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.