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

logarithms to help simplify
certain functions before we

differentiate them.

To start off with, let's
just remind ourselves

about logarithms
themselves. So if we have

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

base E, and if we
differentiate that why by

DX is equal to one over X?

But of course. Inside this
logarithm here this might

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

be why equals the log of a
function of X.

So what happens when we
differentiate this? Well, we

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

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

decks over F of X.

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

But also the properties of
logarithms themselves so.

Let's take our first
example, why?

Equals the natural log.

3X to the 4th plus
Seven all raised to

the fifth power.

And we want to be able to

differentiate this. Well.

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

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

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

inside the log.

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

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

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

logarithms. And now this is now
much easier to differentiate

because this inside the log is
as very straightforward

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

equal to. 5.

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

function inside the log, so
we differentiate that.

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

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

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

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

looked like up here quieter
fearsome derivative that we were

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

let's have a look at another

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

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

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

formula for differentiating
a quotient.

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

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

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

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

the log of the denominator.
Because in logarithms we do

division by subtracting the logs
of the respective parts.

And now these two are both very
easy to differentiate.

So we can have the why by DX

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

differentiate it that's minus

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

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

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

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

1 + 2 X.

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

straightforward
differentiation. What we have

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

them together all over the
same denominator and this is

our denominator. The product
of these two.

So that's 1
- 3 X times

by 1 + 2 X.

So we ask ourselves, how
many times does this

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

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

many times does this go
into this? And the answer

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

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

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

multiply out this bracket minus
three times everything inside

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

everything inside the bracket,
so that's minus.

2 + 6 X.

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

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

Over 1 - 3 X
times by 1 + 2

X. But so long as we're good

with the algebra. This
differentiation is well

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

take another example.

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

Now.

Real problems having the unknown
the variable appear yet again in

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

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

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

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

X to the power sign X
and if we remember.

By using our log laws.

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

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

function by the power.

And now this is a
straightforward product. It's

sign X times by Log X.

We know that we can
differentiate this. What

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

we're differentiating with
respect to X, we can

differentiate this log Y
implicitly, and so the

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

the X equals.

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

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

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

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

the derivative of log X is one.

Over X.

Close the which is log

X. Times the
derivative of U and

you re sign X its
derivative is

therefore cause X.

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

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

Plus and if we're putting
everything over this denominator

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

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

cause X all over X.

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

DY by The X.

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

its X to the power sign X?

So that's sine XX
log X cause X.

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

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

differentiation and again,
something looked impossible at

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

able to differentiate.

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

complicated one. Quotient one
really is quite.

Complex.

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

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

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

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

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

both sides.

And then having taken logs of

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

simplify this side.

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

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

Minus the log of
the denominator.

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

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

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

X by three.

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

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

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

log of one plus X squared.

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

simpler to differentiate.

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

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

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

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

two over 1 - 2 X.

Minus 1/2.

Differentiating the log of
one plus X squared, the

function is one plus X
squared, so its derivative is

2X over one plus X squared.

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

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

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

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

So.

We've got one over Y.

DY by the X
is equal to.

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

give us minus.

6.

All over
1 - 2 X.

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

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

for a common denominator that
must be the product of the

two denominators.

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

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

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

needed multiply the minus X by 1
- 2 X.

Multiply out the bracket minus 6
- 6 X squared.

Minus X +2 X

squared. All over 1
- 2 X one plus X

squared.

We can simplify the top.

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

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

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

out so we have minus.

6 plus X plus
four X squared.

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

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

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

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

one plus X squared square

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

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

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

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

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

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

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

So do you? Why by
DX finally will be.

Let's take this top line.

Minus 6 plus X
plus four X squared.

Times by.

1 - 2 X squared.

1 - 2 X or
squared all over.

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

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

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

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

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

laws of logarithms to help us
simplify certain functions

before we come to
differentiating.

That can make the whole
process of differentiation

so much simpler.