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# Proof: The derivative of πΛ£ isΒ πΛ£

## Video transcript

- [Lecturer] The number e has all sorts of amazing properties. Just as a review you can
define it in terms of a limit. The limit as n approaches infinity of 1 + 1/n to the nth power. You can also define it as
the limit as n approaches 0 of 1 + n to the 1/nth power. What we're going to focus on this video is an amazing property of e, and e has many many amazing properties, but this is the one that's maybe the most relevant to calculus, and that's the notion that if I take the derivative with
respect to x of e to the x, then it is equal to, drumroll, equal to e to the x. That to me is amazing. Let's just appreciate it for a second before we actually prove it. This is part of the
graph of y = e to the x. And so what this says is,
the derivative of e to the x for any x is equal to e to the x! The slope of the tangent line at any point here is equal
to the value of the function. Let's just appreciate that. So right over here the
value of the function is 1, and the slope of the tangent line is 1. Here the value of the function is 2, and the slope of the tangent line is 2. Here the value of the function is 4, and the slope of the
tangent line is equal to 4. So I could just go on, this is just another
amazing thing about e, and we'll see many more in calculus, but let's now prove that
this is actually true. So let's just use our
definition of a derivative. So the derivative with
respect to x, of e to the x, would be the limit of delta
x, or as delta approaches 0, of e to the x + delta x, - e to the x, all of that over, all
of that over delta x. Now let's do some
algebraic manipulation here to see if we can make some sense of it. So this is going to be equal
to the limit as delta x approaches 0 of, let's see, what happens, well I
won't skip any steps here. This is the same thing as e to
the x times e to the delta x, this is just using our
exponent properties here minus e to the x over delta x. Just to be clear, I rewrote
this right over here as this right over here. Now I can factor out an e to the x, and in fact, because e
to the x is not affected as delta x approaches 0, I can factor the e to the
x out of the entire limit. So let's do that, let's take e to the x out of the entire limit, let's
factor it completely out. It does not get affected by the delta x. So this is going to be equal to, factor that e to the x out, e to the x times the limit
as delta x approaches 0, of e to the delta x - 1,
all of that over delta x. So now we're gonna get a little
bit fancy with our limits. We're gonna do what's known
as a change of variable. So I'm gonna say, well
let's see I don't know how to directly find this
limit right over here, but maybe I can simplify it, and who knows, maybe I can get it into one of these forms up here. So what if I were to
make the substitution, and let me do it over here. Let's say I would make
the substitution that n = e to the delta x - 1. So what would this be if we
were to solve for delta x? Well let's see, we could
add 1 to both sides, n + 1 = e to the delta x. To solve for delta x we could just take the natural logarithm,
log base e of both sides. And we would get the natural
log of n + 1 = delta x. And so, we can make that substitution, this could be replaced with this, and what we have in
the numerator over here can be replaced with n, right over there, and what would happen to the limit? Well, as delta x approaches 0, what does n approach? So delta x approaches 0
implies that n approaches what? Let's see, as delta x approaches 0, this would be e to the 0 which is 1 - 1, so looks like n approaches 0. And you can look at it over here, as n approaches 0, right over here, natural log of 0 + 1, natural log of 1, that is 0. So as each of them, as
delta x approaches 0, n approaches 0, as n approaches 0, delta x approaches 0. So then you can replace, if
you make the change of variable from delta x to n, you could still say as n approaches 0. So let me rewrite all of this. That was the fanciest
step in this entire thing that we're about to do. So this is gonna be e to the x times the limit, since we changed our variables now gonna be as n approaches 0, because as delta x
approaches 0, n approaches 0, and vice versa. And this numerator here, we said hey, that's gonna be equal to n over, delta x is now the natural
log of n + 1, ln(n+1). Now what does that do for us? Well what if we were
to divide the numerator and the denominator by n? So let's multiply down here by 1/n, and let's multiply up here by 1/n. Well our numerator's
just gonna be equal to 1. What does our denominator equal? Well here we can just use
our exponent properties. All of this is gonna be equal to, this is gonna be equal to, we
have our e to the x out front, e to the x, and then we have the limit, as n approaches 0, our numerator is now 1, over, and now I'm gonna rewrite this, just using our logarithm properties. If I have, I'll do it right over here. If I have a times the natural log of b, this is the same thing as the natural log of b to
the a power, the ath power, this is just natural log properties. So we have down here, this
would be the same thing as the natural log of n+1,
actually let me write it the other way around, 1+n,
1+n, I just swapped these two, to the 1/n power, so the
natural log of that whole thing. Now, you might be getting
that tingly feeling, because something is
starting to look familiar. What I just constructed
inside the logarithm here, looks an awful lot like what
we have right over here. And the limit as n approaches 0, limit as n approaches 0. And in fact we can use
our limit properties, this one isn't affected,
what's really affected is what's inside the logarithm. So we could say that this
is going to be equal to, and we're approaching our drumroll, e to the x times 1 over,
1 over the natural log, I'll do that blue color, natural log of, I'll give myself some space, the limit, the limit as n approaches 0 of this business right over here, which I could just write
as, 1+n to the 1/nth power. So this is really interesting. What is this? What did I just, what do I
have there in the denominator with this limit? What is this thing equal to? Well we already said,
that is a definition of e. That is this right over here, which is equal to this over here. So this is equal to e. So what does this all boil down to? I think you see where this is going, but this is fun, we're downhill from here. This is e to the x times 1/ln(e). Well, ln(e), what power
do I have to raise e to get to e? Well I just have to raise it to 1. So this gets us e to
the x, and we're done. We've just proven that the
derivative with respect to x of e to the x is indeed
equal to e to the x. That's an amazing finding, that shows us one more dimension of the
beauty of the number e.