| Start | Todays documents, I think I put up some more recently.
I made an announcement to that effect. Yes, antiderivatives. This is too loud. I put up antiderivatives, l'hospital's rule, and applied maximum minimum problems. For those of you who are focused on part two, you should be sure you can do that stuff. |
| 0:31 | Okay, you should be able to do the curve sketching and related rates.
What else So curve sketching, related rates, applied max/min, l'hospital's rule, and antiderivatives. Okay and implicit differentiation. Alright, we will have mock final practice problem stuff up before the end of the week. |
| 1:02 | That will be on the official webpage.
Somewhere in here we will have sample final exams. Plus, I don't know when the solutions will go up but we will put them up before the final. I wrote the rough draft of the final but Professor Sutherland will give it the massage and between the two of us we will have the final, okay? |
| 1:31 | So I have a pretty good idea of whats on the final.
Well you know he's in charge of the course so I said well how about I write a rough draft and he said great so I wrote him a draft of the final. He said this is much too easy there won't be enough crying. No, I'm kidding! We're not talking a big part but we want to make sure you guys are ready for 126. Thats the idea. Okay, so there we go. Stuff you should be able to do... |
| 3:06 | Alright, lets see if you guys could do those derivatives.
And on part two what will happen is the problem will still require you to be able to take derivatives you will just have to do something to the derivatives. Okay? Who can't read my handwriting? That says tan(3x), but you can make it a tan(2x) if you want, but that says tan(3x) plus tan^-1 (5x). |
| 3:55 | Alright, that's long enough.
Lets practice our derivatives. |
| 4:03 | f prime of x, you do lo d hi minus hi d lo. You guys should be good at this by now.
lo times the derivative of the high which is (2x+3) minus hi times the derivative of the bottom which is (15x^2) Over the bottom squared. |
| 4:30 | Now we might ask you to find f prime at one, or f prime at zero or f prime at two.
Then you would have to plug in, okay? But other wise you can just leave it like that. So many of you messed up on this because when you plugged in, I think it was at zero or one, you just didn't plug in properly. So then when you did it you just did the quotient rule totally wrong. You've got to have some faith in yourselves. And as I just heard some of you got a little lost because of the capital x instead of a lower case x, or you put in the wrong letter. |
| 5:03 | Don't use t if its x, don't use x if its t. Pay attention to that stuff, okay?
Alright derivative of this. The derivative of tangent is secant squared, so this is secant squared of 3x times 3. This is because you have the chain rule, okay? What happened to my plus sign? There was a plus sign there. Do you want to do it as product rule? We can do it as product rule, sure. |
| 5:35 | I'll leave it as product rule that's fine. So times tan inverse of 5x
plus tan(3x) times the derivative of tan inverse of 5x. What's the derivative of tan inverse of 5x?
Okay, 1 over 1 plus 5x squared times 5. |
| 6:01 | Many of you will put (5x)^2 but not put 5 in parenthesis or will not multiply by 5 or put a square root that just doesn't belong there.
Okay so remember, what is the derivative of tan inverse of x? 1 over (1+x^2). So if this is 5x then that is 5x all in parenthesis, squared. What is the derivative of sin inverse of x? |
| 6:33 | 1 over the square root of (1-x^2).
So if it was sin inverse of 5x, you would have 1 minus 5x in parenthesis, squared. Then you have to multiply by the derivative of 5x, which is 5. Okay? Good? Bad? Not so much? Okay. |
| 7:00 | What is the derivative of the square root of x?
Well you could write this as (1/2) x^(-1/2), or you could have memorized it is 1 over (2 root x). I recommend memorizing that because it makes life easier. And the derivative of 2 over x you do not use the quotient rule. Many of you used the quotient rule on that and then you did it wrong. Okay? The derivative of 2 over x, well this is the same as 2 times x to the -1. |
| 7:36 | Right? 2 over x is 2 times x to the -1, so the derivative would be -2 x to the -2.
Or 2 over x squared. So you can leave it like this. Okay. |
| 8:05 | What's the derivative of log(4x) cubed?
