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CHRISTINE BREINER: Welcome
back to recitation.

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In this video I'd like us to
work on the following problem.

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What values of b will make
this vector field F a gradient

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field, where F is determined by
e to the x plus y times x plus

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b i plus x*j?

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So the e to the x plus y is in
both the i component and the j

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component.

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And then once you've determined
what values b will make that

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a gradient field, for this
b-- or I should've said these

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b's-- find a potential function
f using both methods from

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the lecture.

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So why don't you pause
the video, work on this,

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and then when you are
ready to look at how I

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do it bring the video back up.

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OK.

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Welcome back.

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So I'm going to
start off working

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on the first part of
this problem, which

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is to find the values of b that
will make this vector field

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F a gradient field.

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And to clarify
things for myself,

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I'm going to write down what M
and what N are based on F. So

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just to have it
clear, M is equal to e

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to the x plus y times x plus
b and N is equal to x times

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e to the x plus y.

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So those are our
values for M and N.

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And now if I want f to
be a gradient field, what

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I have to do is I have to
have M sub y equal N sub x.

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So I'm going to
determine M sub y

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and I'm going to
determine N sub x

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and I'm going to compare them
and see what value of b I get.

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So M sub y, fairly
straightforward because this

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is a constant in y.

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And the derivative of this in
terms of y is just this back.

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Right?

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It's an exponential
function with the value

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that it has in y is linear.

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So you get exactly
that thing back.

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So it actually is just e to
the x plus y times x plus b.

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So the derivative of M
sub y is just itself.

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The derivative of M with
respect to y, sorry.

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Not the derivative of M sub y.

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OK.

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That's an x.

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Let me just rewrite that.

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OK, now N sub x is
going to have two parts.

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N sub x, the derivative with
respect to x of this is 1.

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And so I have an
e to the x plus y.

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And the derivative with respect
to x of e to the x plus y

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is just e to the plus
y, for the same reason

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as the derivative with
respect to y was the same.

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So then I'm just going to get
a plus x e to the x plus y.

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So that means if I factor that
out, I get an e to the x plus y

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times 1 plus x.

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And we see that if F is going
to be a gradient field then

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b has to equal 1.

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Because it can only have one
value, and so b has to equal 1.

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To get N sub x to equal M
sub y, b has to equal 1.

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So now what I'm going to do
is I'm going to erase that b,

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put in a 1, so that the rest of
my calculations refer to that.

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So now the second
part said for this

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b find a potential function
f using both methods

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from the lecture.

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So we're going to go
through both methods

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and hopefully we get the
same answer both times.

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So let me come back here.

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The first method is where
I'm integrating along a curve

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from (0, 0) to (x_1, y_1).

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So I'm going to do it
in the following way.

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I'm going to let C_1--
so here's (0, 0)--

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I'm going to let C_1 be the
curve from (0, 0) up to (0,

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y_1).

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And then C_2 be the curve--
so it's parameterized

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in that direction-- C_2 be
the curve from (0, y_1) to

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(x_1, y_1).

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OK?

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So that's what I'm
going to do, and I'm

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going to let C equal
the curve C_1 plus C_2.

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So I'm going to have
C be the full curve.

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And what I'm interested in
doing is finding f of x_1,

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y_1, which will just equal the
integral along C of F dot dr.

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So now we need to figure
out some important things

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about C_1 and C_2.

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What's happening on C_1 and
what's happening on C_2.

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And the first thing I want
to point out-- actually,

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before I do that, let me remind
you that this is going to be

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the integral on C
of M*dx plus N*dy.

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So this will be helpful
to refer back to.

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That's really what
we're also doing here.

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So on C_1, what do I notice?

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On C_1, x is 0 and dx is 0.

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And y goes between 0 and y_1.

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And then on C_2,
y is equal to y_1.

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So dy is equal to 0 and x
is going between 0 and x_1.

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So those are the values
that are important to me.

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So if you notice from
this fact and this fact,

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we see that if we look at
the integral just along C_1,

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there's going to
be no M*dx term.

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And if we look at the
integral along C_2,

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there's going to be no
dy term because of that.

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So let me write down
the terms that do

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exist, and we'll see some
other things drop out also.

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If I look along
first just C_1, I'm

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only going to get-- I said
the dy term, which-- let

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me just make sure-- dx is 0.

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I'm only going to get the
dy term, which is-- well,

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x is 0 there.

