Start  How are we doing in the homework problems by the way?
Good? Alright so let's practice a similar problem. Okay? We're going to practice a bunch of these today. 
0:31  Before we get into new stuff.
So let's say find the derivative of f(x) = 3x^2 + 4x +1. You guys wanna do it on your own first or should we do it as a team? Team? Okay. 
1:00  So we're going to need f(x+h) and we're going to need f(h) because remember the formula is
lim h>0 of f(x+h)  f(x) /h.
So we have f(x). That's this. We're going to need f(x+h) and then we're gonna need to do some algebra. The algebra is the annoying part. 
1:30  So f(x+h) = 3(x+h)^2 + 4(x+h) +1.
And then before we put it in what they call the difference quotient before we put it in the derivative formula, let's simplify this. So if we foil out (x+h) we multiply that out 
2:11  w'ere going to get 3(x^2 + 2xh + h^2).

2:31  Everybody see that?
So we're going to multiply out the (x+h)^2, that gives us this term. We distribute the 4 and we get the 1. Now let's just multiply through by 3. You get 3x^2 + 6xh + 3h^2 + 4x + 4h + 1. 
3:04  Okay so now let's put in the formula so we do the lim h> 0 of this  f(x) all over h.
So (3x^2 + 6xh + 3h^2 + 4x+ 4h + 1)  (3x^2+ 4x + 1) /h. 
3:37  You see how these start to get kind of tedious.
And cumbersome. You just don't really want to do all that work. So Monday we'll learn the fast way to do these. Some of you probably already know it but the rest of you will be very excited. 
4:08  Okay the 3x^2 here cancels with the 3x^2 there because we're subtracting.
4x4x and 11. Now we have 6xh + 3h^2 + 4h all divided by h. 
4:41  We factor out h from the top.
And the h's cancel. Now if you plug in 0 we get 6x+4. 
5:07  That's the derivative. So as in the previous problem where we had 6x^2 + x
so the x term is what's going to contribute 4h so in that one you would've had just plain h and when you factor out the h you're left with a 1.
So here we factored out the h, we're left with a 4. 
5:30  Okay so that's what contributes to this term.
So when you do your initial expansion there are a couple of clues you could look for. Any terms that don't contain h get canceled. Any terms that contain a single power of h, you'll be able to pull an h out and then you'll be let with something when you do the limit. Anything with more than a power of h, even if you take one h out it's still gonna have an h in it which means it's gonna become 0 when the time comes. Okay so there's three types of terms. They have no h in them, that means they're going to cancel. 
6:02  They have a single h in them, they're going to be left over when you do the limit.
Or they have more than one power of h which means they're going to become 0. Okay so let's have you guys try one. That looks fun. 
10:23  How are we doing?

12:12  The cube is annoying I know. You have to practice what (x+h)^3 is.

12:41  You multiply this out. (x+h) times (x+h) and you get x^2 + 2xh + h^2.
When you multiply out (x+h) first you get x times x, then you get x times h, 
13:14  and another x times h, so thats 2 x times h and 2 x times h and then h times h.
So that's where the 2xh comes from. So when you do x^3 you're gonna get x^3 + 3x^2h + 3xh^2 + h^3. 
13:41  If you multiply it out. Okay?
You all should memorize that if you have not. You should memorize it or you should be able to do it quickly. The coefficients go 1, 3, 3, 1. 
14:07  Alright I think everybody's had enough pain.
Let's do this one. So first let's figure out what f(x+h) is. Because that's where the real work comes from these problems. So f(x+h) is 2(x+h)^3 + 5(x+h)^2 + 8. 
14:33  So if you wanted to get partial credit because you say I'm never going to get this all correct, if you just show that you're looking for the limit of f(x+h) minus f(x) 
15:04  that'll be worth 3 or 4 points out of the total of 10.
Just demonstrating that you can set it up is worth something. Okay so we'll come back to this in a minute. So let's work this out. (x+h)^3 = x^3 + 3x^h + 3xh^2 + h^3, which I just wrote that there. 
15:33  (x+h)^2 = x^2 + 2xh + h^2 because I also wrote that there.
Plus 8. And then distribute these coefficients. So you get 2x^3 + 6x^2h + 6xh^2 + 2h^3 + 5x^2 + 10xh + 5h^2 + h. 
16:14  So that long messy expression is going to be the left side of the difference quotient.

16:43  (2x^3 + 6x^2h + 6xh^2 + 2h^3 + 5x^2 + 10xh + 5h^2 + 8) 
17:08   (2x^3 + 5x^2 + 8).
