| Start | If we have,
Question? Conversation? Ok.
So we have, f(x) = 5x^7/14 -6x^4 +3x-5. Find the derivative. Alright, so this is straight, we did a bunch of these polynomials, I hope that we got this right, |
| 0:31 | You bring the 7 down and you get 5x^6/2-24x^3+3.
2 points, pretty much either 2 or 0. And the TA's can decide if they want to give partial credit or not. Second one, I wrote it wrong in my answer key.. so... there you go. The second one's a power rule, you have f(x) = x^3, I'm sorry this is a product rule. |
| 1:02 | f(x) = (x^3+2)(sqrt(x)-4x)
If you wanted, you could distribute that first, but it's kind of annoying with the square root.
So I wouldn't bother, I would just do product rule. You take the first equation, the first function, multiply it by the derivative of the second. Plus the reverse. And you're done. |
| 1:37 | Ok? Not so hard.
Third one, they get sort of progressively harder. This one you can do with the quotient rule. So f'(x) is the bottom function * the derivative of the top, |
| 2:05 | Minus the top function, * the derivative of the bottom, all over the bottom squared. f'(x) =((9-x)(3x^2+9)-(x^3+9)(-1))/(9-x)^2
So far so good.
Ok, now D. Eh, we'll do it over here. Oh, so let's see. That one was worth 4 points. So 1 = 2 points, 2 = 4 points, 3 = 4 points. |
| 2:49 | This one is worth 5 points.
Ok? So we're going to have to use quotient rule. So we take the bottom function, multiply it by the derivative of the top, which is (sqrt(2x-5))(3(x^2+6x-1)^2(2x+6)). |
| 3:11 | Minus, (x^2+6x+1)^3*(1/2)(2x-5)^(-1/2)*2 |
| 3:30 | All over the bottom squared.
You could right the square root squared. There's a variety of ways to write these, ok? You did not have to simplify if you knew. That was worth 5 points, that was the messiest one. I'm just seeing if you can take derivatives. If you can take derivatives, you can get lots of points on this part of the test. And then the last one we had, f(x) = (x^3+ (1/x^3))^3. |
| 4:06 | The derivative is 3(x^3+(1/x^3))^2(3x^2
And the derivative of this, this is x^-3, so that's -3x^-4.
So if you treated this as a quotient and did the quotient rule, you don't need to do quotient rule for 1/ something. Ok? That's just a negative power. |
| 4:35 | How do we feel so far? So if you feel pretty good on these, that's 20 points already on the exam.
And you're flying high and say I'm gonna rock this and then you get to the other questions, and then you don't feel so good but that's ok. We're very sympathetic, many of you, I see how hard you're trying and you show up and you do your stuff, you know. You're doing okay. Can't guarantee you're all getting As, but you're doing fine. |
| 5:05 | Alright. Question 2, a tangent line question.
It's funny, you guys say to professors 'When are we going to actually use this stuff? Oh, we lost one. Well, since you're not allowed to use calculators or computers in a class like this, I can't really show you a lot of the applications you can use with this, because the numbers are dumb if you can't use a calculator. |
| 5:37 | You have to do tiny numbers, simple numbers, and small examples. I really wish that in 122 you could use a computer.
But you can't. Oh well. But for those of you who are going to take me for business 215, We use a computer almost every class. If you're taking me in the spring, make sure you have a laptop that runs excel. Download the Stony Brook version of office 360. |
| 6:03 | Use that excel, not whatever you have right now on your computer.
Alright, number 2. Find the equation of a line tangent to, y = (3x-5)/(3x^2-10) at x=2. Equation of the tangent line. So first, lets find the y value at 2. y(2) = (3(2)-5)/(3(2)^2-10) = 1/2 |
| 6:43 | That was worth a couple points.
Just getting the y coordinates. Now you find the derivative. Derivative we do, (low*d-high - high*d-low)/(low)^2 |
| 7:24 | Now we have to evaluate it at x=2, so we get the slope.
|
| 7:30 | That's going to equal 3(4) = .12, -10 =2.
