In this video, you're going to learn how to differentiate the
natural logarithm function.
F of X is Ellen of
X. And we're going to differentiate this from first
principles. And then we're going to move to use the result from
this to differentiate the exponential function F of X
equals E to the X, where you may recall that E is an irrational
number. It's approximately equal
to 2.718. And we call the the exponential constant.
Now before we can start to do the differentiation of the
natural logarithm, we need a particular result concerning the
exponential constant. So I'm going to derive that first. What
I'd like you to like you to do is I'd like you to think about
this expression. One plus T all raised to the power one over T.
And what we're going to do is we're going to evaluate this
expression for different values
of tea. And see what happens.
I'm going to drop a table of values, so I'm going to have two
columns. T values and then I'm going to evaluate one plus T to
the power one over T.
And let's start with a very simple T value. Let's suppose T
is one. Let's work this out.
20 is one. This bracket becomes 1 + 1 which is 2 and we want to
raise it to the power one over T&T is one, so we're raising it
to the power one over one, which
is still 1. And two to the power one is just two.
Let's pick a different value for T and see what happens. And this
time I'm going to choose a smaller value for T. I'm going
to let TB not point, not one.
Let's do the same thing.
Put T in here will get one plus North Point N 1, which is one
point N 1.
And we're going to raise it to
the power. 1 divided by T, which is North Point N 1.
And this is where you need a Calculator, so I want
one point, not one raised to the power 1 divided by
nought point nought one.
And that evaluates to two point 705.
So we see as TI has been made
smaller. The value of this expression one plus T to the one
over tease increased from 2 to
just over 2.7. Now let's see what happens when T gets even
smaller and this time my value for T. I'm going to choose to be
not .3 knots and 01 N point, not
one. And let's evaluate this
expression again. So I want one plus T which is going to be one
point not not not one.
And I want to raise it to the
power. One over T. That's one over nought point nought nought
nought one. And again we need a Calculator for this, so in
the Calculator goes one point naut naut naut one I'm going
to raise it to the power 1 divided by nought. Point
nought nought nought one.
And this time I get the value 2.718.
Now, this is an exercise that you could continue for yourself.
It would be a very useful thing
for you to do. And see what happens as T continues to get
smaller and smaller and smaller closer and closer to 0. And what
we find is that as T gets smaller, this expression over
here one plus T to the power one over T gets closer and closer
and closer, surprise, surprise. So this value over here 2.718
the exponential constant. Now that's a very important result
that will need when we want to differentiate the natural
logarithm of X from first
principles. I want to formally write that down in a
mathematical sort of way. What we've done is we've taken the
expression one plus T raised to the power one over T, and we've
evaluated this as TI gets closer and closer to 0, and
mathematically we write this like this. We right limb for
limit. T tends to 0.
Of this expression.
Is E the exponential constant? In other words, the limit as T
tends to zero of one plus T raised to the power one over T
is the exponential constant.
And that is a key result that will need very soon.
We're now ready to do a bit of differentiation from first
principles, so I'm going to remind you a little bit about
the formula for differentiation from first principles.
What I'd like you to imagine.
Is that we have a function?
F of X.
And what we're going to do is we're going to try to calculate
the derivative of this function at a particular point here.
Let's call the point where interested in, let's call
that point a.
And A is the point.
With X Coordinate X.
When we try to calculate the derivative of F of X at this
point. That's when we try to calculate F, dash, devex. What
we're doing is calculating the gradient of the tangent to the
curve at this point, so we're working out the gradient of the
line, which is just touching the curve at this point.
Now, the way we do this mathematically is we move from
our fixed point A to a nearby point a little bit further
around the curve.
Let's call this nearby point B.
Now The X coordinate at B is just a little bit more than the
X coordinate at A.
So the X coordinate here is X plus a little bit more of X, and
there's a little bit more of X in here we write as Delta X.
Delta X stands for a small change or small increment as we
call it in X.
So if this distance is Delta X.
The X coordinate at B must be X.
Plus Delta X.
What about the Y coordinates at A&B? Well, at a The X coordinate
is X. And a lies on the graph of F of X.
So the Y coordinate at a is just the function evaluated at X, so
the Y coordinate at A.
Is just F.
Evaluated attacks or F of X?
The Y coordinate at B in a similar fashion is just the
function evaluated at this X value, which is X Plus Delta X.
So the Y coordinate at B.
Will be just the function evaluated at X Plus Delta X.
Knowing all these values F of XF of X Plus Delta XX&X Plus Delta
X, we can calculate the slope of the line through A&B.
What's the slope of the line through A and be? Well, we do
that by forming a little right angle triangle in here.
And then the slope of the line
through A&B. Is this distance?
Divided by. This distance.
Now this distance is very straightforward to workout. It's
the larger distance from the X axis up to be.
Subtract the smaller distance from the X axis up to a. In
other words, this distance in here is F of X Plus Delta X.
Minus F of X.
This distance in here.
