What does the derivative of a function at a point describe?

Question

I understand that the derivative of a function $f$ at a point $x=x_{0}$ is defined as the limit $$f'(x_{0})=\lim_{\Delta x\rightarrow 0}\frac{f(x_{0}+\Delta x)-f(x_{0})}{\Delta x}$$ where $\Delta x$ is a small change in the argument $x$ as we "move" from $x=x_{0}$ to a neighbouring point $x=x_{0}+\Delta x$. What confuses me is how to interpret its meaning correctly, that is, what does the derivative $f'(x_{0})$ actually describe?

On Wikipedia it says that "the derivative of a function quantifies the rate at which the value of the function changes as we change the input" (or words to that effect). However, the function has a particular constant value, $f(x_{0})$ at a given point $x=x_{0}$ so how can one meaningfully discuss the rate at which the value of the function is changing at that point?

Would it be correct to interpret the derivative of a function at a point as describing how "quickly" it's value changes as we move from that point to (infinitesimally close) neighbouring points? (As such in the example above, in moving from the point $x_{0}$ to $x_{0}+\Delta x$ the value of the function $f$ changes by an amount $f'(x_{0})\Delta x$ for infinitesimally small change $\Delta x$). Is it then simply that the value of the derivative at that point equals the slope of the tangent line to the the function (curve) at that point? (In general then, the derivative of a function is itself a function whose value at each point equals the slope of the tangent line to the curve at that point).

The derivative evaluated at a specific point measures the rate of change of a given function at an instant. — Edward Evans, Mar 30 '16 at 19:50
I don't see the difference between what you are describing in your last paragraph and the Wikipedia quote in italics. — John Douma, Mar 30 '16 at 19:57
Your intuition is correct altought ordinary textbooks wouldn't mention "infinitesimals" (even if Newton and Leibniz actually did) — Marco Disce, Mar 30 '16 at 20:00
Instantaneous rate of change seems intuitive from our experience watching the speedometer as we're driving. If you look at a car's speedometer at 1 pm, and it tells you that your velocity at that instant in time is 45 miles per hour, you wouldn't object that your car's position had a particular constant value at 1 pm. — littleO, Mar 30 '16 at 20:11
Unless it is an inflection point, then the derivative would give the slope of the line tangent to the curve at that point. — MaxW, Mar 30 '16 at 20:14
You pretty much have it. smooth functions have instaneous rates of change that might be constantly changing. The derivative is a formula to find the instantaneous rates of change. You sort of run into a Xeno paradox like problem of how can you have "change" if you have no "time" to observe the change, but that's the entire point of calculus. I, personally, like the idea of a tangent line as a solution to the paradox. The derivative is a formula for the slope of these tangent lines. Your intuition is good. — fleablood, Mar 30 '16 at 20:26
@littleO - Would saying "the derivative of a function $f$ evaluated at $x_{0}$ measures the rate at which the value of the function changes with respect to a change in its argument at the point $x_{0}$" (denoted $\frac{df}{dx}\vert_{x=x_{0}}$) be okay as well? — Taylor Rendon, Jan 04 '21 at 01:26
I think that's fine, but the phrasing "with respect to a change in its argument" might not be quite standard. But yes, that's the idea: If the input to $f$ increases by a small amount (starting from $x$), then does the output of $f$ increase by a relatively large amount? If so, then I'd say that $f$ is increasing rapidly at $x$, or that the instantaneous rate of change of $f$ at $x$ is large. — littleO, Jan 04 '21 at 01:57

score 2 · Answer 1 · answered Mar 30 '16 at 20:15

2

I think your hung up on the idea that a rate needs an associated time interval that it applies to. This isn't true, I think most people probably intuitively think this but consider the following scenario.

Think about a ball being dropped from a building. At every possible moment the ball is traveling a different velocity because it is constantly accelerating due to the force of gravity. Velocity, of course is a rate, specifically, its the rate at which the position of the ball is changing. So, no matter how hard you look, there is no interval, no matter how small, at which the velocity of the ball is a specific number. It will only be traveling at a specific velocity at one moment in time. So, the derivative of the balls position, at some time t is that one exact point in space where the ball will be traveling at that velocity. See? no interval needed.

