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CHRISTINE BREINER: Welcome
back to recitation.

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Today we're going to talk about
some rules of logarithms

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that you're going to
need to remember.

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We're going to prove why one of
them is true, and then I'm

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going to ask you to use these
rules to take a derivative of

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a function.

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So let's just look at these
rules first. So I want to

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point out as I'm talking about
these rules, the first three

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are written with natural log.

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But one can also write them in
any base as long as the base

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is the same all the
way across.

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So in any legitimate base that
one is allowed to use, so with

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a positive base, one can use it
all the way across instead

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of the natural log.

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So the first one says that the
natural log of a product is

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equal to the sum of
the natural logs.

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So natural log M times N is
equal to the natural log of M

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plus natural log of N.

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The second one says the natural
log of a quotient is

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equal to the difference
of the natural logs.

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So natural log of M divided by N
is equal to natural log of M

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minus natural log of N.

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This third one says that the
natural log of something

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raised to a power is that power
as a coefficient times

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the natural log of
the something.

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So natural log of M to the
k is equal to k times the

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natural log of M.

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And what I want to point out
is that there's a distinct

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difference where the power is.

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So if the power is inside the
argument then this rule holds,

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but if the power is outside
the argument--

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so if it's natural log
of M, the whole

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thing raised to a power--

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this does not work.

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This is not equal to what's
written above.

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And then the third-- the
fourth one-- sorry.

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The fourth one is a change
of base formula.

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So if I have, if I have log base
something b, that maybe I

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want to change the base
of M, I can rewrite

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that in the base e.

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I can write that as natural
log of M divided by

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natural log of b.

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And I want to point out a common
mistake people make is

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sometimes they confuse the
second and the fourth because

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they both have quotients.

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But notice that the second one
is the natural log of a

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quotient, and the fourth one
is about the quotient of

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natural logs.

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So that's a distinct difference,
and hopefully then

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you see that they are not, these
two statements are not,

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in fact, the same statement.

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So now what I'd like to do is,
using what we know about

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exponential and log
functions--

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I want to prove number one.

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So let's set out to do that.

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Well, in order to make this top
line make sense we know

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that M and N have
to be positive.

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And so I can find--

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actually, let me write first
what we're doing.

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We're going to prove one.

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So with M and N both positive I
can find values a and b such

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that e to the a equals M and e
to the b is equal to N. And

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let me just write out also
what that means, because

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exponential and log functions
are inverses of one another.

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This means that a is equal to
natural log of M and b is

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equal to natural log of N.
So these are equivalent

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statements.

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This statement and this
statement are equivalent.

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This statement and this
statement are equivalent.

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So now let's use that
information to try and solve

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the problem.

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To try and prove number one.

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So the natural a log of M times
N, well, what is that?

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M is e to the a, N
is e to the b.

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So I can write this as natural
log of e to the a

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times e to the b.

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What's e to the a times
e to the b?

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This is where we use our
rules of exponents.

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e to the a times e to the
b is e to the a plus b.

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So this is natural log
of e to the a plus b.

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And now, what's the point?

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The point is that natural log
in exponential functions are

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inverses of one another,
or natural log of

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e to the x is x.

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So natural log of e to the a
plus b is just a plus b.

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And I've already recorded
for you what those are--

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it's natural log of M plus
natural log of N. So notice

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we've done when we
set out to do--

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natural log of the quantity M
times N is equal to natural

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log of M plus natural
log of N.

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And in a similar flavor one
could immediately do number

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two, and number three follows
quite similarly, as well.

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It uses, obviously, these are
going to use different rules

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for exponents besides the
product of two exponential

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functions is equal to the
sum of the powers.

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It's going to use some
of those other rules.

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And I believe that some of
these other things might

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actually also be proven
in a later lecture

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in the actual course.

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So you'll see these.

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But I would say, you might want
to try and prove two and

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three, at least, on your own--
might be helpful to look at

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how those work using the same
kind of rules here.

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So now what I'd like us to do is
using these rules, I'd like

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us to take a derivative.

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So what I want us to look at is
y equals the square root of

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x times x plus 4.

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And we'll just assume that
x is bigger than 0.

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And I want you to
find y prime.

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Now you could do this just brute
force, cranking it out.

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But I'd like you to try and use
the log differentiation

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technique in order to find
this derivative.

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I'll give you a moment to do it
and then I'll come back and

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I'll show you how I do it.

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OK.

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Welcome back.

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So I'm going to use the log
differentiation and the rules

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I have on the side of the
board there to take a

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derivative to find y prime.

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So first what we do is we take
the log of both sides and then

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we use some of the rules of
logarithms to simplify the

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expression on the
right hand side.

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So I will take natural log y is
equal to natural log of the

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square root of x
times x plus 4.

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Now square root--

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wow, sorry--

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square root is the power of
something raised to the 1/2.

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Right?

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That's what it means to
take a square root.

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You can take this whole
product and

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raise it to the 1/2.

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So I'm going to use rule number
three and I'm going to

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bring that 1/2 that is a power
out in front of the log.

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So I can rewrite this
expression as

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1/2 log of this product.

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That's one too many parentheses,
but that's OK.

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OK.

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So I have 1/2 the natural log of
the product x and x plus 4.

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So now I'm going to use rule
number one which changes the

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product, the natural log of a
product into the sum of the

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natural logs.

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And I can rewrite this as 1/2
natural log x plus 1/2 natural

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log, its quantity x plus 4.

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Essentially what I'm doing here
is I have to distribute

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this 1/2 because I had one term,
and then I'm going to

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have two terms that are added
together, but the 1/2 applies

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to both of them.

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So now I have this nice setup.

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I have natural log
of y is equal to

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something in terms of x.

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And now I can take the
derivative of a both sides.

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Now remember, I want to find
y prime, so there's some

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implicit differentiation
going on.

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So let's just be careful
when we do that.

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If I take the derivative of this
side I don't just get y

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prime, I get y prime over y.

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Where does that come from?

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Well, d dx of this expression
is the derivative of the

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natural log evaluated
at y then times the

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derivative of y.

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You've seen this, I think, a lot
by now, but just to make

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sure you understand where
both of those come from.

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So when I take the derivative
here I get y prime over y.

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When I take the derivative here
with respect to x, well,

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derivative of natural log
of x is just 1 over x.

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So I get 1 over 2x.

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And then the derivative of
natural log of x plus 4, if I

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use the chain rule I get 1
over x plus 4 times the

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derivative of x plus 4, which
is still just 1--

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so I get 1 over 2
times x plus 4.

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So now I wanted us
to find y prime.

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So to find y prime I'm
going to move over

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a little bit more.

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And just notice that y prime
is going to equal y

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times all of that.

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Well, I know y.

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So I'm going to write
what y is.

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y is the square root
of x times x plus

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4 times this quantity.

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1 over 2x plus 1 over
2 times x plus 4.

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So that's actually one way to
write the derivative of y

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prime now--

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or sorry-- the derivative
of y.

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Now I could combine these two
fractions into a single

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fraction and try and make it
look a little bit nicer, or I

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could just leave it this way.

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This is technically
a derivative.

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So if I started trying to
combine things I might find

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out that I could have
just taken the

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derivative the long way.

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So this is a nice short way to
just get to a place where I

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can start to find out
something about the

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derivative of y.

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So I guess I'll stop there.

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