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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--

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or that implicit
differentiation is important,

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is that sometimes you have a
function to find implicitly

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and you can't solve for it.

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You don't have any
algebraic method

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for computing the 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

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by the equation that w of
x plus 1 quantity times

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e to the w of x 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

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is going to be negative 1.

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

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so the only way to
get this side to be 0

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is if w is negative
1 if this term is 0.

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So some of its values
are easy to compute,

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but some of its values aren't.

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

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it's very hard.

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There's no algebraic way you can
manipulate this equation that

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will let you figure that out.

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So in that situation
you might still

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care about what the
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 value

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

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to the curve-- yeah.

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A linear approximation of
the function w of x in order

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to compute this value.

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So as a hint, I've given you--

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so you're trying to
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

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

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equals 0 on the
left hand side, you

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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

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base your linear approximation.

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So why don't you
pause the video,

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take a few minutes to
work this out, come back,

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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

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on this 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

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and 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

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you need in order to construct
a linear approximation.

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So we have a good
candidate for a base point

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here, which is the point (1, 0).

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So this curve,
whatever it looks like,

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it passes through
the point (1, 0).

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

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So we're going to use
the linear approximation

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

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plus w of 1 when x is
approximately equal to 1.

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So this is the
linear approximation

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we're going to 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 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 derivative of w.

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OK?

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

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we have only one piece
of information about w.

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Which is we have
that it's defined

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by this implicit equation.

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So in order to get
the derivative of w

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we have to use 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

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is going to be 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 it's going to be

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a little more complicated.

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We have a product
and then this piece,

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we're going to 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 derivative

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and on the left-- so OK,
so the product rule first.

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We take the derivative
of the first part,

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so that's just w prime of x
times the second part-- that's

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e to the w of x-- plus
the first part-- that's

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w of x plus 1-- 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 of x and I

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get 1.

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OK, good.

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So now I've got
this equation and I

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need to solve this
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,

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from the point of
view of this w prime

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it's only involved in the
equation in a very simple way.

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

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but 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

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doesn't 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

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is w prime of x times-- looks
like w of x plus 2 times e

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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-- well,

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just, you know-- it's equal
to 1 over w of x plus 2 times

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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,

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

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So I put in x equals 1--
well, let me do it over

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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

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is in 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 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 the-- that
was my hint to you.

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Right, this is why
we're using this point

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as a base point, which
is we know the value of w

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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--
well, 0 plus 2 is 2,

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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 equation that I had here.

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

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prime of 1 times x minus 1.

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So w of x is approximately
equal to-- w prime of 1, we saw

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is 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 the
linear approximation.

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And I asked for the linear
approximation, its value

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at the particular
point x equals 3/2.

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So w of 3/2 is approximately
1/2 times-- well, 3/2 minus 1

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is also 1/2-- 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

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could try iterating
this process.

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Now you might have a, you
know-- or choosing some base

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point even closer if you could
figure out the value of w

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and x near that, near this point
that you're interested in, 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

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where we couldn't
compute it explicitly.

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So what we did was we did our
normal linear approximation

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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

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is that we had to use
implicit differentiation.

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In order to compute
the derivative that

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appears in the
linear approximation,

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we implicitly differentiated.

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OK?

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

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we plugged in the values
that we were interested in,

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to actually compute
the particular value

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of that approximation.

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So I'll end there.