Do you have questions? You were just playing with your hair, okay. This is 3 times log(4x) squared. What's the derivative of the natural log of 4x? 4x on the bottom and 4 on top. I forgot the plus sign again, this is crazy. |
| 8:30 | Alright.
I don't love product rule this much. Okay. The derivative of the natural log of 4x. 4x on the bottom and 4 on top. What's the derivative of the natural log of 4? It is not 1 over 4. It's not 1 over 4x. It's not 4 over 4x. It's zero. Why is it zero? Because the natural log of 4 is a number. It's a constant. Okay it's about 1.1. |
| 9:03 | What's the derivative of the log of 4? Zero. Okay?
We good? So far so good? Okay, lets make it more fun. |
| 10:01 | Find the equation of the tangent line to y= 3sin(2x) at x= pi/6.
Are we ready? Do we need another 30 seconds? Okay. So first we are going to need the y coordinate. |
| 10:31 | Remember when you do the tangent line (y - y1) is slope times (x - x1).
So we need a slope and we need a y coordinate because we already have the x coordinate that's pi/6. So to get the y coordinate you plug in pi/6. And y1 is simply 3sin of 2pi/6. |
| 11:03 | Well we know what the sign of 2pi/6 is because thats the sin of pi/3 and you all memorized the unit circle. So that is 3 radical 3 over 2.
Now we have to get the slope. To get the slope we take the derivative. Y prime is going to be 3 cos(2x) times 2 or 6 cos(2x). |
| 11:35 | Now to get the slope we have to plug in pi/6.
So it's 6 cos(2pi/6). Cos(pi/3) is 1/2 so this equals 3. So now the equation is y minus 3rad3 over 2 equals 3 times (x- pi/6) |
| 12:08 | Okay?
Lets do some more. Let's do some of the good stuff. Find f prime of t where f(t) equals 4/5 t^5 - 3t^4 + 8t +e^3. |
| 12:57 | Or |
| 13:16 | Okay, that's long enough you should be able to take those derivatives.
Okay, 4/5 t^5. By the way many of you, |
| 13:44 | many of you do not seem to realize that this is the same as this.
You do not need to use the quotient rule here. But if you're going to do the quotient rule, |
| 14:02 | The derivative of the bottom is zero. You don't leave it out.
When you leave it out you get it wrong. You don't do this. You have to put the zero in so that it cancels this. And then you say oh right I didn't need to do the quotient rule after all. Okay if you do the quotient rule and one of the two parts of the quotient is zero then you didn't need to do the quotient rule. Alright so here the derivative is 4t^4 - 12t^2 + 8 because what's the derivative of e^3? |
| 14:35 | Zero. What is the derivative of the log of 3? Zero.
What is the derivative of pi? Zero, good. Alright. This one is 5 times this stuff to the fourth so you leave that alone, times the derivative of the inside. Which is 3x^2 minus 6 over 2^x plus 0. |
| 15:06 | Some of you when you did this you forgot these parenthesis.
That is incorrect. Okay? Order of operations that is incorrect. You have to do this, okay? If you do not do that you get it wrong. Okay? |
| 15:30 | Good! Now lets do some of the newer stuff.
|
| 16:42 | Okay on part two of the final you might be expected to graph a function.
So why don't we see if you can graph that. Alright, I've had enough of this one. |
| 17:00 | You ready?
So when you want to do graphs you need to do two things you need the first derivative and you need the second derivative. So the first derivative, y prime is -12x^2 +36x +120. And the second derivative is -24x +36. Now I have to find where these are 0. |
| 17:32 | So we're going to take -12x^2 +36x +120 and set it equal to zero.
You divide by -12 and you get x^2 -3x -10 = 0. And does that factor? Why yes it does. Oh my gosh. x minus 5 and x plus 2 |
| 18:00 | So our critical values are x=5, x= -2.
So if this is a 10 point question, when you get to here you are about 4 points in already. Okay, now you set the second derivative equal to zero. And you're going to get x= 3/2. Okay now you're probably about 5 points in. So you're doing okay at this point, alright? |
| 18:37 | Remember it's all about the partial credit.