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So I'm going to get 0 times
e to the 0 plus y, dy.

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From 0 to y_1.

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Well that's nice and easy to
calculate, thank goodness.

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That's just 0.

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So all I have to do for this
one is just leave it at 0.

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That's everything that
happens along C_1.

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That's what I'm interested in.

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I just get 0 there.

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And if I integrate along
C_2, as I mentioned, dy is 0.

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So we don't have any
component with N*dy.

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We just have the component
M*dx that we're integrating.

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OK, so if I integrate along
C_2, I just have M*dx and M is e

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to the x plus y times x plus 1.

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And y is fixed at y_1.

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So it's e to the x plus
y1 times x plus 1 dx.

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And I'm going from 0 to x_1.

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I'm going to make sure I
didn't make any mistakes.

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I'm going to check my work here.

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Yes, I'm looking good.

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OK.

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So this one is 0.

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So all I have to do is find
an antiderivative of this.

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And the term-- if
I multiply through,

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I see that here I get
exactly the same thing when

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I look for an antiderivative.

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And here I get, I
believe, two terms when

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I look for an antiderivative.

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But I'm going to get
some cancellation.

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And ultimately, when I'm all
done I'm going to get this.

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x e to the x plus y_1
evaluated at 0 and x_1.

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You could do this.

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This is really now a
single variable problem.

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So I'm not going to work
out all the details,

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but you might want
to do an integration

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by parts on that first
part of it, if that helps.

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Or an integration by
parts on the whole thing.

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That would also do the trick.

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So what do I get here?

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Then I get x_1 e to
the x_1 plus y_1.

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And then when I put in 0 for
x here, I get 0, so that's it.

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So this, plus possibly a
constant, is equal to my f.

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So I see that in general I
get f of x, y is equal to x e

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to the x plus y plus a constant.

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So that's what I get
in the first method.

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So now let's use
the second method.

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So I should say f of x_1, y_1.

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In the second method, what
I do is-- M, remember,

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is equal to f sub x.

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So f sub x is equal to
M which is equal to e

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to the x plus y times x plus 1.

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So if I want to find
an antiderivative--

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if I want to find f, I should
take an antiderivative, right?

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With respect to x.

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And so notice I already
did that, actually.

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If I just put this as y,
I already did that here.

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And so I should
get something that

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looks like this: x
e to the x plus y

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plus possibly a function
that only depends on y.

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And the reason is when I take
a derivative with respect

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to x of this, obviously
this would be 0.

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So it doesn't show up over here.

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So this, we make sure that-- oh,
that, I shouldn't write equals.

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Sorry.

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That, I shouldn't write equals.

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OK?

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This would imply that
this is equal to f.

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Sorry about that.

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f sub x was equal to
M was equal to this.

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That implies-- when I take
an antiderivative of an x--

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that x e to the x plus y
plus g of y is equal to f.

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So I apologize.

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That wouldn't have been an
equals because obviously

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those two things are not equal.

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That would imply,
I think-- yeah,

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that would imply something
very bad mathematically.

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So make sure you understand
I put an equals sign where

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I should not have.

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This is actually a
derivative of that.

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So this is
antiderivative of this.

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So now I have a candidate for f.

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And so now I'm going to
take the derivative of that.

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And what's the derivative
of that with respect to y?

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So f sub y based on this
is going to be equal to x e

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to the x plus y
plus g prime of y.

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So the prime here indicates
it's in a derivative in y.

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And now that f sub y
should also equal N.

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And N equals x e
to the x plus y.

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So what do I get here?

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I see x e to the x
plus y has to equal

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x e to the x plus y
plus g prime of y.

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Which means g prime
of y is equal to 0.

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Which means when I take an
antiderivative of that I just

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get a constant.

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That means g of
y was a constant.

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So that implies that this
boxed expression right here

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is f of x, y if g of
y is just a constant.

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So let me go through
that logic one more time.

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I had f sub x.

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I took an
antiderivative to get f

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but it involved a constant in
x that was a function of y.

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I take a derivative
of that in y.

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I compare that to what I
know the derivative is in y.

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That gives me that this is 0.

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So its antiderivative, which
is g of y, is just a constant.

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And so altogether this implies
that f of x, y is equal to x e

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to the x plus y plus a constant.

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Which is exactly
what I got before.

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Fortunately, I got two
answers that are the same.

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So that's it.

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I'll stop there.