Is just Delta X.
So now we're in a position to write down a formula for the
slope of the line through A&B.
We say that the slope.
As a baby.
Is equal to this vertical distance which we've got here F
of X plus Delta X.
Minus F of X.
All divided by the horizontal
distance. Which is just Delta X.
What's this that we found here? This quantity is the slope.
Of this line through A&B, what we're really interested
in is the slope of the tangent at A and we achieve
that by bringing B closer and closer into point A.
What we do is we let be get closer and closer to A and as
that happens the slope of this line will become eventually the
slope of the tangent today.
Now as we bring B round to a, what's happening is
that Delta X is getting closer and closer to 0.
So what we want to do is we want to workout this expression in
the limit as Delta X approaches 0. So we want to workout the
limit as Delta X tends to zero of all this.
When we work all
that out. That will give us a value for the
derivative. Of the function at point a, so that will define F
dash of X.
That is a very important result that will need as well when we
come to differentiate the natural logarithm from first
principles. Now I've rushed through this rather quickly.
There's a whole unit on differentiation from first
principles, and if you want a great deal more detail, I would
go back and look at that video.
OK, but now into the real business of the video. We want
to learn how to differentiate this function F of X is
the natural logarithm Ln of X.
Now to do that will need lots of different things. Lots of
results from different branches of mathematics, so let's
summarize what will need first.
One thing will need is the formula for the derivative that
we just wrote down F dash of X is the limit as Delta X tends to
zero of F of X Plus Delta X.
Minus F of X.
All over Delta X.
That's the formula for differentiation from first
principles and will apply that to this function in a minute.
Will also need the result that I gave you right at the beginning
of the video, which was that the exponential constant E can be
written as the limit.
As T tends to 0.
Of one plus T or raised to the power one over T, so that's
another result that will need.
We need a bit more actually. We need something about logarithms.
You need to know that the log of A minus the log of B is the log
of a divided by B. That's one of the laws of logarithms.
And another law of logarithms that will need is that if you
have a multiple M times a logarithm, then that multiple
can come inside as a power. So we can write M log a as the
logarithm of A.
Bring it in as a power to the
power N. These are two laws of logarithms, and if you're not
familiar with those then you should go back and look at the
unit on laws of logarithms.
1 final result we need to know that the natural logarithm of
the exponential constant E is one and again you'll find
information about that.
On the unit, on logarithms. So all those things will need as we
start to differentiate this
function here. OK, let's dive in. Here's our function F of X
is log X.
We want to put this this function log X into this formula
for differentiation from first principles. To do that, we're
going to have to evaluate F at X Plus Delta X.
Let's do that.
I remember this function. The logarithm takes the log of the
input, so when the input is X Plus Delta X, the output will be
the logarithm of X Plus Delta X.
So now we've got F of X Plus
Delta X. And we've got F of X.
We can plug them into the top line of the formula.
F of X Plus Delta X.
Minus F of X.
What will that be?
It will be the logarithm of X Plus Delta X.
Subtract F of X so it'll be. Subtract the logarithm
of X. Ah, with
a logarithm. Minus a
logarithm. And this is where I'm going to use this law of
logarithms here. Which says that if you've got the log of a
quantity minus the log of another quantity, that's the
same as the log of the first term divided by the second term.
So In other words, we can simplify these two logarithms to
a single logarithm. The Ln of X Plus Delta X. That's the first
one divided by the second one, which is just X.
And because I know where this is going, I want to simplify this a
little bit more. I'm going to recognize that within this
bracket here. We've X divided by
X. Which is just one.
And we've Delta X divided
by X. OK, so just a bit of tidying up there.
Also, I'm keeping in mind that I've got this result here about
the exponential constant and in this limiting process here
you'll remember we have the quantity one plus T.
In order that I can use this result, I'm going to make a
substitution in here and I'm going to write Delta X over X.
As simply T so I can start to
use this result. So I'm going to let Delta X over X equal T.
So I'll have a log of one plus T.
We still going to use the formula for differentiation from
first principles. Let's go to that. We want F dash of X is the
limit of all this quantity. Let's take this quantity and
divide it by Delta X.
So F of X Plus Delta X.
Minus F of X divided by Delta
X. Well, it's the top line. Is this quantity in here the
logarithm? And we want to divide it all by Delta X, but I've made
a substitution here. I've let Delta X over X equal T. In other
words, Delta X is X times T. So instead of the Delta X here I'm
going to write this as one over X times T down there.
OK, so far so this is the expression for our derivative.
The only thing we have to do.
Is take the limit as Delta X tends to 0. Let me write
that expression down again.
So that's where we've got to and what we want to do is let Delta
X 10 to 0 now.
Remember that Delta X we've said was X times T.
Remember also that X is a fixed point, but remember that from
right back at the beginning when we came up with our formula for
different station from first
principles. The point a was a fixed point. It was believed
that was the variable point. So a was fixed, X was fixed.
Now if Delta X tends to note an ex is fixed.
