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PROFESSOR: Hi.

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Welcome back to Recitation.

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In lecture you've been computing
derivatives of

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functions from the limit
definition of derivative.

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So today we're going to do
another example of that and do

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some graphing, as well.

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So I've got a problem written
here on the board.

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So we're defining a function f
of x to be 1 over the quantity

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1 plus x squared.

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So what I'd like you to do is
graph the function of the

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curve y equals f of x and to
compute the derivative f prime

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of x from the definition.

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So why don't you take a couple
of minutes to do that

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yourself, then come back, and
we'll work it out together.

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All right.

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Welcome back.

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So to start off, let's try
graphing this function f of x.

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So one thing you can always do
when you start out graphing a

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function, is to just
plot a few points.

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And that'll give you a very
rough sense of where the

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function is, at least
around those points.

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So, for example, when x is
equal to 0 we have f

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of 0 is 1 over 1.

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So that's just 1.

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So we've got this point
here, 0, 1.

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And when x is equal to 1, well,
x squared is 1, so the

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denominator is 2, so the
function value is 1/2.

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So, all right I'm not going
to draw this to scale.

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I'm going to put x
equals 1 here.

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And the function value is 1/2,
so this is the point 1, 1/2.

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And, OK.

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We could do one more.

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When x is equal to 2,
this function--

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2 squared is 4, so that's
5-- so it's 1/5.

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So I don't know.

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1/5 is smaller than
1/2, right?

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So that's maybe--

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down here--

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so this is something like
the point 2, 1/5.

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All right.

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But this is a very rough idea
we're getting, so we can use

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some more sophisticated analysis
to get a better idea

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of what this graph is
going to look like.

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So the first thing we could
notice, for example, is that

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this is an even function.

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Right?

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If I change the sign of x, if I
replace x by minus x, well,

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x squared and minus x squared
are both equal to x squared.

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So if you replace x by
minus x, the function

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value doesn't change.

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So this is an even function
that has symmetry

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across the axis here.

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So, you know, for example,
I could just

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mirror image these points.

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So these points also have to
be on the graph, the points

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minus 1, 1/2 and minus 2, 1/5.

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And any other part of the curve
that I draw will be

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perfectly mirror imaged.

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Another thing to observe is
that x squared is always

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greater than or equal to 0.

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So 1 plus x squared is
always positive.

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So 1 over 1 plus x squared
is also always positive.

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Also, 1 plus x squared, it
reaches its minimum when x is

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equal to 0 and then as x gets
large, either in the positive

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direction or in the negative
direction, this gets larger

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and larger.

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this, just the 1 plus
x squared part.

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So the denominator is getting
larger and larger, while the

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numerator stays constant.

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The whole fraction gets
smaller and smaller.

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So as x gets bigger, either
bigger positive or bigger

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negative, the function value
will diminish off to 0, and it

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has its maximum value
here at 0.

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Because that's when 1 plus x
squared has its minimum.

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So the function sort of has
its maximum here at 0, and

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then it flattens out.

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And as x gets larger and larger
and larger, this goes

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to infinity, so the whole
fraction goes down to 0.

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But it never reaches
it, right?

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Because we said it's
always positive.

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And similarly, on
the other side.

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OK.

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So that's the graph of
the curve y equals 1

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over 1 plus x squared.

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Roughly speaking.

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OK.

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So now let's talk about
computing the derivative.

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So right now, to compute a
derivative, all you have is

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the limit definition
of the derivative.

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So when I ask you to compute the
derivative what you've got

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to do is write down what
that definition says.

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That's the limit of a
difference quotient.

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So we have, by definition, that
f prime of x is equal

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to-- well, it's the limit--

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is delta x goes to 0 of some
difference quotient.

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So on the bottom of the
difference quotient we just

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have delta x, and on the top
we have f of x plus delta x

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minus f of x.

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In our case, we have a nice
formula for f of x.

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So this is equal to the limit as
delta x goes to 0 of 1 over

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the quantity 1 plus x plus delta
x quantity squared--

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oh, I guess I didn't need
that parenthesis there--

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minus 1 over 1 plus x squared,
and the whole thing

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is over delta x.

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So, what would be really nice,
of course, is if this were a

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limit where we could just plug
in the value delta x equals 0

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and evaluate it.

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But the way the definition of a
derivative works, that never

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works, right?

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You're always left with
the numerator.

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As delta x goes to 0, that top
is always going to be f of x

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minus f of x and it's
going to be 0.

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And the bottom is always going
to be delta x going to 0,

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which is 0.

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So you always, when you have a
differentiable function, you

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always have a derivative that's
going to be a limit of

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a 0 over 0 form.

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So you need to do some sort of
manipulation in order to, in

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order to get into a form
you can evaluate it.

