the following content is provided under a Creative Commons license your support will help MIT OpenCourseWare continue to offer high quality educational resources for free to make a donation or to view additional materials from hundreds of MIT courses visit MIT opencourseware at ocw.mit.edu okay so I'd like to begin the second lecture by reminding you what you we did last time so last time last time we define the derivative as a slope of a tangent line so that was our geometric point of view and we also did a couple of computations we worked out that the derivative of 1 over X was minus 1 over x squared and we also computed the derivative of X to the nth power for N equals 1 2 etc and that turned out to be X sorry NX to the N minus 1 so that's what we did last time and today I want to finish up with other points of view on what a derivative is so this is extremely important it's almost the most important thing I'll be saying in the class but you'll have to think about it again when you start over and start using calculus in the real world so again we're talking about what is a derivative and this is just a continuation of last time so as I said last time we talked about geometric interpretations and today what we're going to talk about is rate of change as an interpretation of the derivative so remember we drew graphs of functions y equals f of X and we kept track of the change in X and here the change in Y let's say and then from this new point of view a rate of change keeping track of the rate of change of X and the rate of change of Y it's the relative rate of change we're interested in and that's Delta Y over Delta X and that has another interpretation this is the average change usually we would think of that if X were measuring time and so the average and that's when this becomes a rate and the average is over the time interval Delta X and then the limiting value is denoted dy/dx and that is the so this one is the average rate of change and this one is the instantaneous rate okay so that's the point of view that I'd like to discuss now give you just a couple of examples so let's see well first of all so maybe some examples from physics here so Q is usually the name for for a charge and then DQ DT is what's known as current so that's one physical example a second example which is probably the most tangible one is we could denote the letter s by distance and then the rate of change is what we call speed so those are the two typical examples and and I just I want to illustrate the second example in a little bit more detail because I think it's important to have some visceral sense of of this notion of instantaneous speed and I get to use the example of this very building to do that probably you know or maybe you don't that on Halloween there's an event that takes place in this building or really from the top of this building which is called the pumpkin drop so so let's illustrate this idea of rate of change with the pumpkin drop so what happens is this building well let's see here's the building and here's the here's the dot that's the beautiful grass out on this side of the building okay and then there are some people up here and very small objects well they're not that small when you're close to them they'd get dumped over the side there and they they fall down that's you know everything at MIT or a lot of things at MIT or physics experiments so that's that's the pumpkin drop so roughly speaking actually this is the buildings about 300 feet high we're we're down here on the first useable floor and so we're going to use instead of 300 feet just for convenience purposes we'll use 80 meters because that makes the numbers come out simply okay so so we have the height which is starts out at 80 meters at time 0 and then the acceleration due to gravity gives you this formula for H this is the height so at time T equals 0 we're up at the top H is 80 meters so this is the met the unitive units here are meters and at time T equals 4 you notice five times 4 squared is 80 so I picked these numbers conveniently so that we're down at the bottom okay so so this notion of average change here so the average change or the average speed here maybe we'll call it the average speed since that's over this time that it takes for the pumpkin to drop is going to be the change in H divided by the change in T which is it starts out at what does it start out as it starts out as a T right and it ends at zero so actually we have to do it backwards we have to take zero minus 80 because the first value is the final position and the second value is the initial position and that's divided by four minus zero time four seconds - times zero seconds and so that of course is the minus 20 meters per second so the average speed of this guy is 20 meters a second now so so why did I pick this example because of course the average its although interesting is not really what anybody cares about who actually goes to the event all we really care about is the instantaneous speed when it when it hits the pavement and so that's can be calculated at the bottom so what's the instantaneous me that's the derivative or maybe I should to be consistent with the notation I've been using so far that's d by DT of H all right so that's D by DT of H now remember we have formulas for these things we can differentiate this function now we did that yesterday so we're going to take the rate of change and if you take a look at it it's just the rate of change of a T is zero minus the rate of change for this minus 5t squared that's minus 10t alright so that's using the fact that D by DT of a T is equal to zero and D by DT two squared is equal to 2t the special case well I'm cheating here but there's a special case that's obvious I didn't throw it in over here the case N equals two is that second case there but the case N equals zero also works