Delta y}\cdot\lim_{\Delta x\to0}{\Delta y\over \Delta x}={dz\over dy}\cdot{dy\over dx}$

thus proving that ${dz\over dx}={dz\over dy}\cdot{dy\over dx}$

but ohh would you look at that! that looks as if we reduced a fraction!

this is so very far from being a full rigorous proof but I do hope some of it helped to your understanding.

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  • $\begingroup$ Can you give an example and follow it through logically as you did, but this time it "doesnt work"? I think it would be very helpful in illustrating a tangible difference $\endgroup$ – AlanSTACK May 14 '16 at 8:46
  • $\begingroup$ in order for it not to work we need the process to fail when we break apart the one limit into two different limits. which means that one the limits does not exist , which is to say one of the functions (f or g) is not differentiable at the point. for example : if f(x)=x and g(z)=sqrt(z) thus gf(x)=sqrt(x). gf(x) is not differentiable at 0 since g(z) isn't. which means that you can not use the "chain rule" ( that's the rule we are discussing) for x=0, and it becomes obvious if you try to plug 0 into 1/(2*sqrt(x)) which is the derivative of gf(x) at any other point :-) $\endgroup$ – Rubenz May 14 '16 at 9:13
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Even though this is already closed, I am not pleased with any of these answers personally, and I think I can still say something that might be useful.

First off: when we write something like $$\frac{\text{d}y}{\text{d}x},$$ the symbols $\text{d}y$ and $\text{d}x$ do NOT represent differential forms. A differential form is a linear, antisymmetric $n$-form: a linear opperator that takes in $n$ vectors and spits out a number. These symbols need not represent that, they can represent actual real numbers. Let's make this formal.

Imagine we have a real function $f\space\colon\space\mathcal{D}\subseteq\mathbb{R} \longrightarrow \mathbb{R}$ which is differentiable ($\iff$ derivable) in $\mathcal{D}$. If we have $y=f(x)$ and we make a change $\Delta x$ to the independent variable, the dependent variable will change by $\Delta y$, so $$y+\Delta y=f(x+\Delta x) \implies \Delta y=f(x+\Delta x)-y=f(x+\Delta x)-f(x)$$ Let's define a function $\eta(\Delta x)$ just because: $$\eta(\Delta x) \equiv \frac{f(x+\Delta x)-f(x)}{\Delta x}-f'(x)$$ Here, $f'(x)$ stands for the derivative function of $f$, which we know exists because $f$ is differentiable. This function $\eta(\Delta x)$ has the following property:

$$\lim_{\Delta x \to 0}{\eta(\Delta x)}=f'(x)-f'(x)=0$$

This will be useful in a second. Let's solve for $\Delta y$ in the definition of the function $\eta(\Delta x)$:

$$\Delta y = f'(x)\Delta x + \eta(\Delta x)\Delta x$$

If we make $\Delta x$ go to $0$, we find that the first term goes to $0$ linearly (at the same rate than $\Delta x$ itself does), while the second term, as a consequence of the limit of $\eta(\Delta x)$ being $0$, goes to $0$ faster than $\Delta x$ does. This means that the first term ($f'(x)\Delta x$) is the linear part of $\Delta y$. You might recognise this as the first term of the Taylor expansion of $f$ around a point $x$.

We DEFINE the differential $\text{d}y$ of a variable $y$ as the linear part of its variation (the one that changes like $\Delta x$ and not like $(\Delta x)^k$ for some $k \neq 1$). And you can see from this definition that it makes sense to say that $\text{d}x=\Delta x$ because $\Delta x$ obviously changes like $\Delta x$ does.

So, we have:

$$\text{d}y = f'(x) \text{d}x \implies \frac{\text{d}y}{\text{d}x} = f'(x)$$

Notice that we have always taken $x$ as a fixed point in which $f(x)$ is differentiable, which means that we should really write:

$$\left(\frac{\text{d}y}{\text{d}x}\right)_{x=a}=f'(a)$$

Knowing this, it's not at all surprising that, given $u=\varphi(x)$, $w=\phi(x)$ and $y=f(x)$ differentiable at the points we're interested in, we can prove...

$$\left(\frac{\text{d}y}{\text{d}u}\right)_{u=\varphi(a)}\left(\frac{\text{d}u}{\text{d}x}\right)_{x=a}=\left(\frac{\text{d}y}{\text{d}x}\right)_{x=a}$$

... which we call the chain rule; and, by extension...

$$\left(\frac{\text{d}y}{\text{d}u}\right)_{u=\varphi(a)}\left(\frac{\text{d}w}{\text{d}x}\right)_{x=a}=\left(\frac{\text{d}y}{\text{d}x}\right)_{x=a}\left(\frac{\text{d}w}{\text{d}u}\right)_{u=\varphi(a)}$$

... and its many implications (the inverse of a derivative is the derivative of the inverse function, if there exists one in the first place, etc.)

Notice that this manipulations are ALWAYS TRUE, GIVEN THAT a derivative exists in the first place. Basically, this is saying "manipulating differentials is perfectly okay, but you are supposing that these things are differentiable instead of proving it". If you are doing a proof you won't generally do this kind of manipulations, but it's perfectly fine to cancel, move and do anything with first-order total differentials when solving differential equations or doing any physics problem.

This isn't new, though. Treating derivatives as fractions is just as dangerous as treating good old fractions as fractions. Just like with differentials, doing a manipulation like $$\frac{x}{y}\frac{y}{z}=\frac{x}{z}$$ is implying that you know that all these fractions exist (i.e. you know that $y,z\neq0$, specially $y$ as it's the one you're "cancelling" here).

