www.mathcentre.ac.uk/.../8.2a%20Differentiating%20ex%20and%20in%20x.mp4

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.