Alright, now lets look at the first derivative and see where it's increasing and where it's decreasing. So you pick some numbers. You pick a number less than -2 like -3. You plug it into the derivative and you get -108 minus that. |
| 19:06 | This is going to be negative.
You plug in a number like zero, and you get positive. You plug in a number like 6 and you get negative. So your curve is going down, then it's going up, then it's going down. So that's a minimum and that’s a maximum. Yes? |
| 19:32 | Ah, be careful because if you plug into the factor equation you'll lose the minus sign.
There's a minus sign up here. So if you're not sure, leave the -12 out there, okay? Does that help? So that will help you get the sign correct. I even mess that up a lot too. Yes? I can't hear you. |
| 20:00 | That’s 3/2.
You put the 24x to the other side and you get 24. Okay? So far so good? Alright. Now we have to do the second derivative. You try a number like zero and you get 36. That's positive. You try a number like 2 and you get negative. So its concave up here and concave down here. It kind of matches that. Okay so you have a point of inflection at 3/2. |
| 20:36 | So now you know that you have a max at x = 5,
a minimum at x = -2, and a point of inflection at x= 3/2.
Now you have to find the y-coordinates. This is when you're happy you have scrap paper. You take 5 and you plug it in and you get 542. |
| 21:06 | Is that right?
Tell me if I'm wrong. I don't know, you just trust me on that? Okay. I could be wrong. Plug in -2 and you get 32, plus 72 is 104. Minus 240 is -136. |
| 21:30 | So -144.
Is that right? Did anyone check on a calculator? You're saying yes but is it right? Maybe we're both wrong. Okay. And plug in 3/2 and you get? Something. Alright I'll do it in my head. What? 199 Alright we're going to trust you because you have an electronic device that you're holding up. Alright so now we graph this. |
| 22:02 | We know we have a y-intercept at -8 by the way.
Alright, we have a minimum at (-2, -144). We have a maximum at (5, 542). And we have a point of inflection at (3/2, 199). |
| 22:33 | And you get one of those.
Okay? Should we do another one of these? Alright let's make you graph something more annoying. |
| 23:23 | Okay, once we put in the e we lose half the room.
Yes? |
| 23:30 | Oh that's the y-intercept. The -8 is, you don't need it, it's the y-intercept. Okay?
She wanted to know where the -8 came from. How do you find the y-intercept? You plug zero into the original equation and the constant was 8 so that is the y-intercept. Okay. Alright should we do this one as a team? |
| 24:30 | Alright lets take the derivative.
What is the derivative of e^(-4x^2) + 1? It is e^(-4x^2) times the derivative of -4x^2 which I like to put in front but you don't have to, okay? And the derivative of 1 is zero. So that was a kind of easy derivative. Alright now for the second derivative we need the product rule. |
| 25:02 | So we can do -8x
times the derivative of e^(-4x^2) which is -8xe^(-4x^2).
Plus the derivative of -8x which is -8 times e^(-4x^2). Alright lets find where these things are zero. For the first one, -8xe^(-4x^2) equals zero, |
| 25:36 | what is the important thing to remember about the exponential function? About the e term?
It's never zero it is always positive. So this part is always positive so you just care where is -8x is equal to zero? At x = 0. So that is going to be our critical value. Good? |
| 26:00 | Alright now we take the second derivative and set it equal to zero.
Get this cleaned up a bit. This was 64x^2e^(-4x^2) - 8e^(-4x^2). So we can factor out e^(-4x^2) and you will be left with 64x^2 - 8. When you have one of these exponentials you can factor out very nicely, and remember the e term is never zero. |
| 26:36 | So 64x^2 - 8 has to equal zero.
So you get x^2 equals 1/8 or x= +/- sqrt(1/8). So far so good? Any happy faces? |
| 27:00 | Alright.
So lets see. We have to do some sign testing now so lets test the first derivative. We try a number less than zero, like -1. You get 8 times something that's positive. So it's positive so the curve is going up. Try a number bigger than zero, like 1. You get -8 times something positive, so this is negative so the curve is going down. |
| 27:30 | So there's a maximum at x = 0.