So we get, (2(3) - 1(12))/(12-10)^2 = -3/2 Nice easy number. So you get, (y-(1/2)) = -(3/2)(x-2). And of course, leave it in that form, you don't have to simplify. |
| 8:12 | So far so good? Ok, another 10 points!
You're not passing but you're getting closer. Alright 3rd one. |
| 8:44 | Ok, we have y=12+9x-3x^2-x^3
One what intervals is y increasing or decreasing?
|
| 9:01 | If you want to find where it's increasing or decreasing, you take the derivative and set it equal to 0.
You get 9-6x-3x^2 = 0. So many people have issues with the quadratic when the x^2 is not in the front. So, factor out a -3, you get (x^2+2x-3) = 0. |
| 9:38 | x=-3, x=1.
So you got about, um, three points for getting to there. [?]. Alright, then let's sign test this. If you pick a value less than -3, this comes out negative, |
| 10:04 | Then pick between -3 and 1, this comes out positive, and then negative.
So it is increasing when x is between -3 and 1, and decreasing for x<-3, and x>1. You could use interval notation, you weren't required to. |
| 10:38 | Anyone have any questions so far?
Well no, that's not the domain, I don't know what you mean. Right, domain is what you're allowed to plug in. Right, so x can be anything in a polynomial. This is where the graph is going up and where it is going down. |
| 11:01 | So, but if you mean is it the same as interval notation, do you mean this?
You're not required to do that, but you certainly can. And the other one would be... Ok? Is that what you meant? Yeah? Ok. So you don't have to use interval notation, it's nice to know, but since you guys aren't planning to go on in math after this course, Interval notation is really for mathematicians and physicists. It's just a way of compressing notation. |
| 11:33 | So what are the coordinates of any maximums or minimums, maxima or minima.
This is a minimum. At x=-3, y= 12+9(-3)-3(-3)^2-(-3)^3. I am sensitive to the fact that many of you are helpless without a calculator. I am well aware of that. Ok? |
| 12:05 | You have paper, do the best you can, it doesn't cost you a lot of points either way. So that's a minimum,
Cause it's going down and then up, so the minimum is at (-3, -15).
And then x=1, y = 12+9-3-1, which is 17. |
| 12:45 | So there's a maximum at (1,17) And if you want to graph this, we h=know it has a minimum here at (-3,-15) and a max at (1,17), |
| 13:10 | So probably something like that. And by the way, that's the y intercept of 12, you don't need to label it but you could.
Something like that is perfect. And you get 5 points for that. You get 3 points for just getting the shape right. |
| 13:34 | We're feeling good so far? Some of us are feeling better than others.
Are we allowed to GPNC this course? If you are, some of you should have. But of course, you're the one's who show up. The 40 survivors. I don't think you can, I don't think business student can GPNC this course. |
| 14:02 | Alright, we have y=3x^4-54x^2+18. Where is it increasing or decreasing?
So same thing. Dy/dx = 12x^3 -108x. I tried to give you nice numbers. Factor out a 12x, you get (x^2-9) |
| 14:34 | And that's 12x(x-3)(x+3)
And that means you do the sign test. And if you test it, you get that.
|
| 15:04 | So this is increasing, on (-3,0)U(3,infinity). or non integral notation, -3<x<0, or x>3.
|
| 15:33 | And it's decreasing, (-infinity,-3) U (0,3). Or, x<-3, 0<x<3.
So the same as before, now we're going to find the maximum and the minimums. |
| 16:19 | Ok, what are the coordinates of the maximum? So let's see, this curve is going down, up, down, up. So the minimums are at -3, and 3.
|
| 16:36 | So y(-3) = 3(-3)^4-54(-3)^2+18. That is 3(81)-54(9)+18.
Which I recognize was a bit of a chore, but if you stuck at it you'd get -225. |
| 17:03 | And at 3, well, -3^4 and -3^2 are the same as 3^4 and 3^2, now hopefully you notice that that is also -225, so minima at (-3, -225) and (3,-225) |
| 17:30 | And then the maximum is easy, thats (0,18).
So far so good? You guys don't look happy. Ok, where is it concave up or concave down? |
| 18:22 | If you want to find where it's concave up or concave down, we take the second derivative.