right because that's constants the derivative of a constant is zero and then the factor n there is zero and that's consistent and actually if you look at the formula above it you'll see that it's the it's the case N equals minus 1 so we'll get a larger pattern soon enough with the powers ok anyway back over here we have our rate of change and this is what it is and at the bottom at the point of impact we have T is equal to four and so H prime which is the derivative is equal to minus 40 meters per second so twice as fast as the average speed here and if you need to convert that that's about 90 miles an hour which is why the police are there at midnight on Halloween to make sure you're all safe and also why when you come you have to be prepared to clean up afterwards so anyway that's what happens it's 90 miles an hour it's actually the building is a little taller but there's air resistance and I'm sure you can do a much more thorough study of this example all right so now I want to give you a couple of more examples because time is and and and these kinds of parameters and variables are not the only ones that are important for calculus if it were only this kind of physics that was involved then this would be a much more specialized subject than it is and so I want to give you a couple of examples that don't involve time as a variable so the third example I'll give here is the letter T often denotes temperature and then D DT DX would be what is known as the temperature gradient which we really care about a lot when we're predicting the weather because it's that temperature difference that causes air flows and it causes things to change and then there's another theme which is throughout the sciences and engineering which I'm going to talk about under the heading of sensitivity of measurements so let me let me explain this I don't want to be labor it because I just doing this in order to introduce you to the ideas on your problem set which are the first case of this so on problem set one you have an example which is based on a simplified model of GPS sort of the Flat Earth model and in that situation well if the earth is flat it's just a horizontal line like this and then you have a satellite which is over here preferably above the earth and the the satellite is it knows exactly or the system knows exactly where the point directly below the satellite is so this point is treated as known and I'm sitting here ok with my little GPS device and I want to know where I am and the way I locate where I am is I communicate with this satellite by radio signals and I can measure this distance here which is called H and then the system will compute this horizontal distance which is which is L so in other words what's measured so H measured by radios radio waves and a clock or various clocks and then L is deduced from from H and what's critical in all of these systems is that you don't know H exactly there's an error in H which will denote Delta H there's some degree of uncertainty the main uncertainty in GPS is from the ionosphere but there are lots of Corrections that are made of all kinds and also if you're inside a building it's a problem to measure it but it's it's an extremely important issue as I'll explain in a second so the idea is we we then get to the get at Delta L is estimated by considering this ratio Delta L over Delta H which is going to be approximately the same as the derivative of L with respect to H so this is the thing that's easy because of course it's calculus right calculus is the easy part and that allows us to deduce something about the real world that's close by over here so the reason why you should care about this quite a bit is that it's it's used all the time to land airplanes and so you really do care that they actually know to within a few feet or even closer where your plane is and how high up it is and so forth all right so that's that's it for the general introduction to what a derivative is I'm sure you'll be getting used to this in a lot of different contexts throughout the course and now we have to get back down to to some some rigorous details ok is everybody happy with what we've got so far yeah ah good question the question was how did I get this equation for height I just made it up because it's the formula from physics that you will learn when you take 801 and it has to do with the fact that the in fact it has to do with the fact that if you this is speed if you differentiate another time you get acceleration and this is the acceleration due to gravity is 10 meters per second which happens to be the second derivative of this but anyway I just pulled it out of a hat from your physics class so you can just say see 801 all right other questions all right so let's let's go on now now I have to be a little bit more systematic about limits and so let's let's do that now so now what I'd like to talk about is limits and continuity and this is a warm-up for deriving all the rest of the formulas all the rest of the formulas that I'm going to need to differentiate every function you know remember that's our goal and we only have about a week left so we'd better get started so first of all there's what I will call easy limits so what's an easy limit an easy limit is something like the limit as X goes to 4 of X plus 3 over x squared plus 1 and with this kind of limit all I have to do to evaluate it is to plug in x equals 4 because so so what I get here is 4 plus 3 divided by 4 squared plus one and that's just 7 divided by 17 and that's the end of it so those are the easy limits the second kind of limit well so this isn't the only second kind of limit but I just want to point this out it's very important is that derivatives are always harder than this you can't get away with nothing here so why is that well when you take a derivative you're taking the limit as X