 So let's introduce the derivative. Hello, derivative! So our path to the derivative starts with this notion of the average rate of change, after which we define the instantaneous rate of change as the limit of this average rate of change. Since that instantaneous rate of change is always at some particular point, one important feature is that we want to make sure that we have an interval that contains that point, and the easiest way to do that is to either begin or end the interval at the point we want. So suppose our function is f of x and we begin our interval at x equal to a. Let the length of our interval be h. Then the interval runs between a and a plus h, and our average rate of change is going to be our function evaluated at the end minus our function evaluated at the beginning over the difference between the two endpoints, a plus h and a, which is just going to be h. Now our instantaneous rate of change is going to be the limit of the average rate of change as the length of an interval containing a goes to zero. So we're going to take the limit as h goes to zero of this quantity, which is called the difference quotient. And this particular limit is so important that we give it a very specific name. We call it the derivative at a point. And like every definition in mathematics, it is vitally important that you understand and remember this basic definition. So here it is, let f of x be a function. The derivative of f of x at x equals a, written f prime of a, is going to be defined as the limit as h goes to zero of the difference quotient. Except no substitutions. So for example, we might want to use the definition to find f prime of 3 for f of x equal to 5x plus 3. Now a little later in the course, we'll find fast ways of finding the derivative. So this naturally raises the question of how do we know when we're supposed to use the definition to find the derivative? That's a very complicated question to answer. And unfortunately the only way to tell that you have to use the definition instead of some other fast method is that the question says, use the definition. So we'll pull in our definition of the derivative. And so we need to find f of 3 plus h and f of 3. Now generally speaking, calculus is easy. It's the algebra that causes problems. So here's a quick reminder of how to find a function evaluated at a point. Now there's many ways of doing this, but here's one that I strongly recommend. Let's take our function definition. f of x equals 5x plus 3. And what we're going to do is we're going to take every occurrence of x and replace it with an empty set of parentheses. And the important thing to remember here is that whatever goes in any of these parentheses has to go in all of these parentheses. So I want to find f of 3 plus h. So I put a 3 plus h inside this set of parentheses. And I have to put a 3 plus h inside this set of parentheses. Likewise, I want to find f of 3, so I'll substitute that in. So I have my definition of the derivative. I'll substitute in the values I have for f of 3 plus h and f of 3. I'll do a little bit of algebra. I'll do a little bit of algebra. And finally, I get to a point where I have to do some calculus. I want to find the limit as h goes to 0 of 5. And since 5 is a constant, I know that the limit as x approaches anything of any constant is just the constant. And so my limit is going to be 5. Now it's worth pointing out that in all of this, we only use calculus at the very beginning where we set down the definition of the derivative and at the very end where we actually take a limit. All of the intervening steps are pure algebra. So this is a good time to brush up on your algebra skills.