 This lesson is on derivative formulas at a point, a little different from derivative formulas for a whole function. We'll first start off with a point and then we'll go to a whole function in another lesson. Well, first of all, what is a derivative? The definition is the slope of the tangent line at any point on a curve or the slope of the curve at any point. We've been looking at tangent lines, we're also going to start looking at curves. Well, what's some sort of designation that we have? Derivatives can be designated by the following. The first one is f' of x and we usually say the first derivative of the function with respect to x or f' of x as I just state because that's the quick way to say it. Or we can say dy dx, this means the derivative of y with respect to x. A third way to represent this is d of f of x over dx and that means the derivative of the function f of x with respect to x. Let's talk about slope of secant line, slope of tangent line again, which is of course the slope of the tangent line, which is our derivative and see how we can come up with some algebraic ways to find a derivative. So given a graph that looks like that, we can call the first point a and the second point x. So the point here is a f of a and the point here on our function would be x f of x. Now if we want to do the slope of the secant line, then we would say f of x minus f of a, change in y's, over the change of x's, which is x minus a. If we want the slope of the tangent line, m tan, we have to make x get closer and closer to a. Remember this would be the slope of the secant line and as we draw in closer and closer to a, we will get the slope of that tangent line. So we have to use a limiting value on this. So we put the limit as x approaches a of f of x minus f of a over x minus a. This is one way to represent algebraically the slope of the tangent line. And I also have told you before that the slope of the tangent line is a derivative. So we can actually use our terminology to say that this is equal to f prime at the point in question, which is a in this case. One way to represent algebraically our derivative. Second way, another graph. Again, at this point we're going to call a and this point we're going to call a plus h. So the length between a and a plus h is called h. So this makes this point a f of a and this point a plus h f of a plus h. Now, why do we do it? Because we look at a little interval being added on h being a very, very, very small delta x in our case, like point zero, zero, zero, one. So again, we can get the slope of those secant lines if h is very, very small to become the slope of the tangent line. So in this formula, the change in y over the change in x is f of a plus h minus f of a all over h. And that will give us the slope of the secant line, average change. And the slope of the tangent line, which again is f prime at a and that's m of tangent. And that's equal to the limit as h, remember that's a delta x approaches zero. So there's no change in that denominator of f of a plus h minus f of a all over h. So now we have two formulas that we can determine our derivatives algebraically. Let's determine the slope of our curve or the slope of a tangent line at a point. And there are two ways to look at this. When we zoom in on a function, we are finding the slope of the curve. But we are also finding the slope of the tangent line because the tangent line and the curve seem to intersect at a very small spot on the curve, as we well know. So let's look at our function. f of x is equal to x squared minus one and zoom in at x equals one and find the slope of the curve there. And there's our curve and let's zoom in at x is equal to one, which is right there. And you see our curve is already straightening out. So let's try it again and see how much straighter we can get it. It looks like a line there. So let's find the slope of this line, which is the slope of the curve, or can even be considered the slope of the tangent line at the point x is equal to one. Well, we know we have the point when x is equal to one, y is equal to zero. So we need another point to find the slope. So let's trace along our curve to a point really close. Let's use this one. 1.00625 and 0.012539. So when I do this slope, I'll do change in y, which is 0.012539 minus zero over change in x, 1.00625 minus one and that will compute to 2.00624. So that's our computation of our slope. Is it exact? Probably not. The way we find out if it is exact is to find the slope of the tangent line at x is equal to one and that's what we're going to do next. Now we have two formulas for this. The first one was the slope of the tangent line, which, as we begin to know now, is our first derivative at our point and that's equal to the limit as x approaches one of f of x minus f at one over x minus one. And putting in more values limit as x approaches one, we can replace our f of x with our function, which is x squared minus one and the value at f of one we know is zero, so minus zero over x minus one. The zero can go out. We can factor x squared minus one to x minus one times x plus one over x minus one. We can actually reduce this and we get the limit as x approaches one of x plus one. We can now put in the number for x, which is one, so one and one equals two. That is the exact answer for the slope of the curve as well as the slope of the tangent line at x is equal to one. And this is using our first formula. Let's try it again with our second formula because you have to know how to use both. That one says f prime at one, which is the slope of the tangent line is equal to f of a plus h minus f of a all over a. And we want the limit as h approaches zero on that. So that's equal to a is one, so we'll put one in plus h quantity squared and that takes over for our x squared minus one minus f of a, which we know to be zero again, all over h and we need the limit as h approaches zero. Computing that, we get one plus two h plus h squared minus one all over h limit as h approaches zero. Keep going with that, cross those out, factor out an h. We get two plus h over h limit as h approaches zero. Those go out and we get two plus whatever h is, h is zero, so it's equal to two. So we see that two is the absolute answer for the slope of the tangent line at x is equal to one. That concludes our lesson on derivative formulas at a point.