So the second derivative, is 36x^2-108 = 0 |
| 18:36 | You get x^2 = 3, x = +/-(sqrt)3. I told you it's very hard to make them both come out as nice numbers. They'd have to be really big numbers.
So you do the number line, And you get uh, + , -, +, so it's concave up, concave down, concave up. |
| 19:08 | It is concave up on (-infinity, -(sqrt)3), and ((sqrt)3, infinity)
Also known as x< -(sqrt)3, and x>(sqrt)3.
And it is concave down in the middle. |
| 20:12 | Ok, what are the coordinates and points of inflection? Well we have one at -(sqrt)3 and (sqrt)3, and they will be the same because of the squaring.
So y((sqrt)3) = 3((sqrt)3)^4-54(sqrt(3))^2+18. |
| 20:45 | What is ((sqrt)3)^4? Well, (sqrt)3^2 is 3, then you just square it again and you get 9.
So this is 3(9)-54(3)+18 = -117. |
| 21:05 | It's the same for the -(sqrt)3. So, (-(sqrt)3, -(sqrt)117), ((sqrt)3, -(sqrt)117)
And then if you want to graph this, we've done a bunch of these, it's one of those W shaped graphs.
|
| 21:36 | And those are the points we found, our maxima and our minima, and our points of inflection.
Ok? I don't think I'm sick I think I'm allergic to this room. Chalk dust is amazing. |
| 22:06 | Ok, we're almost done.
|
| 22:44 | Ok, some exponential log stuff. You knew this was coming.
Ok, I don't know why I wrote that. Ok, f(x) = (e^x)/(x^5+1) |
| 23:02 | So for the derivative, use quotient rule.
Bottom function, times the derivative of the top, which is just e^x, - the top times the derivative of the bottom, all over (x^5+1)^2. Ok? B. |
| 23:35 | f(x) = x^2(e^(-x^2)). I think we did this one in class. Or we did x^2e^(x^2), I forget, I know we did close.
It's a product rule. First function, times the derivative of the second, which is (-2xe^(-x^2)) because its e^(-x^2) times the derivative of the power. |
| 24:05 | Plus, the derivative of x^2 which is 2x*(e^(-x^2))
Third one, I forgot to write the f(x) on the exam. Nobody caught that.
|
| 24:31 | You just put (1-x^2) on the bottom, and -2x on top. That's the derivative of the log.
Ok, d. We have f(x)=x^5*ln(6x). f'(x) = x^5* the derivative of ln(6x), which is (6/6x), which is 1/x, so you could also write 1/x. |
| 25:10 | Plus, 5x^2*ln(6x).
Good. Almost done. E. f(x) = (e^x-e^-x)/(e^x+e^-x) |
| 25:38 | Also known as the hyperbolic tangent, if any of you ever want to be engineers, you'll learn hyperbolic functions.
Those are the trig functions of imaginary numbers. It's complex. So I have faith in none of you here, but, the derivative. Well I think we'll need a little more room for the derivative. |
| 26:09 | Bottom,
Times the derivative of the top, so thats just (e^x and the - times a - becomes a +, so +e^-x),
minus the top function, times the derivative of the bottom function, which is (e^x-e^-x).
|
| 26:33 | All over (e^x+e^-x)^2.
Ok? The last one. Equation of a tangent line. |
| 27:16 | Ok, y= x^2*e^(4-x) at x=4.
|
| 27:31 | Ok, first, you need a y coordinate.
So y is 16e^0, e^0 is 1, so it's just 16. So we need the derivative, dy/dx = x^2(-e^(4-x))+2x(e^(4-x)). |
| 28:16 | Now we do this at x=4, leave that one like this, and we get 16(-e^0)+8(e^0).
|
| 28:35 | e^0 is 1, so this is, um, -16+8 = -8.
And that means that the equation of the tangent line is (y-16) = -8(x-4). |
| 29:03 | I assume that you're wondering, did I put this I don't remember. Maybe, it's all a blur.
Monday night was a long time ago. Right before halloween, so tequila, Monday night at the bench, Monday night football...etc. Alright, on Friday we start new stuff. Everybody have a nice day. |