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## AP®︎ Calculus BC (2017 edition)

# Proof: the derivative of ln(x) is 1/x

Proving that the derivative of ln(x) is 1/x by using the definition of the derivative as a limit, the properties of logarithms, and the definition of 𝑒 as a limit.

## Want to join the conversation?

- Isn't the definition of e

(1+1/n)^n?

In the video, we used the definition as (1+n)^(1/n). Are those two the same thing?(5 votes)- You could do a change of variables. If you replace every n with 1/n and every 1/n with n. In the original limit, n went to infinity, so 1/n is going to 0. So the limit is the same as saying the limit as n tends to 0 of (1+n)^(1/n).(9 votes)

- At6:20why can we say that 1/x is not affected as n approaches zero?(4 votes)
- Because there is no 'n' in the expression for 1/x. There's just an x.(11 votes)

- at2:23how do we know that we can not stop there ? I mean if I would substitute Delta X approaching zero, then 1 over Delta X would become infinitely large. Natural log[ of 1 plus (delta x over x) ] would become natural log of 1, since delta x over x would be approaching zero. And ln 1 = 0 . That would give us infinity multiplied by zero and the limit would be zero. Why is that not possible? How do we know that we have to continue on here?(2 votes)
- "zero * ∞" is an indeterminate form – because ∞ is not a
*real*number you can't use the rule that anything times zero is zero.

For more information see:

https://en.wikipedia.org/wiki/Indeterminate_form(5 votes)

- there must be another proof for the definition of e independent of derivatives so this proof wont be circular(3 votes)
- At4:57, "n" is substituted for "delta x" in the limit statement. Nowhere have we let n=delta x. We let n= (delta x)/x. I'm sure there's some magic going on, but why don't we have to state the limit as "nx->0"?(1 vote)
- It was stated that as ∆x approaches 0, so does n.

It's not so much a substitution as it is choosing to write n instead of ∆x, because they both approach the same thing.

The limit should then affect the function in an equivalent way but the change allows you to manipulate the expression differently.(3 votes)

- At7:20, the limit as n approaches zero and the natural logoritm change place, so that we get ln(lim (1+n)^1/2 instead of the other way around. Why?(2 votes)
- What is the derivative of lnx^lnx

Plz! With solution(1 vote)- It is not clear what you mean by lnx^lnx, so I will assume that you mean (lnx)^(lnx).

Using laws of exponents, we rewrite (lnx)^(lnx) as e to some function:

(lnx)^(lnx) = e^{ln[(lnx)^(lnx)]} = e^[lnx * ln(lnx)].

Using the chain and product rules, we find that the derivative of (lnx)^(lnx) is

d/dx of [(lnx)^(lnx)] = e^[lnx * ln(lnx)] * d/dx of [lnx * ln(lnx)]

= (lnx)^(lnx) * [(1/x)*ln(lnx) + (lnx)*(1/x)/(lnx)]

= (1/x) * (lnx)^(lnx) * [ln(lnx) + 1].

Have a blessed, wonderful day!(2 votes)

- Your 'definition' of e is not correct, I think. The limit of (1+n)^(1/n) as n approaches 0 does not exist. The limit equals e only from the right hand side. This proof seems a bit hand-wavy to me. Is there something I'm missing?(1 vote)
- I just graphed the function from -0.9 to -0.1 ad 0.1 onward. Clearly the right hand limit equals the left hand limit. My mistake. Sorry about that!(1 vote)

- Find dy/dx given y ln x=xe^y-1(1 vote)
- At7:44, isn't this a 'circular logical proof'? But how do I prove that the limit is equal to e without using L'Hôpital's Rule?(1 vote)

## Video transcript

- [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 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.