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### Course: Class 12 math (India) > Unit 6

Lesson 6: Proofs for the derivatives of eˣ and ln(x)# Proofs of derivatives of ln(x) and eˣ

Doing both proofs in the same video to clarify any misconceptions that the original proof was "circular". Created by Sal Khan.

## Want to join the conversation?

- When using the chain rule in the proof that derivative os e^x=e^x, in9:29, before proving that the statement is correct, I can't say that the derivativo od ln(e^x) = (e^x)(1/e^x). I'm assuming that de derivative o g(x) in the chain rule, in this case, e^x, is equal to e^x, that is just what I'm still trying to prove... I'm not 100% convinced. Is there any other way to prove that?(1 vote)
- He didn't assert that

d/dx [ln(e^x)] = (e^x)(1/e^x). He asserted that

d/dx [ln(e^x)] = (d/dx [e^x]) (1/e^x). Do you see the difference?(29 votes)

- At5:40: "the 1/x term has nothing to do with n", but x is part of the definition of n! I would say there is more needed here to satisfy us that the x term can be safely moved outside the lim(n->∞) term.

We define 1/n = Δx/x, and then claim n->∞ <=> Δx->0. Why are we free to ignore the other possibility: n->∞ <=> x->∞?(11 votes)- As for the relation between x and n, x is just a constant. (It won’t change its value.)

If we rewrite 1/n=∆x/x as n=x/∆x, it is saying “x can be divided by ∆x n times” (n = changing rate between x and ∆x), or, you can see n as a function of ∆x, and x as a coefficient. Since x is a constant, you’ll find that as ∆x gets closer to 0, n would get closer to infinity. That’s how n is defined here.

Remember, I’m just talking about the relation between x and n. x originally, of course, is a variable.

Also, x originally came from ln(x), so it can’t be a non-positive number like Kaio pointed out.(7 votes)

- He mentions that in another video, he says that a definition of e is the lim as n->infinity of (1+1/n)^n.

Which video is he referencing here?(9 votes)- "Introduction to compound interest and e" part 1, 2, ...(7 votes)

- could be explained in another video but: why is ln(ax), where a is any constant, the derivative still 1/x?(3 votes)
- ln(ax) = ln(a) + ln(x) (<-- Basic log rule!), where, as a is a constant, ln(a) is a constant. d(ln(a) + ln(x)) = d(ln(a)) + d(ln(x)), as ln(a) is constant, d(ln(a)) = 0.

Therefore, d(ln(ax)) = d(ln(x))(11 votes)

- @2:25

I don't understand how he can just move the 1/(delta x) to the exponent.

The limit of Ln( 1 + (delta x)/x ) ^ ( 1 / (delta x) ) as delta x approaches 0 is 0 right?(2 votes)- He is using rules of logarithms: nlog(a) = log(a)^n. The same applies to any logarithm, including natural log (ln).

See: http://www.khanacademy.org/video/proof--a-log-b----log--b-a---log-a---log-b---log--a-b?playlist=Algebra

He uses the rule again at5:20, but this time in reverse.(5 votes)

- Can he switch from delta x aproaches 0 to n aprroaches infinity? Because,myes, if n approaches infinity, then delta x will approach zero. But also, if n approaches minus infinity, delta x will also approach zero... I haven't done the calculation, but i am not sure that n approaching minus infinity will give us e....(2 votes)
- n is defined as a non negative integer, so n cannot approach minus infinity. This is a flaw you must understand as it is fundamental to your future understanding of series and the limiting process.

Keep asking questions!(4 votes)

- Is there a video that graphically displays these functions with regard to their limits? I'm having trouble visualizing the limits of n and delta-x with respect to their role in taking the derivative. Thanks(3 votes)
- what is the circularity Sal was talking about?(2 votes)
- Circularity is when you prove something using a definition that you haven't been proved. For example, if you are trying to prove ln x and e^x and you make the assumption that those derivatives are true in the proof, then you really haven't proven anything.(2 votes)

- I'm confused as to how e=limn-->infinity (1+1/n)^n it feels like it should be 1 as 1/n where n was infinity would be zero, i think?, and then 1^infinity would be one.(1 vote)
- The 1/n term itself does tend towards 0, but it never equals 0. The base of your exponent is always larger than one. Also, since the exponent is getting much larger, the small part being added to the one causes it to become quite a bit larger than one. If you want to, try plugging some very large values of n into that formula on your calculator, and you can see it approaching 2.71828....(4 votes)

- Can we define e as a number so that the derivative of e^x is equal to the derivative of e^x? Then can we use that to find the derivative of ln x? Also, can we find the value of e using this method by assuming e^x to be a polynomial in the form a0+a1x+a2x^2... . Then by taking the derivative of that polynomial, we would get-

a0+a1x+a2x^2...=a1+2a2x+3a3x^2...

