- [Instructor] What we're going to do in this video
is prove to ourselves that the derivative
with respect to X
of natural log of X
is indeed, equal to
one over X.
So let's get started.
So just using the definition of a derivative
if I were to say the derivative
with respect to X
of natural log of X
that is gonna be the limit
as delta X approaches zero
of the natural log of X plus delta X
minus the natural log of X.
All of that over
delta X.
So now we can use a few logarithm properties.
We know that if I have the natural log of A
minus the natural log of B
this is equal to the natural log
of A over B.
So we can use that right over here
where I have the natural log over something
minus the natural log of something else.
So all of this is going to be equal to
the limit as delta X
approaches zero
of the natural log
of this divided by that.
So X plus delta X
over X.
X plus delta X
over X
all of that over
delta X.
And actually, let me just write it that way.
All of that over
delta X.
Well jee, all I did, if I have
natural log of that minus natural log of that
that's the same thing as natural log
of this first expression
divided by that second expression.
It comes straight out of our logarithm properties.
Well, inside of this logarithm X divided by X is one.
And then delta X divided by X
just write that as delta X
over X.
So that's another way of writing that.
And then we could put a one over delta X up front.
So we could say this is the same thing as
this is equal to
the limit
as delta X
approaches zero
of, I 'll do this in another color.
So this, I can rewrite as one over delta X
times the natural log
of
one plus
one plus delta X
over X.
Let me close that parenthesis.
So now we can use another exponent property.
If I have, I'll write it out here.
If I have A times a natural log of B
that is equivalent to
the natural log of B
to the A.
And so here, this would be the A in that case.
So I could bring that out and make that an exponent on this.
So this is all going to be equal to
the limit
as delta X approaches zero
of the natural log
of, give myself some space,
one
one plus delta X
over X
to the
one over delta X power.
One over delta X power.
Now this might start to look familiar to you,
it might start to look close to the definition of E.
And we are, indeed, getting close to that.
In order to get there fully
I'm gonna do a change of variable.
I'm going to say
let's let
N equal delta X over X.
Delta X
over X.
Well, in that case,
then if you multiply both sides by X
you get delta X is equal to NX.
Again, just multiply both sides of this equation by X
and swap to the sides there.
And then if you want one over delta X
one over delta X
that would be equal to
one over NX
which we could also write as one over N
times one over X.
Or actually, let me write it this way.
Actually, that's the way I do want to write it.
So these are all the substitutions
that I want to do when I change a variable.
And we also want to say, well look.
As delta X is approaching zero
what is N going to approach?
Well, as delta X approaches zero
we have N will approach zero, as well.
Zero over X, well that's just gonna be zero
for any X that is not equal to zero.
And that's okay because
zero is not even in the domain
of the natural log of X.
So, this is going to be, for our domain, it works.
That is, delta X approaches zero
N approaches zero.
And you could think about it the other way around
as N approaches zero, delta X approaches zero.
So now let's do our change of variable.
So if we make the substitutions
instead of taking the limit as delta X approaches zero
we are now going to take the limit as
N approaches zero
of
the natural log
of
give myself some parenthesis.
And I'll say, one plus
and now this is the same thing as N.
One plus N.
And then all of that is going to be
raised to
the one over N times one over X
that's what one over delta X is equal to.
This is one over delta X right over here.
Which we have over here.
And it's the same thing as one over N
times one over X.
So let me write that down.
So this is the same thing
as one over N times
one over X.
Now we can use this same exponent property
to go the other way around.
Well actually, let me just rewrite this another time.
So this is going to be the same thing
as the limit
as N approaches zero
of the natural log
of
one plus, I'll just write this in orange.
One plus N
to the one over N.
If I raise something to the exponent
and that's times something else,
I can that's the same thing as raising
it to the first exponent.
And then raising that to the second value.
This comes, once again, straight out of our
exponent properties.
And now we can use this property the other way
to bring this one over X out front.
But in fact, the one over X itself is not affected
as N approaches zero.
So we can even take it completely out of the limit.
So we could take it all the way over there.
And this is when you should be getting excited.
This will be equal to
one over X
times the limit
as N approaches zero
of the natural log
of
one
one plus N.
We do that in orange color.
One plus N to the one over N.
And now, what really gets affected
is what's going on inside of the natural log.
That's where all of the Ns are.
And so let's bring the limit inside of that.
So this is all going to be equal to
give myself some space.
A little bit of extra chalk board space.
This is going to be equal to
one over X
times the natural log
times the natural log of
the limit
as N approaches zero
of
one plus N
to the one over N.
Close those parenthesis.
Now this is exciting.
What is inside the natural log here?
Well, this business right over here, this is
a definition of the number E.
So that is equal to E.
Well, what's the natural log of E?
Well that's just one.
So it's one over X times one.
Well, that is indeed equal to
one over X.
Which is exactly
the result that we were looking for.
That the derivative
with respect to X of natural log of X,
is one over X.
Very exciting.