Why does 3blue1brown use the "around a point" to describe a derivative?

Question

In this article (which includes a link to the video version of the article as well), Grant Sanderson aka 3blue1brown describes a derivative. He says at the end of the passage headed "The Paradox", "Since change in an instant still makes no sense, rather than interpreting the slope of this tangent line as an “instantaneous rate of change”, an alternate notion is to think of it as the best constant approximation for rate of change around a point".

However, I'll be the devil's advocate and disagree with him. I particularly take issue with his usage of the word "around a point". I think that the slope of the tangent line is the slope of the curve at that exact point, not around that point. I'll present two quotations in favor of my case:

1. To partially quote myself from my most recent question, Let us consider 2 different points $A$ & $B$ of the above graph. Now, if we find the slope of the secant line $AB$, it'll be an approximation of the slope of $A$. If we pick a point that is closer to $A$ than $B$, $C$, the slope of $AC$ will be a better approximation of $A$'s slope. Now, if we know what the value is that the slopes of the secant lines are approaching as the points are getting closer and closer to $A$, we will be able to find the best approximation and the most correct answer of the slope of that point: the approached value. It is the best approximation because we know that the approximations are getting better and better as the approximations are getting closer and closer to the approached value, so the approached value is the most accurate approximation, and it is the slope of the curve at that exact point, not around that point. We can calculate this approached value by taking the limit:

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

$f'(x)$ is the approached valued, and it is known as the derivative of $f$ at $x$.

2. To quote @Javier's comment to this question, "Maybe you should try not thinking of limits as movement, because then, as you say, you never "get there". Rather, when you see a limit like the derivative, imagine that there is a number that you cannot calculate, but that you can approximate with arbitrarily high precision. This is not some fuzzy thinking that cheats by evading the concept of instantaneous rate of change; if you can approximate a number arbitrarily well, then you know exactly what it is, even if you cannot calculate it "directly". With this point of view, the limit is not a process that will never end. Instead, it's an indirect way of specifying (without ambiguity) a number that you couldn't otherwise calculate. Maybe this will help."

In short, I think a derivative is the slope of a graph/curve at an exact point, not near that point or around that point.

I think 3blue1brown's reasoning can lead to problems. In the passage titled "The Paradox at Time Zero", he essentially argues that at time zero, the car is not static even though the derivative gives us $0$ at that point. He says, "For smaller and smaller nudges in time, this ratio of the change in distance over change in time approaches $0$, though in this case it never actually hits it". However, I'd argue that as we can see that as our approximations approach $0$ we are getting more accurate. As we get arbitrarily close to zero, our approximations become arbitrarily accurate. So, we can understand that the most accurate approximation is the approached value as our approximations are getting better and better as we are getting closer and closer to the approached value. See @Javier's comment above.

Questions:

1. Am I correct or is 3blue1brown correct?

Answer 1

Grant Sanderson says "Since change in an instant still makes no sense, rather than interpreting the slope of this tangent line as an “instantaneous rate of change”, an alternate notion is to think of it as the best constant approximation for rate of change around a point".

However, I take issue with his usage of the word "around a point". I think that the slope of the tangent line is the slope of the curve at that exact point, not around that point. I think a derivative is the slope of a graph/curve at an exact point, not near that point or around that point. Am I correct or is 3blue1brown correct?

You're correct, and not at all disagreeing with Grant. Read carefully: at no point (haha) did Grant claim that the derivative at a point is the slope around the point.

The boldfaced portion is correct: “instantaneous change” indeed makes no sense, however “instantaneous rate of change” makes perfect sense. While rate of change at a given instant is a meaningful concept, it is not meaningful to speak of change at/across an instant, because change has to be measured over an interval, since an instant is infinitesimally/immeasurably small.

(You're in a vehicle moving at 60km/h. At any instant, you undergo no displacement and therefore have no instantaneous change. However, at every instant, your instantaneous rate of change is 60km/h.)

Grant is saying that even though derivative means “instantaneous rate of change”, that's perhaps not the most elucidating way to understand the idea; he's proposing that we instead think about the (multiple copies of) rates of change “around” the point in question, and that the required derivative is the “best” tangent line you can find at that point (based on these multiple copies).

Even though $f$ needs to be defined at $c$ for $f$ to be differentiable at $c,$ notice that the difference quotient $\frac{f(x+h)-f(x)}{h}$ is meaningful only for $x\neq c$ (i.e., $h\neq0$).

To quote Blue:

---

Reply to the OP's comment:

"However, at every instant, your instantaneous rate of change is 60km/h."-this is closer to my line of thought; however, I think Grant would disagree with you; see below: he doesn't agree that the velocity of the car is $0$ at time $t=0.$

he essentially argues that at time zero, the car is not static even though the derivative gives us $0$ at that point.

I gather from your reports/summaries that you're misinterpreting Grant: him saying that the car is non-static is not the same as him saying that its velocity is nonzero!

• zero derivative $\kern.6em\not\kern-.6em\implies$ static car $\quad\longleftarrow$ Grant claims this.
• non-static car $\implies$ nonzero velocity $\quad\longleftarrow$ Grant does not claim this.