The derivative of $f(x)$ is the value of the limit of the average rate of change of $y$ with respect to $x$ as the change in $x$ approaches $0$. This is the value, in other words, that the average rate of change approaches but NEVER hits.

This means that it is NOT the infintesimal rate of change of $y$ with respect to $x$; $dy/dx$ merely approaches the derivative's value. If the rate of change did actually achieve $0$ change in $x$, you'd get $0/0$ which is an indeterminant form.

So if the derivative is the literal rate of change at an exact instant -- a rate of change with an interval of $0$, what does that actually tell you? Can a specific moment in time really have a rate of change? Is that rate of change ever even maintained, even at a specific instant?

I know that a point by itself can't have a rate of change, you need a continuum of points around it to determine one (hence a limit). What does an instantaneous rate of change tell you?

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    $\begingroup$ What do you think a speedometer measures in a car? $\endgroup$ – Bernard Dec 31 '15 at 1:24
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    $\begingroup$ The derivative also represents the slope of the line... Think of it that way, every point has a slope (we can draw it geometrically for many functions!) and the derivative is just the slope of that line. Since every point has a slope, the derivative must be defined at every point (assuming continuity and and other requirements for the derivative) $\endgroup$ – Brevan Ellefsen Dec 31 '15 at 1:40
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    $\begingroup$ Suppose you start driving and speed up steadily so that after 1 minute you are going 100 km/h. A friend asks you how fast you were going 15 seconds after you started driving. You object that there is no such thing as "rate of change at a single instant"! That has not been defined, you say. Only average rate of change has been defined. But your friend insists. He says, if it hasn't been defined, then define it yourself. He really wants a number that describes how fast you were going at that instant. Since he is your friend, and you don't want to say no, you invent the derivative. $\endgroup$ – littleO Dec 31 '15 at 2:06
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    $\begingroup$ "This means that it is NOT the infinitesimal rate of change of $y$ with respect to $x$". No, it does not mean that. This limit by definition is the infinitesmal rate of change of $y$ with respect to $x$. Before this limit was invented, the "infinitesmal rate of change of $y$ with respect to $x$" had no meaning. $\endgroup$ – Lee Mosher Dec 31 '15 at 2:14
  • $\begingroup$ @Bernard: +1; but to be pedantic, it measures rpm on the other side of the gear box ;) $\endgroup$ – Martin Argerami Dec 31 '15 at 12:11

You are certainly not alone in wondering about this! I should ask: in what sense do you mean the question?

a) If your question -- "can a specific moment in time really have a rate of change?" -- is directed towards the physical world, and the words "time" or "moment" are to be taken as referring to those things from our daily experience, then I'd tell you not to forget what math does: it doesn't constitute the real world, it just models it.

Perhaps the space we live in is actually discrete; i.e. if you zoom in close enough, our world is made of atomic "cells", just like a Minecraft world. Suppose each cell is a cube $1.6 \times 10^{-45}$ meters (ten orders of magnitude below the Planck length) on an edge. We don't know if this hypothesis is true or not: what experiment would disprove it? If it were true, then some things about real numbers that we learn in math (i.e. the idea of the limit is based, that for any number you name, I can always name a smaller one*), would be "wrong" for talking about objects on that size scale.

But it would still work just as well, as an approximation, for things that we currently use calculus for -- e.g. to calculate where to aim our spaceships. The rocket equations themselves are never going to fit the situation exactly (have you accounted for that dust particle? and that one?), the numbers we put into them are never going to be measured precisely.

A model cannot be judged right or wrong in itself; only the application of a model to a real-world situation can be judged, and then only in grades -- more appropriate or less appropriate. If speed comes in discrete chunks, then there may be no moment at which the volleyball, whose arc is described by $y = -x^2$, is ever moving at $-4$ meters/second calculus would predict at $x = 2$. Or maybe speed is continuous, and there is such a moment.

There's no way, even in principle, to tell, so we stick with the model we've got and change it only when it predicts the real world incorrectly.

b) But being less high-minded, it's helpful to have several ways to think about these things (and don't let anyone, including me, convince you that you have to think only their way about it).

As others have said, the derivative of a function $f(x)$ is a function $f'(x)$ which gives you the slope of the tangent line at $x$. If you believe that there can be a tangent line at a single point, then you can just think of that when others say "instantaneous rate of change".

*Here's the technical definition of a limit (ripped from Wikipedia), in case it helps. The statement

$$ \lim_{x \rightarrow 0} f(x) = L $$

means that you can make $f(x)$ as close to $L$ as you like by making $x$ sufficiently close to $0$. With variables, that's:

For every $\epsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x| < \delta$, then $|f(x) - L| < \epsilon$.

You can see how this would not work if there was a smallest real number -- then if I choose $\epsilon$ equal to that number, how are you going to make $|f(x) - L|$ smaller than it?

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    $\begingroup$ Excellent answer, but you should probably say "smallest positive real number". Also, there are concrete ways to quantify how good a mathematical model of reality is, such as through kolmogorov complexity with respect to a (sufficiently) large data set. With that we can give quite good mathematical meaning to our usually vague claims that real numbers are a good model of physical coordinates at the classical scale of say 1 nanometer to 1 parsec, namely that large amounts of experimental data can be explained by very short model descriptions. $\endgroup$ – user21820 Dec 31 '15 at 9:10

Given a function $f$ that relates time $t$ in seconds to the position $x$ (in meters) of a car moving along a straight line away from some relative to a point within the line:


We can find the velocity of the car between the interval $x \in [2,10]$ using the definition of velocity:

$$v=\frac{\Delta distance}{\Delta time}=\frac{f(10)-f(2)}{10-2}$$

But suppose we want to find the instantaneous velocity at $t=2$. Surely the interval $x \in [2,2.11]$ will be more representative as our cars velocity does not have as much time to alter from its velocity at $t=2$ than in the interval $x \in [2,10]$. But yet again we can do better, we can increase our accuracy by shortening the time interval. We look at the interval $x \in [2,2+h]$ and we notice that as we shorten our interval the car is less likely to accelerate .