We would get-

a1=a0

a2=1/2a1

a3=1/6a1

And so on... .

Substituting x and a0 for 1, we get-

e=1+1+1/2+1/6...

This is approximately equal to 2.67, which is very close to the actual value of e.

Also by looking at the pattern we find that 1/ai=i!

Therefore we can express e as-

Autosum i=0 to infinite of 1/i!

Can this formula be used to find the value of e? Is this correct?(2 votes)- Yes, those are valid definitions of e, and your reasoning is correct. Good job.

This page shows more definitions of e and how they're related: http://www.proofwiki.org/wiki/Equivalence_of_Definitions_of_Exponential(2 votes)

## Video transcript

In the first version of the
video of the proof of the derivative of the natural
log of x, where the first time I proved this is
a couple of years ago. And the very next video I
proved that the derivative of e to the x is
equal to e to the x. I've been charged with some of
making a circular proof, and I'm pretty convinced that
my proof wasn't circular. So what I want to do in this
video, now that I have a little bit more space to work with, a
little bit more sophisticated tools, I'm going to redo the
proof and I'm going to do these in the same video to show you
at no point do I assume this before I actually show it. So let's start with the proof. So the first thing I need to do
is prove this thing up here. I want to keep track of this. I don't assume this until
I actually show it. So let's start with the
proof, the derivative of the natural log of x. So the derivative of the
natural log of x, we can just to go to the basic
definition of a derivative. It's equal to the limit as
delta x approaches 0 of the natural log of x plus delta x
minus the natural log of x. All of that over delta x. Now we can just use the
property of logarithms. If I have the log of a minus
the log of b, that's the same thing as a log of a over b. So let me re-write it that way. So this is going to be equal
to the limit as delta x approaches 0. I could take this 1 over
delta x right here. 1 over delta x times the
natural log of x plus delta x divided by this x. Just doing the logarithm
properties right there. Then I can re-write this --
first of all, when I have this coefficient in front of a
logarithm I can make this the exponent. And then I can simplify
this in here. So this is going to be equal to
the limit as delta x approaches 0 of the natural log -- let
me do this in a new color. Let me do it in a
completely new color. The natural log of -- the
inside here I'll just divide everything by x. So x divided by x is 1. Then plus delta x over x. Then I had this 1 over delta x
sitting out here, and I can make that the exponent. That's just an exponent
rule right there, or a logarithm property. 1 over delta x. Now I'm going to make
a substitution. Remember, all of this, this was
all just from my definition of a derivative. This was all equal to
the derivative of the natural log of x. I have still yet to
in any way use this. And I won't use that until
I actually show it to you. I've become very defensive
about these claims of circularity. They're my fault because that
shows that I wasn't clear enough in my earlier versions
of these proofs, so I'll try to be more clear this time. So let's see if we can
simplify this into terms that we recognize. Let's make the substitution so
that we can get e in maybe terms that we recognize. Let's make the substitute delta
x over x is equal to 1 over n. If we multiply -- this
is the same thing. This is the equivalent
to substitution. If we multiply both sides of
this by x, as saying that delta x is equal to x over n. These are equivalent
statements. I just multiplied both
sides by x here. Now if we take the limit as n
approaches infinity of this term right here, that's
equivalent -- that's completely equivalent to taking the limit
as delta x approaches 0. If we're defining delta x to be
this thing, and we take the limit as its denominator
approaches 0, we're going to make delta x go to 0. So let's make that
substitution. So all of this is going to be
equal to the limit as -- now we've gotten rid
of our delta x. We're going to say the limit as
an approaches infinity of the natural log -- I'll go back to
that mauve color -- the natural log of 1 plus -- now, I said
that instead of delta x over x, I made the substitution that
that is equal to 1 over n. So that's 1 plus 1 over n. And then what's 1 over delta x? Well delta x is equal to x over
n, so 1 over delta x is going to be the inverse of this. It's going to be n over x. And then we can re-write this
expression right here -- let me re-write it again. This is equal to the limit as
n approaches infinity of the natural log of 1 plus 1 over n. What I can do is I can separate
out this n from the 1 over x. I could say this is to
the n, and then all of this to the 1 over x. Once again, this is just
an exponent property. If I raise something to the n
and then to the 1 over x, I could just multiply
the exponent to get to the n over x. So these two statements
are equivalent. But now we can use logarithm
properties to say hey, if this is the exponent, I can just
stick it out in front of the coefficient right here. I could put it out right there. And just remember, this was all
of the derivative with respect to x of the natural log of x. So what is that equal to? We could put this 1 out
of x in the front here. In fact, that 1 out of x term,
it has nothing to do with n. It's kind of a constant
term when you think of it in terms of n. So we can actually put it
all the way out here. We could put it either place. So we could say 1 over x times
all that stuff in mauve. The limit as n approaches