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What we'd really like is to
manipulate this numerator

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somehow and pull out, say,
a factor of delta x.

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And then that could cancel with
the delta x we have in

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the denominator.

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Something, some trick
like that.

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Some algebraic or other
manipulation to make this into

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a form where we can plug
in and evaluate.

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So alrright, so right here
there's sort of only one

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manipulation that's natural to
do, which is we can add these

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two fractions together.

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So let's do that, and we
can rewrite this limit.

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The limit is delta
x goes to 0.

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All right, I'm going to pull
this 1 over delta x out front

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just to make everything look
a little bit nicer.

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It's 1 over delta x times--

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OK.

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i want to, you know, subtract
these two fractions.

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I want to put them over a common
denominator, so the

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denominator is just going
to be the product of the

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denominator.

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So that's 1 plus x plus delta
x quantity squared times 1

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plus x squared.

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OK.

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And so this fraction is 1
plus x squared over that

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denominator.

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And the second one is 1 plus x
plus delta x quantity squared

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over that common denominator.

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OK.

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So we still haven't got where
we want to be yet because we

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still have this 1 over
delta x hanging out.

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So OK, so we have to, you
know, keep going.

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And so here, I guess there's a,
this is sort of a problem

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that forces us a little
bit in one direction.

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You, know, there's not much we
can do with the denominator,

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but here in the numerator we can
expand this out and start

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combining stuff.

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So let's do that.

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So this is equal to--

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all right, well, the
limit hangs out--

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the limit is delta x goes to
0 of 1 over delta x times--

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OK, so 1 plus x squared
minus--

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all right, so if you
expand out x plus

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delta x quantity squared--

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using your favorite, either foil
or the binomial theorem

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or just whatever you like,
however you like to multiply

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two binomials--

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so we get a minus 1 minus x
squared minus 2x times delta x

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minus delta x squared.

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That's the top.

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Ok, and we haven't changed
the bottom.

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It's still 1 plus x plus
delta x squared

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times 1 plus x squared.

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OK.

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Well, so what?

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OK, so now some nice stuff is
starting to happen, which is

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this 1 and this minus 1 are
going to cancel, and this x

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squared and this minus x squared
are going to cancel.

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And then after we cancel those
terms we see that in the

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numerator here, everything is
going to have a factor of

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dealt x, right?

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These four are going to cancel,
and we'll just be left

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with these two terms,
both of which are

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divisible by delta x.

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So that's where this
cancellation we've been

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looking for is going
to come from.

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So let's keep going.

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So we cancel those, they
subtract, give us 0.

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This limit is equal to the
limit delta x goes to 0.

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OK.

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And then we can divide this
delta x from the denominator

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in, and what we're left with
upstairs is minus 2x minus

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delta x, the whole thing over
the same denominator, still.

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1 plus x plus delta x squared
times 1 plus x squared.

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All right.

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Great.

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So we've done this
manipulation.

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We finally found a delta x that
we could cancel with that

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delta x we started with
in the denominator.

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And now this limit is
no longer this 0

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over 0 form, right?

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When delta x goes to 0, the
top goes to minus 2x.

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And the bottom--

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well let's see, this delta x
just goes to 0-- so it's 1

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plus x squared times
1 plus x squared.

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So that's not 0 over 0.

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We can just plug
in to evaluate.

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So this is, just works out
to a minus 2x over--

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OK, 1 plus x squared times 1
plus x squared is 1 plus x

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squared, quantity squared.

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And so this is the derivative
that we were looking for.

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This is, just to remind you what
that was, that's d over

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dx of 1 over the quantity
1 plus x squared.

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Now, if you wanted, you could
check this a little bit by

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looking at the graph and looking
at this function and

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just making sure that
it makes sense.

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So for example, this function,
this derivative has the

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property that it's
0 when x is 0.

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And that's the only
time it's 0.

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And if we go back and look at
the graph that we drew over

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here, we see that's also
a property that

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this graph has, right?

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It has this horizontal tangent
line there, and then it

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diminishes off to the right and
it, on the left side it

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increases, then it has that
horizontal tangent line, and

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then it decreases.

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And so if we go back to this
function we see, yes indeed,

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when x is negative, this whole
thing is positive.

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And then at 0 it's 0, and then
it's negative thereafter.

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And similarly, you could note
that this function here is an

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odd function.

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If you change the sign of x,
that changes the sign of this

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whole expression, and so OK,
and so that makes perfect

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sense back here.

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The symmetry of this curve is
such that, you know, if we

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look at a tangent line to the
left of 0 and the symmetric

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tangent line to the right of
0, they're mirror images of

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each other.

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So their slopes or exactly
negatives of each other.

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So that's a nice way you can
sort of put the two different

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pieces of this problem together
in order to double

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check your work.

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So that's that.

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