00:01
I can work through these two problems together, since they're very similar.
00:05
In each problem, they give us two functions, or they plot them, and they ask us to look at a function that's the product of these two functions and also the quotient of those two functions.
00:16
And then they ask us for the derivative of these new functions, that different point.
00:21
So u -prime is, using the product rule, is f -prime g plus f -g -priam.
00:28
The v prime, using the quotient rule, is minus fg prime plus f prime g all over g squared.
00:36
So now to figure out what u prime is at 1, we go to the graph.
00:42
So u prime evaluated 1 is f prime evaluated at 1.
00:46
Well, f prime evaluated 1 is the slope of this at 1.
00:51
And so the slope of that is 2 over this whole region, so that's 2.
00:56
G evaluated at 1 is 1.
01:03
Let's see here.
01:04
Well, wait a minute.
01:06
That was these terms here.
01:08
Now, g prime at 1 is minus 1.
01:15
And f at 1.
01:18
Let's see here.
01:19
Now, where am i going? oh, no.
01:22
I was talking to the wrong time.
01:24
So this, yeah, these are these.
01:26
F at 1 is.
01:30
One, f at one, no, f at one is two.
01:37
Yeah, yeah, there we go.
01:39
And g prime at one is minus one.
01:42
I said the slope here.
01:43
All right.
01:44
That turns out added all up to get zero.
01:46
So uprime at one is zero.
01:48
So if we were to multiply these functions together, we would see that we have a horizontal tangent of the product of those two functions at one.
02:00
Now they want v prime at five.
02:03
So here's v prime.
02:06
And so we'll see here, we need f at 5.
02:10
That's 3.
02:12
G prime evaluated at 5.
02:14
So this has a slope of 2 thirds over here.
02:17
We can figure that out.
02:19
Just by a couple points that it goes to 5, 2.
02:23
So we can figure that out looking at the graph in the book.
02:27
So it has a slope of 2 thirds at that point.
02:30
And then we have f prime at 5 is, minus one -third.
02:37
So that's the slope of this line here.
02:40
And g at 5 once is 2.
02:44
And then we need that point to get 5 squared, which is 4.
02:48
So we get v prime at evaluated at 5 is minus 2 thirds.
02:54
Now over here, we have some, you know, just a little bit more complicated function, but the exact same process, right? so now we just, they relabel these capital f, capital, g, and now we have p and q.
03:06
But it's still the product and the quotient.
03:09
And so, you know, the same rules apply.
03:11
So we just need to get for p -prime at 2.
03:15
So we need f -prime at 2...