If it helps you to think about the rate as a difference over some infinitesimal time than go for it, the definition of a limit above says that a specific moment and this infinitesimal difference are the same thing. Thats part of why the results of calc are so cool. I think once you start to really feel like those two are the same thing you wont need to associate rates with an interval to which they apply. Best of luck.

answered Mar 30 '16 at 20:15

Benjamin Whitesell

41

Thanks for the insightful answer. I intuitively understand your example, but I'm struggling to get round the mental block of how one can consider the rate of change in the value of the function at a point when that function has a given fixed value at that point?! – user35305 Mar 30 '16 at 20:48
would it be correct to interpret the derivative of a function at a point as the rate of change in the value of the function in the limit as the input variable approaches that point (i.e. basically putting in words the limit definition of the derivative)? – user35305 Mar 30 '16 at 21:02
mmmmm You can think of like that, but more fundamentally I would say you aren't intuitively understanding the meaning of a limiting rate of change. Think about y=x . At any point the slope of the line is 1, but at each point you consider it also has a fixed value right? – Benjamin Whitesell Mar 30 '16 at 21:27
Obviously, if you pause a process and think of it at one moment in time, its rate of change is zero, but thats not what a derivative at a point is describing. That's what you intuitively are missing. – Benjamin Whitesell Mar 30 '16 at 21:31
Would it be better to think of the derivative at a point as the slope of the line tangent to the curve at that point? Is the whole idea that when taking the derivative of a function one isn't ever considering the value of a function at a given point, one considers the function over a given neighbourhood. Upon taking the derivative one subsequently evaluates it at a given point and this quantifies the instantaneous rate of change in the value of the function at that point.... – user35305 Mar 30 '16 at 22:15
...For a linear function, this rate of change is constant, i.e. the rate at which the value of the function is changing at each point is the same. For a more general (continuous) function, the derivative will itself be a function and its value will change from point to point, however, at each given point its value at that point will still equal the slope of the line (corresponding to a linear function) tangent to the curve (corresponding to the more general function we're considering) at that point?! – user35305 Mar 30 '16 at 22:16
@BenjaminWhitesell Can you please elaborate on "it will be traveling at a specific velocity at one moment in time"? It doesn't make sense to me. Thanks in advance. – user599310 May 20 '20 at 15:20

score 0 · Answer 2 · answered Mar 30 '16 at 20:10

Your understanding of the derivative is correct. The only thing I would slightly reword is instead of saying

"in moving from the point $x_{0}$ to $x_{0}+\Delta x$ the value of the function $f$ changes by an amount $f'(x_{0})\Delta x$ for infinitesimally small change $\Delta x$"

I would write

"in moving from the point $x_{0}$ to $x_{0}+\Delta x$, the amount that the value of the function $f$ changes approaches $f'(x_{0})\Delta x$ as $\Delta x$ approaches 0"

The reason I say this is because in many cases, there is no actual value of $\Delta x$ such that the change in $f$ from $x_{0}$ to $x_{0}+\Delta x$ actually equals $f'(x_{0})\Delta x$, as there is no actual number that is infintesimally small.

score 0 · Answer 3 · answered Mar 30 '16 at 20:31

Your understanding of the interpretation of the derivative as a rate of change is correct but there is a subtlety.

Let's look at $f(x)=x^2$ where $f:\mathbb R \to\mathbb R$.

$f'(x)=2x$ so $f'(1)=2$.

Does this mean that the derivative $f'(1)$ is the real number $2$?

It does not. The $2$ in this case is the function that takes $x$ to $2x$. Only when we speak of the derivative we usually say it takes $dx$ to $2dx$.

In fact, $f'(x_0)$ is itself a linear function whose graph is the tangent line to the graph of $f$ at the point $x_0$. The input to this function is called the tangent space and $x_0$ is the origin of this input space.

What does the derivative of a function at a point describe?

3 Answers3