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In this video, we're going to
be looking at how we can use

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logarithms to help simplify
certain functions before we

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differentiate them.

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To start off with, let's
just remind ourselves

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about logarithms
themselves. So if we have

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Y equals the log of X,
then this means log to

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base E, and if we
differentiate that why by

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DX is equal to one over X?

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But of course. Inside this
logarithm here this might

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not just be X, it might be a
function of X, so it might

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be why equals the log of a
function of X.

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So what happens when we
differentiate this? Well, we

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have a rule that tells us why by
DX is equal to.

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The derivative of F
of X. That's F dash

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decks over F of X.

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Now. We're going to be making
use these results.

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But also the properties of
logarithms themselves so.

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Let's take our first
example, why?

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Equals the natural log.

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3X to the 4th plus
Seven all raised to

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the fifth power.

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And we want to be able to

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differentiate this. Well.

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It looks very complicated in
here, but here we are raising.

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3X to the power 4 + 7 to the
power five and one of the things

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we do know about logarithms is
that if we are raising what's

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inside the log.

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To the power five, that's
exactly the same as multiplying

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the log by 5. So In other words,
we can rewrite this as Y equals

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5 times the log of three X to
the 4th plus 7 using our laws of

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logarithms. And now this is now
much easier to differentiate

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because this inside the log is
as very straightforward

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function. So we can now
differentiate the why by DX is

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equal to. 5.

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Times by now, let's remember
what we do. This is the

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function inside the log, so
we differentiate that.

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So that gives us 12X cubed
and the derivative of the 70s

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zero over the function itself 3X
to the 4th +7.

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And if we just do a little bit
of simplifying five times by 12

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gives us 60 X cubed over 3X to
the 4th plus Seven and what

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looked like up here quieter
fearsome derivative that we were

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going to have to do here, it's
turned out quite simply. So

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let's have a look at another

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example. Supposing we've got Y
equals the log of.

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We've got 1 - 3 X divided
by 1 + 2 X. Now what

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we've got here is the log of
a quotient, and there is a

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formula for differentiating
a quotient.

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That's going to lead to a very
very complicated expression, but

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what we can do is make use of
the laws of logarithms because

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the laws of logarithms tell us
that when we're doing the log of

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a quotient than that is simply
the log of the numerator minus

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the log of the denominator.
Because in logarithms we do

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division by subtracting the logs
of the respective parts.

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And now these two are both very
easy to differentiate.

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So we can have the why by DX

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equals. And this is a function
of X 1 - 3 X, so we

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differentiate it that's minus

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three. Over the function
itself, 1 - 3 X minus that

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minus sign, and now we do
the same for this logarithm.

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Here's the function of X
which we differentiate, so

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the derivative of 1 + 2 X is
2 over the function itself.

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1 + 2 X.

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Now that is a penalty to pay
for getting away with such a

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straightforward
differentiation. What we have

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to do now is put these two
together. We have to bring

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them together all over the
same denominator and this is

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our denominator. The product
of these two.

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So that's 1
- 3 X times

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by 1 + 2 X.

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So we ask ourselves, how
many times does this

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divide into this? And the
answer is 1 + 2 X, so we

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have minus 3 * 1 + 2 X
minus two and then how

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many times does this go
into this? And the answer

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is 1 - 3 X so it's 2
* 1 - 3 X.

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Now we need to simplify this so
we can keep the denominator as

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it is 1 - 3 X times
by 1 + 2 X and let's

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multiply out this bracket minus
three times everything inside

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the bracket, so it's minus 3 -
6 X and then minus two times

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everything inside the bracket,
so that's minus.

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2 + 6 X.

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Now we need to simplify this bit
on top, and we've minus 6X plus

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6X, so that's no axis minus 3 -
2 - 5. So we have minus 5.

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Over 1 - 3 X
times by 1 + 2

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X. But so long as we're good

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with the algebra. This
differentiation is well

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worth the simplifying as we
have done here, so let's

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take another example.

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This time let's take Y equals
X to the power sign X.

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Now.

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Real problems having the unknown
the variable appear yet again in

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the power in the index, so this
sign X is the problem. If it

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wasn't in the index, we might be
able to differentiate it, but

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one way of getting it down, so
to speak out of the index is to

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take logs of both sides so we
have log Y equals the logs.

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X to the power sign X
and if we remember.

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By using our log laws.

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When we raise the function
inside the log to a power, then

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that's just the same as
multiplying the log of the

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function by the power.

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And now this is a
straightforward product. It's

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sign X times by Log X.

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We know that we can
differentiate this. What

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about this at the left
hand side log Y? Well, if

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we're differentiating with
respect to X, we can

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differentiate this log Y
implicitly, and so the

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derivative of the log of Y
is one over Y times DY by

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the X equals.