infinity of the natural log of 1 plus 1 over n to the n. The natural log of
all of that stuff. Or, just to make the point
clear, we can re-write this part -- let me make a salmon
color -- equal to 1 over x times the natural log of the
limit as n approaches infinity. I'm just switching places here,
because obviously what we care is what happens to this term as
it approaches infinity, of 1 plus 1 over n to the n. Well what is -- this should
look a little familiar to you on some of the first videos
where we talked about e -- this is one of the definitions of e. e is defined. I'm just being clear here. I'm still not using
this at all. I'm just stating that the
definition of e, e is equal to the limit as n approaches
infinity of 1 plus 1 over n to the n. This is just the
definition of e. And natural log is defined
to be the logarithm of base, this thing. So this thing is e. So I'm saying that the
derivative of the natural log of x is equal to 1 over x
times the natural log. This thing right here is e. That's what the
definition of e is. I'm not using the definition
of the derivative e, or the definition of the
derivative of e to the x. I'm just using the
definition of e. And the definition of
natural log is log base e. This says the power that you
have to raise e to to get to e, well this is just equal to 1. There we get that the
derivative of the natural log of x is equal to 1 over x. So, so far I think you'll be
satisfied that we've proven this first statement up here,
and in no way did we use this statement right here. I just used the definition
of e, but that's fine. I mean we assumed we know the
definition of e, even when we just talk about natural log,
we assume that it's base e. In no way did I assume
this to begin with. Now, given that we've shown
this and we didn't assume this at all, let's see
if we can show this. So the derivative -- let's do a
little bit of an exercise here. Actually, I could probably
do it in the margins. Let's take the derivative
of this function. The natural log of e to the x. So there's two ways we
can approach this. The first way we could simplify
this and we could say this is the exact same thing
as the derivative. We could put this x out
front of x times the natural log of e. And what's the
natural log of e? The natural log of e we
already know is equal to 1. So this is just the
derivative of x. And the derivative
of x is equal to 1. So that's pretty
straightforward. The derivative, in no way did
we assume this to begin with. We just simplified this
expression to just this is the same thing as the derivative
of x, because this term cancels out. And the derivative
of x is just 1. Or we could do it
the other way. We could do the chain rule. We could say that this could be
viewed as the derivative of this inner function, of this
inner expression, so the derivative of the inner
expression, I don't know what that is. I'm not assuming
anything about it. I just don't know what it is. So I'll write it in
yellow right here. So it's equal to the
derivative with respect to x of e to the x. I don't know what this is. I have no clue what this
is, and I haven't assumed anything about what it is. I'm just using the chain rule. If the derivative of this
inside function with respect to x, which is this right here,
times the derivative of this outside function with respect
to the inside function. So the derivative of natural
log of x with respect to x is 1 over x. So the derivative of natural
log of anything with respect to anything is
1 over that anything. So it's going to be equal to --
so the derivative of natural log of x with respect to e to
the x is equal to 1 over e to the x. Once again, I in no way
assumed this right here. So far in anything we've done,
we haven't assumed that. But clearly, my derivatives,
either way I solve it -- one way I solve it I got 1. The other way, I kind
of didn't solve it. I got this expression
right here. They must be equal
to each other. So let me write that down. This must be equal to that. It's just we just looked at it
two different ways and got two different results. But I still don't know
what this thing is. I just left it kind of open. I just said whatever
the derivative of e to the x happens to be. But we know, since these two
expressions are equal, we know that the derivative with
respect to x of whatever e to the x -- so whatever the
derivative with respect to x of e to the x happens to be, we
know that when we multiply that times 1 over e to the x --
that's when we just did the chain rule -- that we should
get the same result as when we approached the problem
the other way. That should be equal to this
approach because they're both different ways of looking
at the derivative of the natural log of e to the x. So that should be equal to 1. Well, we're almost there. We could just simplify this
and solve for our mystery derivative of e to the x. Multiply both sides of this
equation by e to the x, and you get the derivative with respect
to x of e to the x is equal to e to the x. And I want to clarify this. At no point in this entire
proof, at no point did I assume this. In fact, this is the
first time that I'm even making the statement. I didn't have to assume this
when I showed you that the derivative of the natural
log of x is 1 over x. And I didn't have to assume
this to kind of get to it. So in no way is this
proof circular. So anyway, I didn't want to
appear defensive, but I wanted to clarify this up. Because I don't want to in any
way blame those who think that my original proof was circular. It's my fault because I
didn't explain it properly. So hopefully this should
provide a little bit of clarity on the issue.