 Hello, and welcome to the screencaster. We're going to calculate the derivative of a function at a point using the limit definition of the derivative. So here's our main example here. We're going to take the function g of x equals negative 16x squared plus 30x plus five, and we're going to calculate the exact value of g prime of one. That's the derivative of g at the number one. So how do we go about doing this? Well, when we're faced with a problem like this that involves calculating something, and we know there's a formula for it. The right question to ask is, what is the formula? What is the formula for calculating the derivative of any function at any point? So we can either search our memories or look it up in our books. And remember that the derivative of any function f at a point a is equal to the limit as h approaches zero of the fraction f of a plus h minus f of a divided by h provided that the limit exists. Just to recap the concept behind what the derivative is, the fraction you see there is the average rate of change in f. It's the average velocity of f, if you will, from a to a plus h. H is a small degree of separation between a and another point. And what we're doing is we're seeing what that average rate of change does as the h, the separation value, goes to zero. And that will give us an instantaneous rate of change or an instantaneous velocity if we want to think of it in that way. So let's do a quick concept check here. And we're going to try to map this formula that works in general onto our specific problem here. We're trying to calculate g prime of one for a specific function g and the point x equals one. So in the generic formula for the derivative, which I repeated here, what do we need to modify in order to make this fit our specific problem that we're trying to work out? Do we need to use the formula for g instead of the letter f? Do we need to set a equal to one? Do we need to set g equal to one? Do we need to set h equal to one? What we're doing is to set h equal to a. So I want you to think about this by pausing the video and coming back and selecting all the possible options that you think are correct. There could be more than one right answer here. So there are actually two correct answers that you should be thinking of in this particular concept check, and those are a and b. The formula that you see there centered is a generic formula that works for any function. Now I don't want any function f, I want a specific function g. And remember we defined that in terms of a specific formula over here. So I'm definitely going to be replacing f with my function g. Also, I need to note that when it says a in the definition, what that refers to is the point where I want the derivative. This is the point at which I want the instantaneous rate of change or the point at which I want to see the tangent line if you want to think of it in that way. For me in this problem that is at x equals one. So I'm definitely going to go through everywhere in my generic formula and replace a equal to one. I'm certainly not going to set g equal to one because that doesn't make any sense. g is a function and I don't want to just set it the function equal to the number one. h remember is always a small degree of separation between two points that I'm using to calculate the slope of a secant line. So that is going to be a changeable value and I'm going to change it by letting it go to zero. So I don't want to set h equal to one and nor do I want to set this equal to a because a is going to be equal to one. So d and e are really the same answer. In other words, I'm going to take my generic formula that you see here and make some applications. Here, let's do this on this slide and then we'll go to another place and do the calculation. So here's the generic formula for the derivative. Now I'm going to go through and replace all instances of f with my particular function g and the same will go for a. I'm going to set a equal to whatever point I am calculating the derivative at. Okay, so now I have a skeleton for a formula for g prime of one. And the next thing I might want to do is go through on the right hand side. And I've got two values of g that I'm computing, g of one plus h and g of one. And I just want to go through and actually apply the formula for g to those particular situations. And that's what I've done here on the third line on the right. What you have is g of one plus h minus g of one. But I'm using the particular formula for g in its place. And now I have a lot of calculations to do and it's helpful to think of the strategy and the to-do list of what we're going to do next. We're going to go through and compute each of the terms in the numerator separately. There's two things being subtracted. I'm going to calculate those things and simplify them, then do the subtraction. And then that will leave me with a fraction with a simpler numerator. And I would like to simplify the entire fraction if possible. Once the entire fraction is simplified, I'm going to take the limit as h goes to zero, if all works out well, I should be able to take this limit as h goes to zero just by setting h equal to zero. That may not be possible. If I can't set h equal to zero, I want to still think of the limit exists by using tables or graphs. So let's go to a different space and actually calculate this derivative g prime of one. So here's the formula for g prime of one as we just left it on the slide. So I have to take the limit of this large fraction here. And the strategy is I'm going to go through the numerator here and I have two pieces, this large piece right here and then this slightly smaller piece right here and they're going to be subtracted. So I'm going to simplify each of those two pieces completely using some basic algebra. Then bring it back over here to this page and subtract and hopefully that will give me a fraction that I can simplify a little bit. So over here on the next slide, I have each of those two pieces already set up to start calculating. And this is just basic algebra, okay? So I'm going to go through and let's see what we can do. In this first piece here, I can square the one plus h using the FOIL method. That's going to give me negative 16 times 1 plus 2h plus h squared. And then I can distribute this 30 throughout this group. And then it'll give me 30 plus 30h and then I have a five hanging off the edge. Then I'm going to go through and distribute this negative 16 through the group. It'll give me negative 16 minus 32h minus 16h squared. And then I have all this other stuff left over from the previous line. And now it's time to think about like terms. And I notice the linear term, the constant term, sorry. I have a negative 16 plus a 30 plus a 5, and that all adds up to 19. And then for the linear terms that have just a factor of h on them, I have negative 32h and a plus 30h. So that of course adds up to negative 2h. And then finally there's only one quadratic term and that's this negative 16h squared. So that would be the first half of the numerator that we saw over here. We're going to come back in a second and just put it in its place. But first let's get the other half of the numerator and that's just this number. That's a negative 16, that's a negative sign right there, times 1 squared plus 30 plus 5 and that of course is the same calculation I made up here. The negative 16 plus 30 plus 5 and that gives me plus 19. Okay, so now we're going to go back to the previous slide and just put in what we need. So I still have a limit that I need to take. It's very important to keep this limit here until we actually take a limit. We have to deal with h eventually. Do not omit the limit symbol before the end of the calculation. Otherwise you're not calculating a derivative, you're calculating something else. So the first half of the numerator was 19 minus 2h minus 16h squared. Then I had one to subtract, that's this part right here, subtract and that second bracket turned out to be a 19 as well, this is all divided by h. So that's happy because it's much simpler. It's going to give you a simpler yet because I see I have a 19 minus a 19 and those subtract off and maybe you can even see one simplification step ahead. But let's not get too far ahead of ourselves. The subtracting off the 19 gives me negative 2h minus 16h squared all over h. And I see that I have a common factor of h in both terms of the numerator. So I'm going to do a two step process and that is, well first of all, remember that I have not taken a limit yet, so that's gotta be there. Then I'm going to factor out the common factor and then I'm left with negative 2 minus 16h all over h. Now that I have a common factor divided by itself, those divide off. And let's just put this off to the side. I have the limit as h goes to 0 of the quantity negative 2 minus 16h. Now I'm at the point where I need to take the limit as h goes to 0. And I can do that this time because there's no division by 0 or anything funny happening. If I just simply substitute h equals 0 into this, I'm going to get my limit. And that's going to be negative 2 minus 0 if I just set h equal to 0 here. And that of course is negative 2, okay? So g prime of 1, the derivative of this function g at x equals 1 is negative 2.