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.