So that's what we're going to now simplify. So remember what I keep telling you. So these terms on the right cancel. 2x^3 cancels. 5x^2 cancels. 8 cancels. 
17:32  And you now have the lim h>0 of 6x^2h + 6xh^2 + 2h^3 + 10xh + 5h^2. Whole thing divided by h.
So remember I told you, when you go through this the first thing is any term in the beginning that doesn't have an h in it should be canceled. 
18:02  So everything now should have at least one h in it.
Which it does. And anything that has more than one power of h, this one, this one, and this one, are gonna end up becoming zero. Because you're going to take the limit and they're going to be zero. They're not going to be important. So the only thing that's going to be left is this term and this term. So let's see how we do that. 
18:30  Factor out the h.
You get 6x^2 + 6xh + 2h^2 + 10x + 5h. All divided by h. Okay, now this is the crucial part of calculus, you let h be 0. 
19:00  Because as I said you have people that do both things, you have people that say
you cancel this because h is not 0 and you can plug in 0 for h because it's essentially 0.
See you have to be able to hold both concepts in your head. It's really close to 0 but it's not 0. So sometimes you can treat it like 0 and sometimes you don't. And now when you let h be 0 these three terms will cancel, the terms that contain the h's in them, and you get this. 
19:33  How'd we do on that one?
Good? Not ready for college? Alright let's do a couple more. One that's not quite so messy. The polynomials are messy because there's lots of terms. 
20:08  Alright how about something like 1/x.
Let's do that one as a team. How do we do 1/x? Well let's set it up. I'm gonna do the lim h>0 of 1/x+h, that's the f(x+h) term. 
20:35  1/x all divided by h.
Now what do we do? It's not quite so easy. We can't just cancel. So we have to figure out what we're going to do here so when you have a pair of fractions like that how would you combine them? 
21:03  You use a common denominator.
So why don't we try that first. See if we can turn that into a single fraction on top. So the left fraction, you're going to have to multiply the top and bottom by x. So that's gonna be x/x(x+h). And the right fraction I'm going to multiply the top and bottom by x+h. 
21:38  So far so good?
See what I did? The left fraction the top and bottom and multiplied by x. The right fraction the top and bottom were multiplied by x+h. Okay next step. 
22:20  Okay so now let's put those two fractions together.
You're going to get, distribute the minus sign, (xxh)/x(x+h)/h. 
22:42  See in every step we know we're going to have a problem because every time you want to let h go to 0
you keep getting 0 in the denominator.
So remember that stands for the change in x, slope. And that change in x is still 0 it's still causing us an issue. So you have to find a way to get rid of that h. 
23:00  Okay these x's cancel.
And now we have h/x(x+h) all over h. You may not realize that this h and this h will cancel. So let's show you why. This h down here and this h up here. 
23:32  You can think of this h in the denominator as h/1.
Question: "How'd you get the x and x+h on top on the right side?" Remember multiply this one top and bottom by x, and this one top and bottom by x+h. The common denominator. So if you think of this as h/1 when you take a fraction divided by a fraction you take the bottom one, flip it, and multiply it. 
24:04  So you have h/x(x+h) * 1/h
The h's now cancel.
The h's cancel and leave you with 1, not 0. So you have the lim h>0 of 1/x(x+h). 
24:39  Now you can plug in 0 for h.
Because you don't have that denominator issue anymore. Now if you take the limit h becomes 0 and you get 1/x^2. 
25:03  The good news is on the exam you'll probably have to do one of these.
And I don't think I give you one of these type. They're too hard. I would give you a square or a cube. You're not going to have to do ten of these that's just nasty. 
25:32  Alright let's try one other annoying type.
What if I give you the square root of x? But on the bright side we haven't done any trigonometry, nothing with pi. 
26:00  What do you do with the sin of (x+h)?
It gets tricky right. It would be good if I give like an arc tangent. Alright we'll do this one as a team. So we're going to want the lim h>0 of √x+h  √x / h. 
26:32  So one of the fun things about teaching in this room is the seats are really comfortable.
So we lose a lot of people. Alright let's see if we can figure out how to do this one. Only one is pretty good. Oh wait I need two. 
27:01  Eh, it's hard to tell.
Okay so how would we do this one? Any ideas? You can't just combine those right? It's the square root of something minus the square root of something. You can't quite square it. Almost multiply it by itself. Not quite the reciprocal. The conjugate. So you're going to multiply the top and bottom by the conjugate. Remember when you did that in the 8th grade and you didn't know why? This is why. 
