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| 0:31 | It's like they hide you it.
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| 1:26 | Ok, there's 4 annoying derivatives to do. Product rule, quotient rule and chain rule.
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| 1:34 | Alright, let's do it, so we can move on to other types of problems.
So you should expect on the exam to have to do a bunch of derivatives. So if you can do stuff like this, you'll be in good shape. So another problem with that, you said, alright, I know it's product rule because I have a function multiplied by another function. And the derivative of each of these is going to require the chain rule. So it tests both things at once. Ok, you want to find the derivative, so little stuff you can do to annoy your TA, you don't write dydx, you just write =. |
| 2:08 | And your TA will say but it doesn't equal that the derivative equals that.
So make sure, you have lots of time, make sure you're being very careful with how you layout your answers. Ok, when you do your exams in pen, many of you make a huge mess, if you know you're one of those, do your exam in pencil. Ok? |
| 2:30 | It becomes very difficult to follow what you're thinking otherwise.
Alright, so leave the first function alone, and multiply it by the derivative of the second function. So remember you bring down the 3, and make this to the 2, times the derivative of the inside. Plus, I'm probably going to run out of room, but now I just go the other way. ` Now you leave the second equation alone. |
| 3:02 | Squeeze the font a bit.
Times the derivative of this one. You bring down the 2, make that equation to the 1, times the derivative of the inside which is 3x^2+4. Ok? Were we able to do that? So there's 2 components to this, one you have to demonstrate the product rule, the other is you demonstrate the chain rule inside the product rule. |
| 3:31 | Ok.
This is quotient rule, but no chain rule, just quotient rule. So no matter what you do, it's bottom function * the derivative of the top function - top function * derivative of bottom function/(bottom)^2 |
| 4:18 | Ok? 2 for 2?
Good, alright, third. This is also chain rule. So one thing you could do, is you could rewrite this as a 1/2. |
| 4:39 | Ok?
The derivative is 1/2(1-(1/x))^(-1/2)*(1/x^2) |
| 5:06 | Good? Ok. One more.
So this is quotient rule again, and inside the quotient rule you have chain rule so this is a messy one. Lots of stuff to write. Ok, this is the long fraction part. Bottom function * the derivative of the top function, which is 3*x^2+2x+1)*(4x^3+2) |
| 5:48 | Minus, the other way around, (x^4+2x+1)^3, times the derivative of the bottom, 2(x^2-5)*(2x) |
| 6:08 | The whole thing, over the bottom squared, so that becomes (x^2-5)^4
Ok? How'd we do on that? Feeling good about this? Ok, that makes me happy.
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| 6:33 | I'll give you another type of question in a second.
Get more derivatives. How about, |
| 7:03 | that one
and that one.
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| 7:30 | Those should be relatively simple.
Ok, how do you do the derivative of a log? What do you do, you add a fraction bar, and the function goes in the denominator because it's 1/ that, and the derivative goes on the top. That's all you have to do. That was easy, right? And e is even easier. How do I do the derivative of e^ something? You leave the alone, and then you multiply by the derivative of the inside. |
| 8:11 | So you should be able to do those types of problems very quickly.
Ok, how about this. |
| 8:59 | Ok, find the equation of the tangent line to y=(sqrt)(e^x+3) at x=0.
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| 9:08 | This is calculus class, you only know a couple things. So if you see a question like this and you're not sure what to do, you need to do the derivative. Because that's how we determine slopes and tangent lines.
You can use the equation (y-y1)=m(x1-x) You know x1 is 0, you're going to find y1 and you're going to find the slope by taking the derivative. |
| 9:33 | So first let's find y, so at 0 you get (sqrt)(e^0+3), e^0 is 1, so square root of 4 is 2.