goes to X 0 of f of X well f of X let's try we'll write it all out in all its glory here's the the formula for a derivative now notice that if you plug in x equals x zero always gives zero over zero so it just basically never works so we always are going to need some cancellation to make sense out of the limit now in order to in order to make things a little easier for myself to explain what's going on with limits I need to introduce just one more piece of notation what I'm going to introduce here is what's known as a left hand and a right hand limit if I take the limit as X tends to X 0 with a plus sign here of some function this is what's known as the right hand limit and I can display it visually so what does this mean it means practically the same thing as X tends to X 0 except there's one more restriction which has to do with this plus sign which is we're going from the plus side of X 0 that means X is bigger than X 0 and I say right hand so there should be a hyphen here by hand limit because on the number line if X 0 is over here the X is whoops sorry the X is to the right alright so that's a right hand limit and then this being the left side of the board I'll put on the right side of the board the left limit just to make things confusing so that one has the minus sign here I'm just a little dyslexic and I hope you're not so I may have gotten that wrong but so this is the left hand limit and I'll draw it so of course that just means X goes to X 0 but X is to the left of X 0 and again on the number line here's the X 0 and the X is on the other side of it okay so those two notations are going to help us to clarify a bunch of things it's much more convenient to have this extra bit of description of limits than to just consider limits from both sides ok so I want to give an example of this and also an example of how you're going to think about these sorts of problems so I'll take a function which has two different definitions say it's X plus 1 when X is bigger than 0 and minus X plus 2 when X is less than 0 all right so maybe put commas there those are so when X is greater than 0 it's it's X plus 1 now I can draw a picture of this it's going to be kind of a little small because I'm going to try to fit it down in here but maybe I'll put the axis down below so so at height 1 I have to the right I have something of slope 1 so it goes up like this all right and then to the left of 0 I have something which has slope negative 1 but it hits the axis at 2 so it's up here all right so I have this sort of strange antenna figure here which is my graph maybe I could should draw these in another color to depict that all right and then if I calculate these two limits here what I see is that the limit as X goes to 0 from above of f of X that's the same as the limit as X goes to 0 of the formula here X plus 1 which turns out to be 1 and if I take the limit from the so that's the left-hand limit that's this sorry that's sorry though yeah I told you I was dyslexic which right this is the right so it's that that right half here we go so now I'm going from the left and it's f of X again but now because I'm on that side the thing I need to plug in is the other formula minus X plus 2 and that's going to give us 2 now notice that the left and right limits and this is one little tiny subtlety and it's almost the only thing that I need you to really pay attention to a little bit right now is that this we did not need need x equals 0 value in fact I never even told you what f of 0 was here if we stick it in we could stick it in ok let's say we stick it in on on this side let's let's make it be that it's on this side so that means that this point is in and this point is out so that's a typical notation this little open circle and this closed dot for when you include the so so in that case the value of f of X happens to be the same as its right hand limit namely the value is 1 here and not 2 okay so that's the the the first kind of example question okay so now our next job is to introduce the definition of continuity so that was the other topic here so we're going to define so F is continuous at X 0 means that the limit of f of X as X tends to X 0 is equal to f of X 0 right so so the reason why I spent all this time paying attention to the left and the right and so on and so forth and focusing is that I I want you to pay attention for one moment to what the content of this of this definition is what it's saying is the following so a continuous at X 0 has has a various ingredients here so the first one is that this limit exists and what that means is that there's an honest limiting value both from the left and right and they also have to be the same all right so that's that's what's going on here and the second property is that f of X 0 is defined so I can't be in one of these situations where I haven't even specified what f of X 0 is and they're equal okay so that's the situation now again let me emphasize a tricky part of the definition of a limit this side the left-hand side is completely independent is evaluated by a procedure which does not involve the right-hand side these are separate things this one is to evaluate it you always avoid the limit point so that's if you like a paradox because it's exactly the question is is it true that if you plug in X 0 you get the same answer as if you move in the limit that's the issue that we're considering here we have to make that distinction in order to say that these are two the other otherwise this is just tautological it makes it's a very little it doesn't have any meaning but in fact it does have a meaning because one thing is evaluated separately with reference to all the other points and the other is evaluated right at the point in question and indeed what these things are are exactly the easy limits that's exactly what we're talking about here they're