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

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1. Mar 31, 2016

### Frank Castle

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 $f(x)$ of a variable $x$ is a measure of the rate at which the value of the function changes with respect to the change of the variable". 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 quantifying the slope of the line tangent to the curve (corresponding to the coordinate plot of the function $f(x)$) 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 (with respect to a change in its input variable $x$) at that point. For a linear function, this rate of change is constant, i.e. the rate at which the value of the function is changing (with respect to a change in its input variable $x$) 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 the coordinate plot of a linear function) tangent to the curve (corresponding to the coordinate plot of the more general function we're considering) at that point?!

Apologies for such a basic question, but I've really got a mental block in my head about this and want to clear it up.

2. Mar 31, 2016

### Samy_A

Yes.

3. Mar 31, 2016

### Staff: Mentor

The function has a particular value at a specified point, but the function values are not constant, in general. A different explanation of the meaning of the derivative of a function at a point is that $f'(x_0)$ is the slope of the tangent line at the point $(x_0, f(x_0))$.

4. Mar 31, 2016

### PeroK

It's quite a common question and I think a lot of people have to digest this idea at some stage. You have, in fact, given a fairly good description of what a derivative is.

5. Mar 31, 2016

### Frank Castle

Thanks. How could I improve it though? Are there any deeper insights that I'm missing?

By this do you mean that the function has a different value at each point? I thought I understood the meaning of the derivative of a function, but the description on Wikipedia (that I quoted in my post) has thrown a spanner in the works. I'm unsure how to interpret it correctly?!

6. Mar 31, 2016

### PeroK

In terms of a simple derivative there's perhaps not much more to say. In mathematical terms, though, the derivative is what it is formally defined to be. None of the descriptions actually matter one iota.

The next step is understanding things like the chain rule, the relationship between differentiation and integration (Fundamental Theorem of Calculus) and partial derivatives for a function of several variables; or, of course, functions of a complex variable.

What level are you studying at?

7. Mar 31, 2016

### Frank Castle

Undergraduate level (I'm ashamed to say). I understand how to use the derivative and its associated rules, etc. and I thought I understood intuitively what a derivative is, but now I'm not so sure, the description I read on Wikipedia has really thrown a spanner in the works for me (I'm not happy with simply accepting definitions when I learn maths, I want to have an intuitive idea of what the particular operation is describing). What the sticking point for me is, how do I make sense of the statement:

What exactly does this statement mean? Is it simply that although the function has a fixed value at that point the value of the function will in general be changing from point to point and the derivative evaluated at a particular point describes the rate at which its value is changing at that point in the sense that if a move away from that point (say $x_{0}$) by a small amount $\Delta x$, the value of the function $f$ will approximately change by an amount $f'(x_{0})\Delta x$?

8. Mar 31, 2016

### PeroK

There are a hundred ways to describe something. Personally, I wouldn't worry if you don't understand a particular person's way of describing something, as long as you understand someone else's. For example, you said "the derivative is itself a function", which I think is better than the extract from Wiki. From a maths point of view that is crucial. Never forget that, by the way: the derivative is itself a function.

9. Mar 31, 2016

### Staff: Mentor

I'm not sure why you are puzzled by this. Consider the graph of $y = f(x) = x^2$. The graph is a parabola with its vertex at the origin. The derivative is f'(x) = 2x. At the point (1, 1) on the parabola, f'(1) = 2, so the slope of the tangent line is 2 at this point on the parabola. At the vertex of the parabola, f'(0) = 0, so the tangent line at (0, 0) is horizontal (its slope is 0).

10. Mar 31, 2016

### Frank Castle

I think what confuses me is how one can consider the rate of change in a value at a single point, surely you can only discuss it relative to another point? It would make sense to me that the derivative of a function at a point describes the rate at which its value changes as you move away from that point. Before, I always thought of the derivative at a particular point as quantifying the slope of the line tangent to the function curve at that point...
In the linear case it is easy for me to understand as the derivative is simply the slope (gradient) and is constant $m$, thus the function value $y=f(x)=mx+b$ is changing at a rate of $m$ units of $y$ per unit of $x$ at each point along the line. Can one simply generalise this to an arbitrary continuous function, except that now the derivative is not constant and so will correspond to a different slope (of a different line tangent to the curve) at each point. However, at a given point $x_{0}$, the derivative $f'(x_{0})$ equals the slope of the line that is tangent to the curve $y=f(x)$ at the point $x_{0}$ and hence the value of the function at that point is changing at a rate of $f'(x_{0})$ units of $y$ per unit of $x$?!

Last edited: Mar 31, 2016
11. Mar 31, 2016

### Staff: Mentor

Another way to think about it is this: Imagine that you're a bug travelling along the curve, from left to right. At the point in question, which direction would you be facing?

Do you know what a secant line is? If not, it's a line that connects two points on a curve. The slope of the secant line between $(x_0, f(x_0))$ and $(x_0 + h, f(x_0 + h))$ has a slope of $\frac{f(x_0 + h) - f(x_0)}{x_0 + h - x_0}$, which is change in y values divided by change in x values, or rise over run. As you take smaller and smaller values of h you get secant lines that are closer in slope to the tangent line at $(x_0, f(x_0))$. The derivative of f at $x_0$ is defined by this limit:
$$f'(x_0) = \lim_{h \to 0}\frac{f(x_0 + h) - f(x_0)}{h}$$
Yes, that's what it is (and what I already said).

