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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.