Proofs of derivatives of common functions
Proof: d/dx(ln x) = 1/x Taking the derivative of ln x
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- 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.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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