TITLE: When can we not treat differentials as fractions? And when is it perfectly OK?
QUESTION [50 upvotes]: Background
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)

Example

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.

Question
When exactly can I treat differentials/derivatives as fractions and when can I not?
Please keep in mind that at the end of the day, I am a first year college student. An answer that is easy to understand is preferred over one that is more mathematically rigorous but less friendly to a beginner such as me.

REPLY [3 votes]: 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 $\Delta y = f(\vec{x}+\Delta \vec{x}) - f(\vec{x})$.