So we know this is y-2=m(x-0) Alright, now let's find the derivative. dy/dx =, so this is (e^x+3)^1/2, so it's 1/2*(e^x+3)^(-1/2)*e^x |
| 10:15 | Or in other words, e^x/2*(sqrt)(e^x+3) |
| 10:31 | So at x=0,
You're going to get e^0/(2*sqrt(e^0+3)), it becomes 1/4.
So this is (y-2)=(1/4)(x), and you can leave it like that. |
| 11:05 | You don't have to make it 1/4x+2 or anything fancy.
So far so good? Ok. Find the x value |
| 11:32 | of any maximums or minimums, find the maximum and minimum values, in other words, x-coordinates and y coordinates, and |
| 12:01 | find what intervals is the function increasing or decreasing.
Ok, so, another derivative. So let's take the derivative of this. We want to find where it's increasing, that would be where the derivative is positive. You want to find where it's decreasing, that would be where the derivative is negative. |
| 12:35 | Potential maximums and minimums will be where the derivative is 0.
They're called critical values. So let's see we get 3x^2+x-2, and then we're gonna wanna set this equal to 0. That'll be -2,+1, |
| 13:04 | That works, so you get x=-2/3, and x=-1.
Make a number line, put the appropriate values on the number line, and now test values in each region and see if the derivative is negative or positive. And you check them by plugging back into the derivative. |
| 13:32 | So you check a number less than -1, like -2.
That comes out negative, that comes out negative, so the curve is going up there. Try a number between -1 and 2/3, like 0. And you get a negative times a positive, so this is negative. So it's going down. So that's a max. Maximum. And then try a number bigger than 2/3, like 1. |
| 14:03 | And when you plug it in you get positives, so that will be a minimum.
So let's see. x values of maximums and minimums, that's these. And uh, you have a max at x=-1, and a min at x=2/3 So to find the y value, you take the x value and plug it back into the original equation. |
| 14:38 | You get when x=-1, y=15/2,
2/3 is a little annoying you get 110/27, something like that.
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| 15:25 | That's how I keep the Alzheimer's away. I do that stuff in my head.
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| 15:30 | And then where is it increasing or decreasing? Well it is increasing when x<-1 and when x>2/3
And it's decreasing when x is between -1 and 2/3.
Yes you could write that in interval notation but you don't have to. So far so good? Let's see what other derivative stuff we can hit you guys with. |
| 16:05 | Oh that's right. We didn't do definition of the derivative did we? Oh we did that last class. We did it, ok.
I'll give you one more of these. |
| 16:34 | Ok, let's see. Let us find,
the x coordinates of the maximums/minimums, the x coordinates of any points of inflection, the intervals where it is increasing and decreasing, and the intervals where it is concave up and concave down.
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| 17:46 | Ok, let's see how you do.
We're gonna take the derivative, the derivative is dy/dx=6x^2-6x-36, and set it equal to 0. |
| 18:05 | So you divide that through by 6,
and you get 6(x^2-x-6), and x=3, x=-2.
So if you were finding y values, you have to remember to go plug back into the original equation, but for the derivatives since you're just trying to see where the derivative is 0, you can get rid of the 6 and make life easier. |
| 18:38 | So that means we're gonna have
-2 and 3, and you test some values and you find that it goes +,-,+. So it goes up, comes back down, that makes it a maximum because you go up and then come down.
Then it goes down and comes back up that makes it a minimum. |
| 19:01 | OK? So let's do a second derivative.
The second derivative is 12x-6=0, x=1/2 Pick a number less than 1/2, that comes out negative, pick a number greater this is positive. |
| 19:30 | So concave down, concave up. I know I'm doing this a little fast I want to make sure we get it in.
Ok, so that means we have a maximum, at x=-2, a minimum at =3. I'm sorry, -2 and +3, and a point of inflection at x=1/2 |
| 20:11 | It is increasing when x<-2, and when x>3, and its decreasing when -2<x<3.
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| 20:36 | It is concave up for x>1/2, and concave down when x<1/2.
Alright, more stuff to review on Friday, take a picture of that. |