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

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# Proofs of derivatives of ln(x) and eˣ

## 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.