the ones that you can evaluate this way so we have to make the distinction and these other ones are going to be the ones which we can't evaluate that way so these are the nice ones and that's why we care about them why we have a whole definition associated with them all right so now so now what's next well I need to give you a little tour a very brief tour of the zoo of what are known as discontinuous functions so sort of everything else that's not continuous so the first example here let me just write it down here is jump discontinuities so what would a jump discontinuity be well we've actually already seen it the jump discontinuity is the example that we had right there this is when the limit from the left left and right exists but are not equal okay so that's that's the as in the example right in this example the two limits one of them was one and one of them was two so that's a jump discontinuity and this kind of issue of whether something is continuous or not is may seem a little bit technical but but it is true that that that people have worried about it a lot the Bob Merton who was a professor at MIT when he did his work for the Nobel Prize in Economics was interested in this very issue of whether stock prices of various kinds are continuous from the left or right in a certain model and that was a very serious issue in developing the model that that priced things that are hedge funds use all the time now so left and right really can mean something very different in this case left is the past and right is the future and it makes a big difference whether things are continuous from the left or continues from the right right is it true that the point is here here somewhere in the middle somewhere else that's a serious issue so the next example that I want to give you is a little bit more subtle it's what's known as a removable discontinuity and so what this means is that the limit from left and right are equal so a picture of that would be you have a function which is coming along like this and there's a hole maybe where who knows either the function is undefined or maybe it's defined up here and then it just continues on alright so that the two limits are the same and then of course the function is begging to be redefined so that we remove that hole and that's why it's called a removable discontinuity now let me give you an example of this are actually a couple of examples so these are quite important examples which we'll be working with in just a few in it in a few minutes so so the first is the function G of X which is sine x over X and the second will be the function H of X which is 1 minus cosine X over X so we have a problem at G of 0 G of 0 is undefined on the other hand it turns out that this function has what's called a removable singularity namely the limit as X goes to 0 of sine x over X does exist in fact it's equal to 1 so that's a very important limit that will work out either at the end of this lecture or the beginning of next lecture and similarly the limit of 1 minus cosine X divided by X X goes to zero is zero maybe I'll put that a little farther away so you can read it okay so these are these are very useful facts that we're going to need later on and what they say is that these things have removable singularities sorry removable discontinuity at x equals zero all right so as I say well we'll get to that in a in a in a few minutes okay so are there any questions before I move on yet the question is why is this true is that is that what your question is the answer is it's very very unobvious I haven't shown it to you yet and there's if you were not surprised by it then that would be very strange indeed so we haven't done it yet you have to stay tuned until we do okay we haven't shown it yet and actually even this other statement which maybe seems stranger still is also not yet explained okay so we are going to get there as I said either at the end of this lecture or at the beginning of next other questions all right so let me let me just continue my tour of the zoo of discontinuities and I guess I want to illustrate something with the convenience of right and left hand limits so I'll save this board about right and left hand limits so a third type of discontinuity is what's known as an infinite discontinuity and we've already encountered one of these I'm going to draw them over here remember the function y is 1 over X that's this function here but now I'd like to draw also the other branch of the per app of the hyperbola down here and allow myself to consider negative values of X so here's the graph of 1 over X and the convenience here of the distinguishing left on the right hand limits is very important because here I can write down that the limit as X goes to 0 plus of 1 over X well that's coming from the right and it's going up so this limit is infinity whereas the limit in the other direction from the left that one is going down and so it's quite different it's minus infinity now some people say that these limits are undefined but actually they're going into some very definite directions so you should you should whenever possible specify what these limits are on the other hand the statement that the limit as X goes to 0 of 1 over X is infinity is is simply wrong ok it's not that people don't write this it's just that it's wrong it's not that they don't write it down I mean in fact you'll probably see it and it's because people are just thinking of the of the right hand branch it's not that they're making a mistake not because necessarily but anyway it's sloppy and there's some sloppiness that will endure and others that will try to avoid so here you want to say this and it does make a difference you know plus infinity is a you know an infinite number of dollars and minus and infinity is an infinite amount of debt they're actually different they're not