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Now we need to differentiate
this sign X times by Log X. It's

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a product you times by V and we
know that the derivative of

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product is U times DV by DX Plus
V Times du by DX. So let's write

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that down you which we said was
sign X times divided by DX and

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the derivative of log X is one.

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Over X.

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Close the which is log

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X. Times the
derivative of U and

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you re sign X its
derivative is

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therefore cause X.

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Now we can do a little bit of
simplifying. Here we can put it

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all. Over this denominator of X,
and so we have sine X.

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Plus and if we're putting
everything over this denominator

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of X, we need to multiply
anything that's not over the

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denominator of X already by X,
so that's X Times Log X times

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cause X all over X.

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But this side with one over Y
from what we're really after is

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DY by The X.

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And so we need to multiply up by
why, but we know what? Why is

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its X to the power sign X?

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So that's sine XX
log X cause X.

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OX and we're going to multiply
up by Y, and that's X sign

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X to the power sign X, and
so we finished the

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differentiation and again,
something looked impossible at

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the beginning by making use of
the laws of logs, we have been

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able to differentiate.

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So let's take a look another
example, this time a more

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complicated one. Quotient one
really is quite.

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Complex.

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Now again, we could do this as a
quotient you over V.

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And use the formula which if
we remember is VDU by DX minus

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UDV by the X all over V
squared. But simply having

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said it, that's quite
frightening. So if we have a

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look at this one of the ways
we can do it is to take logs

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both sides.

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And then having taken logs of

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both sides. We can make use of
our laws of logarithms to

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simplify this side.

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So let's begin by remembering
that if we have a quotient when

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we do the logarithm, then that's
the log of the numerator.

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Minus the log of
the denominator.

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But we can simplify this one
even further because we're

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raising 1 - 2 X to the power
three, and in terms of our laws

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of logs, that's exactly the same
as multiplying the log of 1 - 2

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X by three.

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Minus now here we've got a
square root and the square

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root is the same as raising
something to the power 1/2

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so that this is exactly the
same as taking a half of the

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log of one plus X squared.

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Now what we see is that this
that we've got here is much

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simpler to differentiate.

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So we can do that. The
derivative of the log of Y doing

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it implicitly is one over YDY by
X equals 3 Times Now we want

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to now differentiate the log of
1 - 2 X. So the function

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is 1 - 2 X. So we
differentiate that it's minus

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two over 1 - 2 X.

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Minus 1/2.

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Differentiating the log of
one plus X squared, the

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function is one plus X
squared, so its derivative is

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2X over one plus X squared.

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Again, there's a penalty to be
paid. We really ought to put

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these back together again, so
it's all one.

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So let's do that and let's just
simplify a little bit here.

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There's a two will cancel there
with a 2.

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So.

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We've got one over Y.

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DY by the X
is equal to.

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We've got 3 times by minus two
over 1 - 2 X, so that's going to

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give us minus.

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6.

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All over
1 - 2 X.

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We've got minus X over one
plus X squared.

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Now we need to bring these
two terms together so we look

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for a common denominator that
must be the product of the

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two denominators.

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And 1 - 2 X goes into this
one plus X squared times minus

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six times by one plus X squared
minus X times by well one plus X

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squared goes into the whole of
this 1 - 2 X times. So we

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needed multiply the minus X by 1
- 2 X.

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Multiply out the bracket minus 6
- 6 X squared.

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Minus X +2 X

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squared. All over 1
- 2 X one plus X

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squared.

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We can simplify the top.

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Minus 6 minus X minus four X
squared all over 1 - 2 X

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one plus X squared. And now
finally we want DY by the X

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equals, but it's going to be
this. Let's take that minus sign

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out so we have minus.

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6 plus X plus
four X squared.

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All over 1 - 2 X
times by one plus X squared.

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And we need to multiply both
sides by Y and let's just recall

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what why was. Why was 1 - 2
X or cubed?

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So that's 1 - 2
X or cubed all over

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one plus X squared square

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root off. And clearly is a
little bit more simplified, can

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be done here, because we've got
a 1 - 2 X here and a 1 - 2 X

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cubed there so we can cancel
that with one of those.

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Here we've got a one plus X
squared times by the square root

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of 1 plus X squared, and if we
follow our laws of indices,

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that's to the power one. And
this is to the power half. So we

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add them together. That gives us
to the power three over 2.

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So do you? Why by
DX finally will be.

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Let's take this top line.

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Minus 6 plus X
plus four X squared.

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Times by.

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1 - 2 X squared.

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1 - 2 X or
squared all over.

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Just check on that denominator
one plus X squared to the power

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one times by one plus X squared
to the power a half and the one

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and add it to the half. Gives us
1 1/2 or three over 2, so that's

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one plus X squared raised to the
power three over 2.

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So what we've seen in this
video is that we can use the

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laws of logarithms to help us
simplify certain functions

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before we come to
differentiating.

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That can make the whole
process of differentiation

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so much simpler.