So, doing regular fraction cancellation can lead you to problems. For example, if you're given the equation...

$$x^2+5x=0 \implies x(x+5)=0$$

... and you're asked to solve for $x$, you could think you can divide both sides by $x+5$ and get $x=0$ as the only solution, but doing the cancellation of the $x+5$ on the numerator by the one on the denominator implied knowing in advance that $x+5\neq0$, which is why this method didn't give you also the solution $x=-5$.

I hope this helped, thank you for reading. And, oh, as a final comment, the definition of the symbol $\text{d}y$ as a differential form is related to the one I gave here by the following equation (given $y = f(x^1,...,x^n)$ a general scalar function):

$${\text{d}\tilde{y}}(\text{d}\vec{x}) = \text{d}y$$

Here, ${\text{d}\tilde{y}}$ is a differential 1-form called the gradient of $f(\vec{x})$ (the covariant form of $\vec{\nabla}f$) and $\text{d}y$ is the linear part of $\Delta y = f(\vec{x}+\Delta \vec{x}) - f(\vec{x})$.

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I am a layman, so I will put my take on the presented related rates problem in layman's terms. To avoid the deep stuff in a case like this, I try to apply the chain rule directly and not worry about treating derivatives as fractions and differentials as numbers. Volume is explicitly given as a function of radius and since the instantaneous rate of change of volume with respect to time is given, volume would also seem to be a function of time. Equating these two expressions for volume allows that time is a function of radius and radius is a function of time, disallowing negative time.

I like to write $v=v(r)$ and $r=r(t)$ to remind me of what I'm doing, expecting that the quantities on either side of the equals sign should be interchangeable. Then:

$dv/dt=d(v(r))/dt=d(v(r(t))/dt$ and ${dv\over dt}={{d({4\over3}\pi(r(t))^3)\over d(r(t)}}$ ${d(r(t)\over dt}$$=(4\pi r^2){dr\over dt}$

Although the notation might be a little different, I believe this is one way to solve the problem similar to calculus textbooks and the notation aids me in the understanding. But for all I know it may be just as confusing as the rest of it and not even be correct or widely applicable or relevant, in which case a real mathematician can surely throw it out.

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This is something that hasn't been mentioned yet. While I'm not aware of cases where $\frac{dy}{dx}$ fails when treated as a fraction and gives incorrect results, the notation $\frac{d^2y}{dx^2}$ will give incorrect results if treated as such. Keep in mind that this is a shorthand for $\frac{d^2y}{(dx)^2}$. Suppose $$x=sin(t)^2+t$$ $$y=t^2+1$$ and we want to find $\frac{d^2y}{(dx)^2}$ implicitly (that is, without writing y in terms of x in order to do it, which might be hard or not even possible). Finding $\frac{dy}{dx}$ will work fine by treating the differentials as fractions: $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dx}}=\frac{2t}{2sin(t)cos(t)+1}$$ which is the correct result that you will get with a direct approach: $$\frac{dy}{dx} = \frac{d(t^2+1)}{dx} = \frac{dt}{dx}2t$$ Finding $\frac{dt}{dx}$: $$\frac{dx}{dx} = \frac{dt}{dx}2cos(t)sin(t) + \frac{dt}{dx}$$ $$1 = \frac{dt}{dx}(2cos(t)sin(t)+1)$$ $$\frac{1}{2cos(t)sin(t)+1} = \frac{dt}{dx}$$ Substituting $$\frac{dy}{dx} = \frac{1}{2cos(t)sin(t)+1}(2t)$$ Which is the same result. However, if you try to do this for $\frac{d^2y}{(dx)^2}$, you will get the wrong answer: $$\frac{d^2y}{(dx)^2} = \frac{d^2y}{(dt)^2}\frac{(dt)^2}{(dx)^2} = \frac{d^2y}{(dt)^2}(\frac{dt}{dx})^2 = \frac{d^2y}{(dt)^2}\frac{1}{(\frac{dx}{dt})^2}$$ $$\frac{d^2y}{(dt)^2} = 2$$ $$\frac{dx}{dt} = 2sin(t)cos(t)+1$$ hence, by substituting $$\frac{d^2y}{(dx)^2} = (2)\frac{1}{(2sin(t)cos(t)+1)^2}$$ This can be shown to be false by taking the derivative of $\frac{dy}{dx}$ with respect to x in a direct way: $$\frac{d(\frac{dy}{dx})}{dx} = \frac{d(\frac{2t}{2sin(t)cos(t)+1})}{dx} = \frac{2\frac{dt}{dx}(2sin(t)cos(t)+1) - \frac{dt}{dx}(2cos(t)^2-2sin(t)^2)(2t)}{(2sin(t)cos(t)+1)^2}$$ Substitute $\frac{dt}{dx} = \frac{1}{2sin(t)cos(t)+1}$ $$\frac{d(\frac{dy}{dx})}{dx} = \frac{4sin(t)cos(t)+ 2 - 4tcos(t)^2-4tsin(t)^2}{(2sin(t)cos(t)+1)^3}$$ which is not the same. I actually discovered this because I made this mistake when I put aside my skepticism and trusted Leibniz notation to give correct results despite not being well-defined. Leibniz notation for higher derivatives in general does not give correct results when manipulated as a fraction. The notation for higher derivatives must be modified for these manipulations to work. See this paper

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