WEBVTT
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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.