Alright, lets do the second derivative. We've got negative sqrt(1/8) and positive sqrt(1/8) So lets try some values. How about -1. This is always positive. 64 - 8 is positive. So we concave up in this region. |
| 28:03 | Try a number between negative sqrt(1/8) and positive sqrt(1/8) like zero.
This is positive, zero minus 8 is negative so it's negative so it's concave down here. And then lets try a number bigger than 1/8 like 1. This is positive and this is positive so it's positive and so it's back to concave up. Anything else we know about this graph? What is the limit as x goes to infinity of that graph? |
| 28:32 | By the way as x goes to infinity this is zero so you get 1.
When x goes to minus infinity this is zero so you get 1. Why is it zero? Because 1 over e^(4x^2) + 1 So 1 is going to be a horizontal asymptote. That might help with the graph. Okay so we have two points of inflection You have one at x = sqrt(1/8) and one at x = - sqrt(1/8). |
| 29:06 | So now we have to find the y values.
So this will be a hard question of the event. Although it's really not hard. So when you plug in zero, that's 1 plus 1 is 2. When you plug in the square root of (1/8) you get 1 over e times 4 times an (1/8) which is e^ (1/2) +1 |
| 29:35 | so you get 1 plus 1 over sqrt(e)
and same with the negative.
And then you graph it. And you get that, you get the bell curve. That's the point (0,2). |
| 30:02 | That's the point (sqrt(1/8), 1+(1/sqrt e)).
And that's ( - sqrt(1/8), 1+(1/sqrt e)). So if I were doing this on an exam, this would be of minimal value, okay? The important thing is if you get the shape of the graph right I don’t really care about that. Recognizing it has a horizontal asymptote is of some value. |
| 30:30 | Whenever you deal with graphs you should always test the limits to see what's going on at plus and minus infinity.
Okay let's do one more graph and then you'll have had enough for one day. |
| 31:19 | Alright I think we're going to have to do this as a team so I don't run out of time because I have to give the make up at 5:30.
Let's take the derivative. So y prime is 4x^3 -8x. |
| 31:37 | And y double prime is 12x^2 - 8.
So we set these equal to zero. 4x^3 -8x = 0 and you can factor out a 4x and you're left with x^2 - 2. |
| 32:05 | And you get x = 0, x = +/- sqrt(2).
Okay, and for the second derivative you set that equal to zero 12x^2 - 8 =0 you get x^2 = 8/12 also known as 2/3. So x= +/- sqrt(2/3). |
| 32:32 | So let’s do some sign testing.
Okay. Oh and zero. So you try a number less than -sqrt(2) like -2. And that comes out negative. Try a number like negative 1 and it's positive. |
| 33:01 | Try a number like positive one and it's negative.
And positive 2 and it's positive. The curve is going down, up, down, up. So it's a min, max, min. Second derivative |
| 33:35 | We put a negative sqrt(2/3) and a positive sqrt(2/3).
So try a number less than -2/3 like -1. And that comes out positive. Try a number like zero and you get negative, and then positive again. So we have two points of inflection where it goes from concave up to concave down. And from concave down to concave up. |
| 34:01 | Now we need the y-coordinates.
So we have a max at x = 0 We have mins at x = sqrt(2), x = - sqrt(2) And we have points of inflection at x = sqrt(2/3) and x = - sqrt(2/3). Alrighty now we come up with the y-coordinates. |
| 34:31 | Well, squaring these things is easy lets see.
When x equals zero y equals 1 and when x equals sqrt(2) Well sqrt(2) ^4 is 2^2 is 4 Sqrt(2^2) is 2 so you get 4 - 8 is -4 Plus 1 is -3 And it's the same here. And the sqrt(2/3) you get 4/9 - 8/3 which is -23/9 |
| 35:01 | -23/9 + 1 is -14/9.
|
| 35:39 | There you go, that is what the graph looks like.
Now I've got a question, how many of you do not need to take MATH126? So that's it so you don't have to take any more math after this? So first grade, second grade, high school, this is it. |
| 36:03 | One more math class for the rest of your lives!
How exciting is that! I knew you'd be happy. [students cheering] It makes me feel all warm knowing that you know, you can forget the unit circle in exactly a week and 2 days. That's so awesome for all of you. I'll see you on Wednesday. |