00:01
Hi there.
00:02
In this problem, we have a function g of the variables u and v, and we can also think of it as a function f, the variable's x, and x depends on u and v, and so does y.
00:22
Let's keep this in mind as we go through here because the variables can be tricky.
00:26
So if we make a tree diagram, it's in the book, we have f depends on the variables x and y, and x and y each depend on the variables u and v so let's use this to get our answer as we go so first we are asked to find g partial with respect to u at the point zero comma zero now that means when u and v are zero since u and v are the inputs for g so to get this let's use the tree diagram here i want to get down to the variable u so if you go down the left branch, we get f partial x, and then we multiply it by x partial u, plus, let's go down on the other side, f partial y, times y partial u.
01:38
And we want to evaluate this whole thing when u and v are both zero.
01:47
Okay, so let's go carefully, left to right, and evaluate each of these.
01:53
F of x, not f of x, f sub -x here, f partial with respect to x, when u and v are both zero, we need to figure out what f's inputs are.
02:06
In other words, when you and v are both zero, what is x? so let's look, and x, our first input, for f is defined as e to the u plus sine v.
02:25
And so let's keep track of this.
02:29
X equals e to the u plus sine v and y, the second input, e to the u plus cosine v.
02:46
So when u and v are both zero, let's write what x and y are, so we don't get confused.
02:54
X equals well we get e to the 0 plus sign 0 e to the 0 is 1 sign of 0 is 0 1 plus 0 is 1 and y equals let's see e to the 0 plus cosine 0 so e to the 0 is 1 cosine 0 is also 1 1 plus 1 is 2 so x and y or 2 this makes sense looking at the table they gave us pretty much had to be that but it's always good to check.
03:29
So now we know exactly where to look in the table for each of these.
03:33
You want f sub x and this is at the point where x and y are one and two.
03:41
So if you look in the table, we see that f partial x at 1 comma 2, that's the bottom row, would give us the number 2.
03:54
And so that part gives us 2.
03:57
Now x sub u comes next and again x, x, is given in u and v variables.
04:05
So this will be the top row of the table.
04:09
And x sub u, i'm sorry, we actually don't need the table for this one.
04:13
It's staring us right in the face.
04:15
X subu, let's calculate that right here.
04:20
So x sub u.
04:25
Now here's our x over there.
04:29
So the partial derivative e to the u plus sine of v is e to the u.
04:35
That's the derivative of e to the u and sine v has no use in it so that's it um and and at 0 comma 0 u is really just 0 so that will just be the number 1 okay next up f partial of y same thing as before let's look in the table we care about the point 1 comma 2 so that will be the bottom row and we see that's five.
05:12
So again, from the table, f partial y is five.
05:18
And y partial u, well, same thing as over here, let's look.
05:22
Y partial u is again, e to the u.
05:30
And when u and v are zero, this becomes this.
05:34
So we get two times one plus five times one, two plus five equals seven...