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PROFESSOR: Hi.

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Welcome back to recitation.

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In lecture you've been talking
about implicitly defined

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functions and implicit
differentiation.

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So one of the reasons that these
are important is, or

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that implicit differentiation is
important, is it sometimes

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you have a function to find
implicitly and you

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can't solve for it.

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You don't have any algebraic
method for computing the

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function values as
a formula, say.

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So, for example, this function
that I've written on the board

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that I've called w of x is
defined implicitly by the

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equation that w of x plus 1
quantity times e to the w of x

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is equal to x for all x.

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So this function, some of its
values you can guess.

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Like at x equals 0, the function
value is going to be

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negative 1.

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And the reason is that this
can't ever be 0, so the only

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way to get this side to be 0
is if w is negative 1 if

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this term is 0.

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So some of its values are easy
to compute, but some of its

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values aren't.

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So for example, if I asked
you what w of 3/2

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is, it's very hard.

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There's no algebraic way you can
manipulate this equation

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that will let you
figure that out.

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So in that situation you might
still care about what the

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function value is.

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So what can you do?

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Well, you can try and find a
numerical approximation.

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So in this problem I'd like you
to try and estimate the

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value w of 3/2 by using a linear
approximation to the

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function, to the curve--

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yeah.

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A linear approximation of the
function w of x in order to

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compute this value.

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So as a hint, I've given you,
so you're trying to

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compute w of 3/2.

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As a hint I'm pointing out
to you that w of 1 is 0.

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Right?

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If you put in x equals
0 and w of 0 equals--

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sorry-- if you put in x equals
1 and w of 1 equals 0 on the

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left hand side, you do indeed
get 1, as you should.

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So, OK.

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So, good.

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So that'll give you a hint about
where you could base

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your linear approximation.

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So why don't you pause the
video, take a few minutes to

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work this out, come back, and
we can work it out together.

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All right.

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Welcome back.

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So hopefully you've had a
chance to work on this

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question a little bit.

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So in order to do this linear
approximation that we want,

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what we need to know are we need
to know a base point and

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we need to know the derivative
of the function

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at that base point.

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And those are the two pieces of
data you need in order to

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construct a linear
approximation.

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So we have a good candidate for
a base point here, which

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is the point 1, 0.

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So this curve, whatever it looks
like, it passes through

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the point 1, 0.

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And that's the point we're
going to use for our

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approximation.

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So we're going to use the linear
approximation w of x is

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approximately equal to w prime
of 1 times x minus 1 plus w of

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1 when x is approximately
equal to 1.

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So this is the linear
approximation we're going to

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use, and we have that
w of 1 here is 0.

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So this is, this is equal
to w prime of one

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times x minus one.

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Just the w of 1 is 0.

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It just goes away.

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So in order to estimate w of x,
and in particular w of 3/2,

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what we need to know is
we need to know the

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derivative of w.

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OK?

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And to get the derivative
of w, we need to use--

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well, we have only one piece
of information about w.

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Which is we have that
it's defined by

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this implicit equation.

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So in order to get the
derivative of w we have to use

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implicit differentiation.

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OK?

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So let's do that.

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So if we implicitly
differentiate this equation--

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so let's start with the, the
right hand side is going to be

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really easy.

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Right?

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We're going to differentiate
with respect to x.

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The right hand side
is going to be 1.

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On the left hand side is going
to be a little more

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complicated.

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We have a product and then this
piece, we're going to

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have a chain rule situation.

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Right?

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We have e to the w of x.

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So, OK.

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So we're going to
take an implicit

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derivative and on the left--

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so OK, so the product rule
first. We take the derivative

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of the first part, so that's
just w prime of x times the

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second part-- that's
e to the w of x--

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plus the first part-- that's
w of x plus 1--

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times the derivative
of the second part.

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So the second part is
e to the w of x.

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So that gives me an e to the
w of x times w prime of x.

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That's the chain rule.

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So that's what happens
when I differentiate

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the left hand side.

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And on the right hand side
I take the derivative

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of x and I get 1.

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OK, good.

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So now I've got this equation
and I need to solve this

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equation for w prime.

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Because if you look up here,
that's what I want.

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I want a particular
value of w prime.

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And as always happens in
implicit differentiation, from

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the point of view of this w
prime it's only involved in

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the equation in a
very simple way.

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So there's it multiplied by
functions of x and w of x, but

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not, you know, it's just
multiplied by something that

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doesn't involve w
prime at all.

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And then here it's multiplied
by something that doesn't

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involve w prime at all.

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So you can just collect your w
prime's and divide through.

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You know, it's just like solving
a linear equation.

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So here if we collect our w
prime's, this is w prime of x

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times-- looks like--

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w of x plus 2 times
e to the w of x.

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Did I get that right?

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Looks good.

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OK, so that's still
equal to 1.

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So that means that w prime
of x is just--

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well, just, you know--

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it's equal to 1 over w of x plus
2 times e to the w of x.

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OK, so this is true
for every x.

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But I don't need this equation
for every x.

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I just need the particular
value of w prime at 1.

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So that's, so I'm going to take
this equation, then, and

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I'm just going to put
in x equals 1.

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So I put in x equals 1--

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well, let me do it over here--
so I get w prime of 1.

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And I just everywhere I had
an x, I put in a 1.

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So actually, in this equation
the only place x appears is in

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the argument of w.

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So this is w of 1 plus 2
times e to the w of 1.

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OK.

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So in order to get w prime
of 1 I need to

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know what w of 1 is.

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But I had that.

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I had it, it was right
back here.

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There was, that was
my hint to you.

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Right, this is why we're using
this point as a base point,

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which is we know the value
of w for this value of x.

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So we take that value.

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So w of 1 is 0.

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So this is just 1 over--

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well, 0 plus 2 is--

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2, and e to the 0 is 1.

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So it's just 1 over 2.

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Sorry, that's written
a little oddly.

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We can make it 2 times 1.

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So 1 over 2.

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OK.

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So I take that back
upstairs to this

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equation that I had here.

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And I have that w of x is
approximately equal to w prime

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of 1 times x minus 1.

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So w of x is approximately
equal to w prime of 1--

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we saw is--

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1/2 times x minus 1.

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And that approximation was
good near our base point.

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So that's good when
x is near 1.

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All right.

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And then, so this is the
linear approximation.

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And I asked for the linear
approximation, its value at

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the particular point,
x equals 3/2.

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So w of 3/2 is approximately
1/2 times--

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well, 3/2 minus 1
is also 1/2--

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so this is a quarter.

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OK, so this is our estimate
for w of 3/2.

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w of 3/2 is approximately 1/4.

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If you wanted a better estimate
you could try

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iterating this process.

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Now you might have
a, you know--

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or choosing some base point
even closer if you could

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figure out the value of w and
x near that, near this point

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that you're interested in--

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3/2.

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So just to sum up what we did
was we had this implicitly

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defined function w.

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We wanted to estimate its value
at a point where we

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couldn't compute
it explicitly.

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So what we did was we did
our normal linear

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approximation method.

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Right?

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So we wrote down our normal
linear approximation formula.

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The only thing that was slightly
unusual is that we

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had to use implicit
differentiation.

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In order to compute the
derivative that appears in the

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linear approximation, we
implicitly differentiated.

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OK?

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So that happened just like
normal, and then at the end we

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plugged in the values that
we were interested in, to

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actually compute the particular
value of that

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approximation.

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So I'll end there.

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