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I'm now going to show you what I think are probably the two

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coolest derivatives in all of calculus.

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And I'll reserve that.

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None of the other ones have occurred to me right now.

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But these are definitely to me some of the neatest.

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So let's figure out what the derivative of

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the natural log is.

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And just as a review, what is the natural log?

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Well the natural log of something is the exact same

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thing as saying logarithm base e of that something.

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That's just a review.

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So let's take the derivative of this.

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I think I'm going to need a lot of space for this.

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I'm going to try to do it as neatly as possible.

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So the derivative of the natural log of x equals-- well

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let's just take the definition of a derivative, right?

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We just take the slope at some point and find the limit as we

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take the difference between the two points to 0.

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So let's take the limit as delta x approaches 0 of

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f of x plus delta x.

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So I'm going to take the limit of this whole thing.

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The natural log ln of x plus delta x-- right, that's like

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one point that I'm going to take evaluate the function--

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minus the ln of x.

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All of that over delta x.

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And if you remember from the derivative videos, this is just

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the slope, and I'm just taking the limit as I find the slope

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between a smaller and a smaller distance.

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Hopefully you remember that.

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So let's see if we can do some logarithm properties to

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simplify this a little bit.

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Hopefully you remember-- and if you don't, review the logarithm

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properties-- but remember that log of a minus log of b is

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equal to log of a over b, and that comes out of the fact that

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logarithm expressions are essentially exponents, so they

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follow the exponent rules.

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And if that doesn't make sense to you, you should

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review those as well.

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But let's apply this logarithm property to this equation.

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So let me rewrite the whole thing, and I'm going to keep

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switching colors to keep it from getting monotonous.

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So we have the limit as delta x approaches 0 of this big thing.

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Let's see.

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So log of a minus b equals log a over b, so this top, the

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numerator, will equal the natural log of x plus

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delta x over x.

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Right? a b a/b, all of that over delta x.

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And so that equals the limit as delta x approaches 0-- I think

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it's time to switch colors again-- delta x approaches 0

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of-- well let me just write this 1 over delta

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x out in front.

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So this is 1 over delta x, and we're going to take

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the limit of everything.

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ln x divided by x is 1 plus delta x over x.

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Fair enough.

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Now I'm going to throw out another logarithm property, and

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hopefully you remember that-- and let me put the properties

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separate so you know it's not part of the proof-- that a log

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b is equal to log of b to the a.

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And that comes from when you take something to an exponent,

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and then to another exponent you just have to multiply

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those two exponents.

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I don't want to confuse you, but hopefully you

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should remember this.

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So how does apply here?

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Well this would be a log b.

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So this expression is the same thing as the limit.

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The limit as delta x approaches 0 of the natural log of 1 plus

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delta x over x to the 1 over delta x power.

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And remember all this is the natural log of

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this entire thing.

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And then we're going to take the limit as

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delta x approaches 0.

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If you've watched the compound interest problems and you know

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the definition of e, I think this will start to

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look familiar.

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But let me make a substitution that might clean things

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up a little bit.

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Let me make the substitution, let me call it n-- no, no, no,

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let me call u-- is equal to delta x over x.

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And then if that's true then we can multiply both

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sides by x and we get xu is equal to delta x.

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Or we would also know that 1 over delta x is

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equal to 1 over xu.

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These are all equivalent.

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So let's make the substitution.

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So if we're taking the limit is delta x approaches 0, in this

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expression if delta x approaches 0, what

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does u approach?

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u approaches 0.

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So delta x approaching 0 is the same exact thing as taking

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the limit as u approaches 0.

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So we can write this as the limit as u approaches 0 of the

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natural log of 1 plus-- well we did the substitution, delta x

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over x is now u-- to the 1 over delta x, and that same

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substitution told us that's the same thing as one over xu.

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Remember we're taking the natural log of everything.

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And we know this is an exponent property, which I'll now

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do in a different color.

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We know that a to the bc is equal to a to the

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b to the c power.

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So that tells us that this me is equal to the limit as u

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approaches 0 of the natural log of 1 plus u to the 1/u, because

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this is one over xu, right?

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1/u, and then all of that to the 1/x.

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And how did I do that?

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Just from this exponent property, right?

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If I were to simplify this, I would have 1/x times 1/u, and

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that's where I get this 1 over xu.

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Well then we can just do this logarithm property in reverse.

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If I have b to the a I can put that a out front.

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So I could take this 1/x and put it in front

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of the natural log.

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So now what do I have?

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We're almost there.

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We have the limit as u approaches 0.

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Take that 1/x, put it in front of the natural log sign.

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1/x times the natural log of 1 plus u to the 1/u.

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Fair enough.

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When we're taking the limit as u approaches 0, x, this term

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doesn't involve it at all.

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So we could take this out in front, because the limit

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doesn't affect this term.

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And then we're essentially saying what happens to this

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expression as the limit as u approaches 0.

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So this thing is equivalent to 1/x times the natural log of

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the limit as u approaches 0 of 1 plus u to the 1/u.

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And by now hopefully you would recognize that

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this is the definition.

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This limit comes to e, if you remember anything

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from compound interest.

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You might remember it as the limit-- as n approaches

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infinity of 1 plus 1 over n to the n.

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But these things are equivalent.

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If you just took the substitution u is equal to

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1/n, you would get this.

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You would just get this.

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So this expression right here is e That expression is e.

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So we're getting close.

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So this whole thing is equivalent to 1/x times

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the natural log, and this we know, this is one of

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the ways to get to e.

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So the limit as u approaches 0 of 1 plus u to the 1/u.

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That is e.

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And what is the natural log?

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Well it's the log base e.

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So you know this is equal to 1/x times the log base e of e.

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So that's saying e to what power is e.

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Well e to the first power is e, right?

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This is equal to 1.

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So 1 times 1/x is equal to 1/x.

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There we have it.

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The derivative of the natural log of x is equal to 1/x, which

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I find kind of neat, because all of the other exponents

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lead to another exponent.

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But all of a sudden in the mix here you have the natural log

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and the derivative of that is equal to x to the

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negative 1 or 1/x.

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Fascinating.

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