Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The average rate of change can be modeled as a function: $f:D^2 \to \mathbb{R}$ where $D$ is the domain of the primary function in consideration. It maps two variables - the ends of the interval- to the average rate of change of that particular interval.

The derivative is $f$ restricted to the set: $\{(x,x): x \in D\}$, essentially making it a "slice".

Is this a valid interpretation?

share|cite|improve this question
When you say "3D" and "2D" it's unclear what you mean. From the rest of your question I assume that you mean average rate of change is a function $f:D^2\to \mathbb R$ where $D$ is the domain of whatever other function you're considering, and the derivative is $f$ restricted to the set $\{(x,x):x\in D\}$. But in general it's best to use standard notation, and if you don't know the notation to look it up. – Alex Becker Jul 15 '12 at 3:30
@AlexBecker I'll keep that in mind. I specified my definitions of a "2D" and "3D" functions, which should help clarify any previous vagueness. – user26649 Jul 15 '12 at 3:35
What is an independent/dependent variable? The question is vague. – copper.hat Jul 15 '12 at 3:39
@copper.hat… – user26649 Jul 15 '12 at 3:49
Why would the usage of "independent/dependent variable" be considered vagueness? Upto my knowledge they are terms with concrete mathematical definitions. – user26649 Jul 15 '12 at 3:52
up vote 1 down vote accepted

If you have a function $f\colon I\to\mathbb R$ defined on an interval $I$, and you consider its "average rate of change" as a function $g\colon I^2\to\mathbb R$ defined as $$g(x,y) = \frac{f(y)-f(x)}{y-x},$$ hoping to say that $f'(x) = g(x,x)$, then you immediately run into the problem that $g$ is not even defined at $(x,x)$ because you'll be dividing by zero. You'll have to consider the limit of $g(u,v)$ as $(u,v)$ tends to $(x,x)$ instead. That should be fine -- I'd wager that the limit is defined and equal to $f'(x)$ exactly where $f$ is differentiable. But really, you've just replaced the one-variable limit in the definition of the derivative with a two-variable limit (and those are much more complicated), so I don't see this as a very useful interpretation.

share|cite|improve this answer
Thanks for the help. – user26649 Jul 15 '12 at 16:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.