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Video transcript

in the first version of the videos of the proof of the derivatives of the natural law natural log of X natural log of X where the first time I prove this is a couple years ago and I'd the very next video I prove that the derivative of e to the X is equal to e DX 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 and 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 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 the log of a over B so let me rewrite it that way so this is going to be equal to the limit 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 and then I can rewrite 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 when we do this in a new color let me do 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 and then plus Delta X over X and then I had this 1 over Delta X sitting out here and I could 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 you know just from my 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 and I've become very defensive about these claims of circularity and you know 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 the e and maybe terms that we recognize let's make the substitution Delta x over X is equal to 1 over n if we multiple this is the same thing this is the equivalent 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 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 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 is n approaches infinity of the natural log I'll go back to that move color 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 just rewrite this expression right here let me rewrite it again this is equal to the limit as n approaches infinity of the natural log of 1 plus 1 over N and what I can do is I can separate out this end from the 1 over X so I could say this is to the N and then all of this to the 1 over X and once again this is just an exponent property right if I raise something to the N and then to the 1 over X I could just multiply the exponents and get to the N over X so these two statements are equivalent but now we can use logarithm properties to say hey if I this is the exponent I can just stick it out in front of the coefficient right here I could put it out right there so we get and just remember we this was all the derivative with respect to X of the natural log of X so what is that equal to we could put this one out of X in front here in fact that one 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 could 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 and move 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 write rewrite this part let me make that salmon color n 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 end well what is this should look for 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 and I'm just being clear here I'm not I'm still not using this at all I'm just stating that e is the definition of e e is equal to the limit as n approaches infinity of 1 plus 1 over N to the end this is just the definition of E and natural log is defined to be the logarithm of base this thing so if I'm taking the net 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 e the definition of e is I'm not using the definition of the derivative of 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 so obviously when this is asking this says the power that you have to raise e to to get to e well this is just equal to 1 and 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 use the definition of e but that's fine and I mean we assumed we know the definition of you 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 a little bit of an exercise here and let me get make some 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 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 it's equal to one so this is just the derivative of X and the derivative of X is equal to one so that's pretty straightforward the derivative we just this is just you know we no way did we assume this to begin with we just simplify this expression to just this is the same thing as the derivative of X because this term cancels out and the derivative X is just 1 or we could view 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 the 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 it's 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 1 over e to the X once again i knowwe assumed this right here in no way have so fired anything we've done we haven't assumed that but clearly my derivatives either way I solve it and one way I solved it I got one 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 look 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-ended 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 of 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 get we should get the same result as when we approach the problem the other way that should be equal to this approach because they're 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 you know and at no point in this entire proof at no point did I assume this did I assume this in fact this is the first time that I'm even making that I'm making the statement I didn't have to assume this when I showed you that the GNAT derivative of the natural log of X is 1 over X and I didn't have to show it 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 you know 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