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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,

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and then I'm going to ask you
to use these rules to take

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a derivative of a function.

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So let's just look
at these rules first.

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So I want to point out, as
I'm talking about these rules,

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the first three are
written with natural log.

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But one can also
write them in any base

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as long as the base is the
same all the way across.

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So in any legitimate
base that one

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is allowed to use, so
with a positive base,

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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

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is equal to the sum
of the natural logs.

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

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of M plus natural log of N.

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The second one says the
natural log of a quotient

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is equal to the difference
of the natural logs.

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So natural log of
M divided by N is

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equal to natural log of
M minus natural log of N.

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This third one says that the
natural log of something raised

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to a power is that
power, as a coefficient,

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times the natural
log of the something.

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

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the natural log of M.

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

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a distinct 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 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,

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b-- maybe I want to
change the base-- of M,

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I can rewrite that
in the base e.

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I can write that
as natural log of M

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divided by natural log of b.

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And I want to point out,
a common mistake people

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make is sometimes they confuse
the second and the fourth

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because they both
have quotients.

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But notice that
the second one is

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the natural log of a
quotient, and the fourth one

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is about the quotient
of natural logs.

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So that's a distinct difference,
and hopefully then you

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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

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we know about 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

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we know that M and N
have to be positive.

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And so I can find--
actually, let

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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.

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And let me just write
out also what that means.

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Because exponential
and log functions

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are inverses of
one another, this

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means that a is equal
to natural log of M

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and b is equal to
natural log of N.

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So these are
equivalent 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

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to try and solve the problem.

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To try and prove number one.

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So the natural 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 times e

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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

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are inverses of one another, or
natural log of 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

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what those are-- it's natural
log of M plus natural log of N.

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So notice we've done
what we set out to do.

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Natural log of the
quantity M times N

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is equal to natural log of
M plus natural log of N.

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And in a similar flavor
one could immediately

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do number two, and number
three follows quite similarly,

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as well.

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It uses-- obviously,
these are going

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to use different
rules for exponents

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besides the product of
two exponential functions

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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

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might 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

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and three, at
least, on your own--

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might be helpful to look
at how those work using

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the same kind of rules here.

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

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I'd like us to
take a derivative.

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So what I want us
to look at is y

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equals the square root
of 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

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and 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 I

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have on the side
of the board there

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to take a derivative
to find y prime.

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

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and then we use some of
the rules of logarithms

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to simplify the expression
on the right-hand side.

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

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of the square root
of x times x plus 4.

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Now square root-- wow,
sorry-- square root

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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 raise it to the 1/2.

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So I'm going to use
rule number three

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and I'm going to bring that 1/2
that is a power out in front

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of the log.

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So I can rewrite this expression
as 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 of x and x plus 4.

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

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changes the product-- the
natural log of a product

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into the sum of
the natural logs.

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And I can rewrite this
as 1/2 natural log

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x plus 1/2 natural log
the quantity x plus 4.

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Essentially what
I'm doing here is

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I have to distribute this
1/2 because I had one term,

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and then I'm going to have two
terms that are added together,

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but the 1/2 applies
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 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,

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so there's some 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 prime,

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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

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of the natural log
evaluated at y,

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then times the derivative of y.

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You've seen this, I
think, a lot by now,

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but just to make sure you
understand where both of those

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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,

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if I use the chain
rule I get 1 over x

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plus 4 times the 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 a little bit more.

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And just notice that y prime
is going to equal y times

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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 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

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the derivative of y prime now--
or sorry, the derivative of y.

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Now I could combine
these two fractions

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into a single fraction
and try and make

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it look a little bit nicer, or
I 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

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I might find out that
I could have just taken

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the derivative the long way.

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

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where I can start to
find out something

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about the derivative of y.

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So I guess I'll stop there.