27:33  In fact if you learned common denominators you could do these.
Among other reasons. So we're going to need to multiply the top and bottom by what's called the conjugate. The conjugate's purpose is to get rid of the radicals. The conjugate of a+b is ab. 
28:04  That's the conjugate. What's nice is when you multiply those together you get a^2  b^2.
So when you get the squares that gets rid of the square root. That's what you're trying to do. So multiply the top and bottom by √(x+h)+x will get rid of the radicals on the top. It's going to put one on the bottom though. So we have to hope that's not an issue. But our issue really is we have to find a way to get rid of that h. 
28:31  Alright so let's multiply the top.
The lefthand term is √(x+h) * √(x+h) that's (x+h). And then as I said with the conjugate you get a^2  b^2 because you're gonna get √x * √x+h  √x * √x+h so those are going to drop out. And then you're going to get x. 
29:02  So again, this square root times this square root is the left term.
The middle terms cancel and √x * √x is x. In the denominator is h * √x+h +√x. Okay. Cancel the x's. 
29:46  ?? one of your questions.
On the exam are you expected to write limit at every step? What do you think should I say yes or no? You're expected to but I'll let you get way with it if you don't. You should try to remember to because the TAs will be looking for it. 
30:03  And then they'll come up to me and they'll say "What do we do with a student who doesn't write limit all the way through?"
And I'll say "Well, we could execute the student."
But that's probably a bit too much of a lesson.
So I think it's simpler just to say we'll let you get away with it this time. But the next calculus class you take you won't get away with it. Of course how many of you are taking another calculus class? There you go, alright. 
30:32  So we can cancel these h's.
When we cancel the h's what are we left with on top? Not zero. 1 right? So the top is 1/√x+h + √x. Now let h be 0. 
31:00  And you get 1/√x + √x
or 1 / 2√x.
And you get really sadistic teachers who give you cube root of x. Or 1 over cube root of x. 1 over the cube root of x^2, you can get really nasty. So what's gonna happen is we're going to learn some shortcuts. Soon. But let's just have you practice one more without me. 
31:44  Is it a hard one? Nah.
There you go. See how you do on that one. 
37:13  I think people are just about ready.
Ready for the weekend. You guys going to Brookfest tonight? Some of you. Are we excited for Post Malone? Is there a Pre Malone? 
37:30  I gave it a shot. And what's the other guy? Slushi?
Isn't that something you get at 711? Anyway i hope it's good! We've had some pretty good concerts over the years. Jimmi Hendrix. Janis Joplin. I don't think these guys are at that level yet. The Doors. Led Zeppelin. Not there yet. We had Bruno Mars a few years ago. He was pretty good. 
38:03  Anyway well I won't be there. Just incase you were wondering.
So let's find the lim h>0 of 5/x+h+1  5/x+1 all over h. Everybody gets that step. That's the partial credit step. 
38:30  Alright, common denominator.
So you multiply this fraction, top and bottom, by x+1. And this fraction, top and bottom, by x+h+1. Question: "Should we put parentheses around the x+h?" You can. You can have parentheses around x+h. It doesn't really matter, but you can. That was interesting what we ?? Lim h>0 okay so multiply this by x+1 over x+h+1 * x+1. 
39:05  Minus 5(x+h+1) / (x+h+1)(x+1). Whole mess over h.
Okay now combine them to one fraction. On top. 
39:31  So let's see we've got some distributing to do.
The denominator is (x+h+1)(x+1) and then this whole thing divided by h. So far so good? 
40:09  Everybody with me? Alright.
So let's do some canceling. The 5x and the 5 cancel with the 5x and the 5. And you get the lim h>0 of 5h on top, (x+h+1)(x+1), all over h. 
40:34  Does everyone know what you can do with this h and this h?
You can cancel. Okay? This h cancels with that h and you get the lim h>0 5 on top, (x+h+1)(x+1). 
41:02  Okay now just let h=0.
And you get 5/(x+1)(x+1) which you could leave. Or you could make 5/(x+1)^2. Doesn't really matter. This last step isn't important. I mean obviously you should be able to simplify it there. Anybody able to get there? 
41:39  Why am I adding h here?
Well the original expression is 5/x+1. So if I want f(x+h) then x has to be replaced with x+h. I understand this is painful. The good news is this is as hard as it's going to get. 
42:00  Alright everybody have a nice weekend. Look for an assignment sometime between today and tomorrow.
See you Monday. 