That means that team must tend to not as well. So in this
process over here, we must let T
10 to 0. So what we really
want is the
limit. As TI tends to
0. Of one over XT.
Natural logarithm of 1 plus T and when we work this out,
that's actually going to give us F dash devex, the derivative
that we're looking for.
Now we're nearly finished. I just want to tidy this up a
little bit more, take the limit, and then will be will be all
done. You remember one of the laws of logarithms that I just
wrote down was that M Log A?
Was logged A to the power N.
The power. The power can be written outside or a number
outside can go inside as a
power. Well, with the one over T
outside here. And that one over T I'm going to take inside the
logarithm as a power. So I'm going to write this as the
limit. As TI tends to 0.
Of one over X, I'm going to leave that bit outside, but
the one over T bit will go inside and have logarithm
one plus T and one over T inside as a power and give
me a one over T up there.
Now this is looking much more like that result that we had at
the very beginning. Remember the result we had at the beginning
about the exponential constant. It said that E was the limit as
T tends to zero of one plus T to the power one over T.
Now you see I've got the one plus T to the one over T in here
I've got a limit as T tends to
zero there. The X is fixed, So what we can do is we can leave
the one over X there. We want to take the limit as T tends to
zero of this expression in here, which is the logarithm.
Of one plus T to the one over T.
But we've already said that as T tends to 0, this quantity just
tends to the exponential
constant E. So we end up with just simply Ln of E.
Final result we had was that the natural logarithm of the
exponential constant was one.
So this bit in here is going to simplify to just one.
In other words, we've deduced that the derivative F Dash Devex
is simply just one over X Times one or one over X.
It's a very important result. Let me summarize it for you.
We're saying that if F of X is the natural logarithm of X.
Then F dash to vex. The derivative is simply one over X
and we've proved that using differentiation from first
principles. I'm well aware that to do that we require lots of
different bits of mathematics from lots of different areas.
Differentiation from first principles, properties of
logarithms, and so on, so you may need to look back at lots of
other bits of work that you've done in the past in order to
pull all this together, but we've got a very important
result now that we can use in lots of situations. What we've
shown is that if you want to differentiate with respect to X
and natural logarithm function, Ellen of X, the result that
you'll get is one over X.
We can use this result whatever the letter is in here. So for
example, if we had D by DY.
Of Allen of Y.
The result will be one over Y.
If we had D by DTLN of T, the result will be one over
T. In other words, when you differentiate the natural
logarithm function, you'll get one over whatever the
variable is.
OK, very important result.
We're going to use that result now to move on and differentiate
the exponential function.
So now I'm going to look at the function Y is a function of X,
is equal to Y to the X.
Or just simply Y equals E to the X and we're going to
learn how to differentiate this exponential function.
Let me remind you something
about logarithms. If A equals B to the power, see.
An alternative way of writing
this expression. Is that the logarithm to base B of A?
Is C. This is something that's been discussed at great length
in the unit on logarithms, but it's a result that will need
here in order to differentiate the exponential function.
What I want to do is take this Y equals E to the power
X and write it in its log rhythmic form.
And the log rhythmic form using the results over here is that
the log to base E.
Of Y is X, so that's just the log arhythmic form of
the same expression here.
Now log base E of X is what we call Ln of X
the natural logarithm.
What I want to do is deal with this equivalent form instead of
the original form.
To differentiate this, what I want to do is differentiate both
sides with respect to X.
So I want deep IDX of the left
hand side. Will equal D by DX of the right hand
side. Now this is easy if you differentiate X with respect to
X, you just get one.
The left hand side needs a bit
more thought. What we've got here is that why is a function
of XY is E to the X?
And we're taking another function of it. We want the
logarithm of Y. So we've got a function of a function we're
trying to differentiate a function of a function.
So we need a rule for differentiating functions of
functions, and that's called a chain rule or the rule for
differentiating function of a function.
The rule says that if you want to differentiate this function
of a function with respect to X, you differentiate it with
respect to Y.
And then multiply by DYDX.
So that's the function of a function rule applied there.
And if you've not met that before, there's a unit on
differentiating function of a function using the chain rule.
On the right hand side we have one.
Nearly finished all we have to do now is differentiate log Y
with respect to Y, but we've just done a lot of that. We've
just realized improved that if you want to differentiate, log Y
with respect to Y. You'll just get one over Y, so following on
from here will get one over Y.
Multiplied by DYDX.
And that's got to equal 1.
So rearranging all this dyd X.
Will equal just why times one which is just Y.
But why is E to the X?
So we've shown that dyd X is E to the X.
So we've another very important result now, and the result is
that if Y is E to the power X.
Then the derivative dyd X.
Is the same function E to the power X, and that's a result
which is well worth remembering. It will crop up
over and over again in lots of different calculations. Lots
of different applications. In fact the exponential function
or multiples of it is the only function which when you
differentiate it, it gives you the same function that you
started with, so that's something worth remembering.