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

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# Proof: d/dx(ln x) = 1/x

## Video transcript

I'm now going to show you what
I think are probably the two coolest derivatives
in all of calculus. And I'll reserve that. None of the other ones have
occurred to me right now. But these are definitely to
me some of the neatest. So let's figure out
what the derivative of the natural log is. And just as a review,
what is the natural log? Well the natural log of
something is the exact same thing as saying logarithm
base e of that something. That's just a review. So let's take the
derivative of this. I think I'm going to need
a lot of space for this. I'm going to try to do it
as neatly as possible. So the derivative of the
natural log of x equals-- well let's just take the definition
of a derivative, right? We just take the slope at some
point and find the limit as we take the difference between
the two points to 0. So let's take the limit as
delta x approaches 0 of f of x plus delta x. So I'm going to take the
limit of this whole thing. The natural log ln of x plus
delta x-- right, that's like one point that I'm going to
take evaluate the function-- minus the ln of x. All of that over delta x. And if you remember from the
derivative videos, this is just the slope, and I'm just taking
the limit as I find the slope between a smaller and
a smaller distance. Hopefully you remember that. So let's see if we can do some
logarithm properties to simplify this a little bit. Hopefully you remember-- and if
you don't, review the logarithm properties-- but remember that
log of a minus log of b is equal to log of a over b, and
that comes out of the fact that logarithm expressions are
essentially exponents, so they follow the exponent rules. And if that doesn't make
sense to you, you should review those as well. But let's apply this logarithm
property to this equation. So let me rewrite the whole
thing, and I'm going to keep switching colors to keep it
from getting monotonous. So we have the limit as delta x
approaches 0 of this big thing. Let's see. So log of a minus b equals log
a over b, so this top, the numerator, will equal the
natural log of x plus delta x over x. Right? a b a/b, all of
that over delta x. And so that equals the limit as
delta x approaches 0-- I think it's time to switch colors
again-- delta x approaches 0 of-- well let me just write
this 1 over delta x out in front. So this is 1 over delta x,
and we're going to take the limit of everything. ln x divided by x is 1
plus delta x over x. Fair enough. Now I'm going to throw out
another logarithm property, and hopefully you remember that--
and let me put the properties separate so you know it's not
part of the proof-- that a log b is equal to log
of b to the a. And that comes from when you
take something to an exponent, and then to another exponent
you just have to multiply those two exponents. I don't want to confuse
you, but hopefully you should remember this. So how does apply here? Well this would be a log b. So this expression is the
same thing as the limit. The limit as delta x approaches
0 of the natural log of 1 plus delta x over x to the
1 over delta x power. And remember all this
is the natural log of this entire thing. And then we're going
to take the limit as delta x approaches 0. If you've watched the compound
interest problems and you know the definition of e, I think
this will start to look familiar. But let me make a substitution
that might clean things up a little bit. Let me make the substitution,
let me call it n-- no, no, no, let me call u-- is equal
to delta x over x. And then if that's true
then we can multiply both sides by x and we get
xu is equal to delta x. Or we would also know
that 1 over delta x is equal to 1 over xu. These are all equivalent. So let's make the substitution. So if we're taking the limit is
delta x approaches 0, in this expression if delta x
approaches 0, what does u approach? u approaches 0. So delta x approaching 0 is the
same exact thing as taking the limit as u approaches 0. So we can write this as the
limit as u approaches 0 of the natural log of 1 plus-- well we
did the substitution, delta x over x is now u-- to the 1
over delta x, and that same substitution told us that's the
same thing as one over xu. Remember we're taking the
natural log of everything. And we know this is an exponent
property, which I'll now do in a different color. We know that a to the bc
is equal to a to the b to the c power. So that tells us that this me
is equal to the limit as u approaches 0 of the natural log
of 1 plus u to the 1/u, because this is one over xu, right? 1/u, and then all of
that to the 1/x. And how did I do that? Just from this exponent
property, right? If I were to simplify this, I
would have 1/x times 1/u, and that's where I get
this 1 over xu. Well then we can just do this
logarithm property in reverse. If I have b to the a I can
put that a out front. So I could take this 1/x
and put it in front of the natural log. So now what do I have? We're almost there. We have the limit
as u approaches 0. Take that 1/x, put it in front
of the natural log sign. 1/x times the natural log
of 1 plus u to the 1/u. Fair enough. When we're taking the limit as
u approaches 0, x, this term doesn't involve it at all. So we could take this out in
front, because the limit doesn't affect this term. And then we're essentially
saying what happens to this expression as the limit
as u approaches 0. So this thing is equivalent to
1/x times the natural log of the limit as u approaches
0 of 1 plus u to the 1/u. And by now hopefully you
would recognize that this is the definition. This limit comes to e, if
you remember anything from compound interest. You might remember it as the
limit-- as n approaches infinity of 1 plus
1 over n to the n. But these things
are equivalent. If you just took the
substitution u is equal to 1/n, you would get this. You would just get this. So this expression right here
is e That expression is e. So we're getting close. So this whole thing is
equivalent to 1/x times the natural log, and this
we know, this is one of the ways to get to e. So the limit as u approaches
0 of 1 plus u to the 1/u. That is e. And what is the natural log? Well it's the log base e. So you know this is equal to
1/x times the log base e of e. So that's saying e
to what power is e. Well e to the first
power is e, right? This is equal to 1. So 1 times 1/x is equal to 1/x. There we have it. The derivative of the natural
log of x is equal to 1/x, which I find kind of neat, because
all of the other exponents lead to another exponent. But all of a sudden in the mix
here you have the natural log and the derivative of that
is equal to x to the negative 1 or 1/x. Fascinating.