Surely (unless your car is very unearthly) our car cannot significantly change instantaneous velocity in the matter of a $0.001$ second time period. Then we realize that we can let $h$ get as close as $0$ as possible and by doing that, we further close in on the moment $t=2$.

Now we have realized that if we look at the interval $x \in [t_0,t_0+h]$ and allow $h \rightarrow 0$ we get closer to studying exactly what happens at the time $t=t_0$. And we know from the definition of velocity that in this time interval our velocity will be:


(And you may realize that here we are calculating some sort of slope)

By choosing $h$ more and more close zero we realize we get a better and better representation of the instantaneous velocity at the moment $t=t_0$. (we also get closer to calculating a specific type of slope: the slope of the tangent line at $t_0$) .And when dealing with functions that we represent in terms of algebra, it is in our power to allow $h$ to get as arbitrarily close to $0$ as we like, getting an answer as arbitrary close to the actual instantaneous velocity as we like. Can we get a perfect answer, that just depends on wether or not you think:

An arbitrarily close answer (say $99.99..9%$ percent accuracy) is perfect.

And I think you'll agree that to some degree the above holds true. But in the real world finding the function that exactly predicts $x$ in terms of $t$ is the hard part, almost always too hard that we stick to models, which are nothing but approximations.


The derivative $f^\prime(x)$ of a function $f(x)$ gives you the slope of the tangent line at any point $x_0$ where both $f(x_0)$ and $f^\prime(x_0)$ are defined. Let the function for the tangent line at $x_0$ be $g(x)$. At a point $x_1$ a sufficiently small distance away from $x_0$, the function $f(x_1)$ has almost the same value as $g(x_1)$. So the instantaneous rate of change tells you how a function would approximately behave if you "zoomed in" on it close enough for it to appear linear.

Another way of understanding this is by imagining the function $f(x)$ representing the position vector of a car. If at a given point $x_0$ the car suddenly had constant acceleration equal to $0$, the car would continue moving with a rate of change equal to the derivative $f^\prime(x_0)$.


There's actually a hidden and more fundamental question, namely the completeness of the real numbers, that arises most simply in the intermediate value theorem. In layman terms, this theorem says that if you draw a curve on a piece of paper that begins in the top half and ends in the bottom half, then somewhere along the way you must have been exactly on the horizontal line between the two halves, if you didn't lift up your pencil throughout. Is this true? If you say it's obvious, say by repeatedly halving the interval of time that contains the crossing point, then you somewhat already accept the physical existence of limits of position, in this case the limit of the upper and lower bounds of these nested decreasing intervals.

Better still, the completeness property implies that there is a first crossing point. We can then say that there is an exact point in time before which we have never reached the middle line. In the same manner, velocity is the limit of change in position over change in time, if it exists. If a particle has instantaneous velocity $v$ at time $t$, we can plot its position with respect to time, and given any (positive) error margin $ε$ we can find some (positive) window size $δ$ such that the position-time graph for time from $t-δ$ to $t+δ$ lies strictly between two lines with gradients $v+ε$ and $v-ε$ respectively. This can be considered a physical barrier that the particle can never reach. Other than that, there is no concrete physical entity that corresponds to the instantaneous velocity.

Note that depending on the model we adopt for the real world, the above may be only partially or even totally inapplicable. If we take Einstein's relativity, then indeed velocity has to be defined as $\frac{dx}{dt}$ and you will see that velocity has by itself no direct meaning, in contrast to Newtonian mechanics where the (measurable) momentum is $mv$. If we go all the way to quantum mechanics, then it's no longer meaningful to talk about a particle having a point position, because it simply does not! Every particle's position is more accurately described by its wavefunction as a whole, which is never a point distribution but spread out over space. There is still instantaneous velocity, not of a point per se, but defined rather as the instantaneous rate of change of its wavefunction's mean position. An analogy would be the surface of a lake changing over time, where the particle corresponds to all the water in the lake and so the height of the water corresponds to the density at that point. It is easy to understand that while the mean position of the water molecules changes over time, it does not mean that all the water molecules are moving in any one direction in particular.

For the mathematical pedantic

Yes a curve that can be drawn on paper according to classical mechanics is only a special kind of continuous curve. Even so, whatever restrictions you impose, it's not mathematically necessary for the intermediate value theorem to hold if it wasn't Euclidean space.


Note that $\frac {dy}{dx}$ does not approach the limit, it is the limit. Physical measurements in the real world generally approximate the limit. The limit is part of an ideal model of the real situation.

It happens that the limit represents a useful tool within the models we have of the physical world, and that it behaves in useful ways within the model, which enable us to draw conclusions about the behaviours of the systems we study.

Mathematicians, of course, also study such models for their own sake.


Well, "instantaneous rate of change" is a polite way of talking about infinitesimals. In one of Newton's approaches to the calculus, he used infinitesimals but the atmosphere in the 17th century tended to be hostile to such creatures partly for reasons having to do with religious dogma. Therefore Newton had to develop a more palatable language that does not use dreaded things like the "infinitely small". George Berkeley specifically made fun of Newton over this issue, and claimed, correctly, that it amounts to the same thing. Where Berkeley went wrong was in thinking that there was a substantive problem with it. This had to do with his own philosophical prejudices, namely empiricism. For more details see this publication.

The instantaneous rate of change is the ratio of two infinitesimals.


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