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

[Frank Castle]

For a function that is differentiable at a:
limh→0f(a+h)−f(a)h=f′(a)limh→0f(a+h)−f(a)h=f′(a)\lim_{h \to 0}\frac{f(a + h) - f(a)}{h} = f'(a)
For a function that is continuous at a:
limh→0f(a+h)=f(a)limh→0f(a+h)=f(a)\lim_{h \to 0} f(a + h) = f(a)

The important distinction between continuity and differentiability is that the latter is a stronger condition: every function that is differentiable at a is automatically continuous at a. A function that is merely continuous at a does not have to be differentiable there; for example, f(x) = |x|.

Right. Yes, I understand that. What I meant was that $f'(a)$ is defined as the limiting value of $\frac{f(a+h)-f(a)}{h}$ in the limit as $x$ tends to $a$ ($h$ tends to zero). It is the limit that defines what the value of the derivative is at a given point, whereas for the continuity case, if $f(x)$ is continuous at $x=a$ then it's limiting value as $x$ tends to $a$ ($h$ tends to zero) is equal to $f(a)$, i.e. $\lim_{h\rightarrow 0}f(a+h)=f(a)$. However, if $f$ is discontinuous at a point, then the limit $\lim_{x\rightarrow a}f(x)$ can exist, but it's limiting value won't be equal to $f(a)$, as shown in your example.
What I was trying to highlight was that if the limit $\lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}$ exists, then by definition it's limiting value is $f'(a)$, whereas the limit $\lim_{x\rightarrow a}f(x)$ can exist and have some limiting value, but this isn't necessarily equal to $f(a)$ (of course, if $f$ is continuous, then by definition it is equal), in other words, the value $f(a)$ is not defined as the limiting value of the limit $\lim_{x\rightarrow a}f(x)$, but when it is equal to the limiting value of this limit, then the function is continuous at that point.

 

[Mark44]

However, if $f$ is discontinuous at a point, then the limit $\lim_{x\rightarrow a}f(x)$ can exist, but it's limiting value won't be equal to $f(a)$, as shown in your example.

Another possibility is that this limit doesn't exist at all, such as for f(x) = 1/x, with a = 0.

 

[Frank Castle]

Another possibility is that this limit doesn't exist at all, such as for f(x) = 1/x, with a = 0.

True, good point.

Would the rest of what I put in my last post be correct? I think I'm understanding the derivative definition correctly now, would you say that's fair?

 

[dustball]

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 $f'(x_{0})\Delta x$?

Correct.

 

[rrogers]

To give an alternate non-calculus viewpoint we can use a very old fashioned statement. The derivative is the slope of the line/plane that intercepts a curve/surface at exactly one-point. If the line isn't unique then the "derivative" isn't unique. It is presumed that the line is as short as necessary to avoid bumps. This leads into the whole idea that tangents/derivatives can be said to live in a different tangent space associated with a curve and not related to the space the surface/curve is defined in.
This is from some very old books that tried to avoid mensuration and wanted "intrinsic" definitions (I presume).
In addition: a derivation is defined by $D(f(x)*g(x))=f(x)*D(g(x))+D(f(x))*g(x)$
I add this to show that the derivative operator is a member of a broader class of operators having specific properties.

 

[dustball]

To give an alternate non-calculus viewpoint we can use a very old fashioned statement. The derivative is the slope of the line/plane that intercepts a curve/surface at exactly one-point. If the line isn't unique then the "derivative" isn't unique. It is presumed that the line is as short as necessary to avoid bumps.

Sorry, this definition does not work, consider the derivative of $x^3$ at $x=0$ or $x^2sin(x^{-2})$ (taken to be 0 at 0) at $x=0$.

 

[dustball]

But endeed, the full power of classical analysis and limit theory is not needed to understand differentiation. It can be done on a more elementary level. Unfortunately not too many people are aware of it.