12. Mar 31, 2016

### Frank Castle

Thanks for the info. Sorry to repeat what you'd already said.
Also, would the description that I added to the end of my last post be correct? (i.e.
)

13. Mar 31, 2016

### Staff: Mentor

Yes

14. Apr 1, 2016

### HallsofIvy

That is a crucial question and is, largely, the reason "Calculus" was created. Both Newton and Leibniz were concerned with the problem of finding how gravity determined the orbits of planets. They knew, of course, that "force equals mass times acceleration". But that caused a conceptual problem! Since it was clear to both that gravitational force depended on the distance from the sun. At a specific instant, the distance from the sun to a planet is a fixed number (not "constant"- that should only be applied to functions, not individual numerical values) while "speed" is "change in distance divided by change in time" while "acceleration" is "change in speed divided by change in time". They require "change in time" so can't be at a given instant. So how can acceleration, which requires a change in time, depend on distance, which does not?

The solution to that problem was to use a "limiting process" so that we can define "instantaneous" speed and acceleration.

15. Apr 3, 2016

### Frank Castle

So is the idea that by defining the derivative in terms of a limit we can consistently quantify the rate of change in the function at a particular point? Is the limit definition basically stating that we can make $\Delta x$ arbitrarily close to zero, without actually equaling zero, such that within this arbitrarily small interval $(x_{0},x_{0}+\Delta x)$, the ratio $\frac{f(x_{0}+\Delta x)-f(x_{0})}{\Delta x}$ is equal to the value $f'(x_{0})$, i.e. the slope of the tangent line to the point $x_{0}$, and quantifies the rate at which the value of $f$ is changing due to a change in $x$ at the particular point $x_{0}$?!

If the value of $\frac{f(x_{0}+\Delta x)-f(x_{0})}{\Delta x}$ as $\Delta x= x-x_{0}$ approaches 0 is independent of the manner in which $\Delta x$ approaches 0, i.e. independent of whether $\Delta x$ approaches 0 from the left hand side of $x_{0}$ or the right hand side of $x_{0}$, then it's limiting value exists and we say that is exactly equal to the rate of change in $f(x)$ relative to $x$ at the point $x_{0}$.

Would this be correct at all?

Last edited: Apr 3, 2016
16. Apr 3, 2016

### Staff: Mentor

No, not equal. For a specific nonzero value of $\Delta x$, the ratio $\frac{f(x_{0}+\Delta x)-f(x_{0})}{\Delta x}$ gives the slope of the secant line, the line that joins $(x_0, f(x_0))$ and $(x_0 + \Delta x, f(x_0 + \Delta x))$. The smaller that $\Delta x$ is, the closer the slope of the secant line will be to the slope of the tangent line at $(x_0, f(x_0))$; i.e., $f'(x_0)$. But as long as $\Delta x$ is different from zero, the slope of the secant line will be different from $f'(x_0)$.

17. Apr 3, 2016

### Frank Castle

But I thought that the whole point of a limit was that the input variable is never actually equal to the value it is approaching? (At least that's what I've read in a couple of introductions to limits)

18. Apr 3, 2016

### Staff: Mentor

Right, and I didn't say that it was.

My objection was to what you wrote:
$f'(x_0)$ is NOT equal to the ratio above at each point in some interval $[x_0, x_0 + \Delta x]$.

It might be helpful to actually work with numbers and a specific function. Let y = f(x) = x2 and find the derivative of this function at the point (1, 1), using the difference quotient with several values for $\Delta x$.

$\Delta x$ = 0.1
Ratio - $\frac{f(1.1) - f(1)}{.1} = \frac{1.21 - 1}{.1} = 2.1$

$\Delta x$ = 0.01
Ratio - $\frac{f(1.01) - f(1)}{.01} = \frac{1.0201 - 1}{.01} = 2.01$

Continue in this fashion for smaller values of $\Delta x$. You should see that the difference quotient (identified above as the ratio, and which gives the slope of the secant line) will be slightly larger than the value of f'(1), which is 2. However, smaller values of $\Delta x$ give values of the difference quotient that are closer to, but not equal to the value of f'(1).

19. Apr 3, 2016

### Frank Castle

Sorry, very bad (and incorrect) phrasing on my part. I was trying to reword the definition of the limit, specifically the part $0<\lvert\Delta x\rvert <\delta$, in an intuitive way for the specific case of the derivative.

To check my understanding: the concept of the derivative arises naturally when one poses the question, "what is the slope of a curve at a given point?". Obviously it doesn't make sense to calculate the slope at a single point as one needs two points along the curve to do so, and in particular, the ratio would be indeterminate (as we would have $\frac{0}{0}$). Therefore, in order to define the rate of change in the value of a given function at a particular point we employ the properties of the limit. This is useful as $\Delta x$ can be allowed to approach zero, but not actually equal zero, and as such one can quantitatively define the derivative of a function at a given point $x$ as the limit, $$\lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$$ which equals the slope of the line tangent to the point $x$. We therefore say that the rate of change in the value of the function $f$ with respect to a change in $x$ at a given point is defined as the limit given above, and is equal to the derivative of $f$ at $x$, $f'(x)$, if the limit exists.

Would this be correct at all?

20. Apr 3, 2016

### PeroK

The beauty of mathematics is in its brevity, precision and unambiguity. You can describe the derivative in as many words as you like, but you can't say more than is expressed in those 10 or so symbols.