One of the subjects I see commonly misunderstood is the concept of the derivative, the foundation of Calculus. This blog is the first in a three-part series that will attempt to demonstrate how it can be taught in a much more simple fashion than it is in most schools. In order to understand the concept of the derivative, you only need to have a basic understanding of algebra, graphs, and an understanding of what the slope of a line is (and how to find it: (y2-y1 over x2-x1)).

Before we dive into the derivative, we need to first understand the concept of "limits" and the notation involved. The main notation that you will need to understand is that the "function of x", denoted f(x), is the same as "y", or for example, f(x) = x + 3 means the same thing as y = x + 3. So in this case, f(4) = 7, since when x is 4, y = 4 + 3 = 7, so we have the point (4, 7). Whatever is inside the parentheses replaces the x in order to find the function's value at that point (the "y") just like you learned for graphing.

A "limit" means that we want to find what the value of a function should be at a given point. Sometimes, the limit is straightforward. For example, the "limit when x approaches 4 for f(x) = x + 3" is 7, since the graph approaches 7 from either side (and f(4) = 7 as we found above). Now, sometimes, we cannot directly compute the function value (such as when we have a denominator of zero), so we use the "limit" to describe the situation instead. So instead of looking at the point itself, we look very close to the point to see what the value should be.

I have detailed more about the limit below, but if you understand that the limit means that we are looking very close to a point, that is the main thing you need to know to understand the derivative. Also, please understand how we use function notation, where f(x) just replaces "y" in functions, and that when there is a number inside the parenthesis, we are looking for the function's value (the "y") at that point.

For more on limits:

Up until you study Calculus, most of your functions have been continuous (meaning no breaks in the line), so the limit at a point is the same as the function's value (the y value). Where limits first come into play is in hyperbolas, such as f(x) = 1 / (x^2).
In this case, the graph is "discontinuous" at 0, since we cannot divide by zero. However, the limit of this graph at 0 does exist, since the graph (I would suggest using a graphing calculator to see it) shoots upward on either side of the y-axis (where x =
0). So the "limit when x approaches 0 of f(x) = 1 / (x^2)" is positive infinity. In general, limits are used to determine what the function of x (a.k.a. "y") is doing when the value itself does not exist. (Please note that the "limit when x approaches 0 of
f(x) = 1 / x does NOT exist since one side goes to positive infinity and the other goes to negative infinity.)

Limits have another very important role in times when there is a "hole" in the graph. To understand what I mean, try to graph the line f(x) = (x^2 - 1) / (x + 1) on a graphing calculator. The graph should look like the graph of f(x) = x -1. However, use the "trace" key (if using a TI calculator) to find the value of y when x = -1. The calculator cannot find it. What happened? We can use algebra to find out what happened. Remember that (x^2 - 1) can be factored to (x + 1)(x - 1). Then, using the property of cancellation, f(x) = (x^2 - 1) / (x + 1) = ((x + 1)(x - 1))/(x + 1) = x - 1. However, at the point where x = -1, we originally had a divide by zero case, which gives us the error. From the graph though, we can see that the "limit as x approaches -1 for f(x) = (x^2 - 1) / (x + 1)" is -2. Limits can tell us what the function value (y) should be even at times where there are holes.