the same so you know this is sloppy and this is actually more correct okay so now in addition I just want to point out one more thing remember we calculated the derivative and that was minus 1 over x squared but I want to so I want to draw the graph of that and make a few comments about it so I'm going to draw the graph directly underneath the the graph of the function and notice what this graph is it goes like this it's always negative and it points down so now this may look a little strange that the derivative of this thing is this guy but that's because of something very important and you should always remember this about derivatives the derivative of function looks nothing like the function necessarily so you should just forget about that as being an idea some people feel like if one thing goes down that other thing has to go down just forget that intuition it's wrong what we're dealing with here if you remember is the slope so if you have a slope here that's corresponds to just a place over here and there's a slope gets a little bit less steep that's why we're approaching the the horizontal axis we're getting a little the numbers getting a little smaller as we close in now over here this slope is also negative it's going down and as we get down here it's getting more and more negative as we go here the slope this function is going up but its slope is going down alright so the slope is down on both sides and the notation that we use for that is well suited to this to this left and right business namely the limit as X goes to 0 of negative 1 over x squared that's going to be equal to minus infinity and that applies to X going to 0 plus and X going to 0 minus so so both add this property in it finally let me just make one last comment about these two graphs this function here is an odd function and when you take the derivative of an odd function you always get an even function and that's closely related to the fact that this 1 over X is an odd power in this to X to the first power is an odd power and x squared is an even power so all of the your intuition should be reinforcing the fact that these pictures look right ok now there's one last kind of discontinuity that I want to mention briefly which I will call other ugly discontinuities and there are lots and lots of them and so one example would be the function y is equal to sine 1 over X as X goes to 0 and that looks a little bit like this back and forth and back and forth it oscillates infinitely often as we tend to zero and there's there's no left or right limit in this case so there is a very large quantity of things like that fortunately we're not going to deal with them in this in this course a lot of times in real life there are things that oscillate as time goes to infinity but but we're not going to worry about that right now okay so that's our final mention of a discontinuity and and now I need to do just one more piece of groundwork for our formulas next time namely I want to check for you one basic fat one limiting tool so this is going to be a theorem but fortunately it's a very short theorem and has a very short proof so the theorem goes under the name differentiable implies continuous and what it says is the following it says that if F is differentiable in other words it's a derivative exists at X 0 then F is continuous at x0 so we're going to need this as a tool a key step in the product and quotient rules so I'd like to prove it right now for you so here's the proof fortunately the proof Asst is just one line so first of all I want to write in just the right way what it is that we have to check so what we have to check is that the limit as X goes to X 0 of f of X minus f of X 0 is equal to 0 so this is what we want to know we don't know it yet but we're trying to check whether this is true or not so that's the same as the statement that the function is continuous because the limit of f of X is supposed to be f of X 0 and so this difference should have limits 0 and now the way this is proved is just by rewriting it by multiplying and dividing by X minus x0 so I'll rewrite the limit as X goes to X 0 of f of X minus f of X 0 divided by X minus x0 times X minus x0 ok so I wrote down the same expression that I had here this is just the same limit but I multiplied and divided by X minus x0 and now when I take the limit what happens is the limit of the first factor is f prime of X zero that's the thing we know exists by our assumption and the limit of the second factor is zero because the limit is X goes X 0 of X minus x0 is clearly zero so that's it the answer is zero which is what we wanted all right so that's that's the proof now there's something exceedingly fishy looking about this proof and let me just point to it before we proceed namely you're used in limits to setting X equal to zero and this looks like we're multiplying dividing by zero exactly the thing which makes all proofs wrong in all kinds of algebraic situations and so on and so forth you've been taught that that never works right but somehow these limiting tricks have found a way around this and let me just make explicit what it is in this limit we never are using x equals x 0 that's exactly the 1 value of x that we don't consider in this limit that's how limits are cooked up and that's sort of been the theme so far today is that we don't have to consider that and so this multiplication and division by this number is legal it may be small this number but it's always nonzero and so this really works and it's really true and we've just checked that a differentiable function is continuous so I'm going to have to do these carry out these limits which are very interesting zero over zero zero limits next time but let's hang on for one second to see if there any questions before we stop yeah there is a question I'll repeat this this proof right here okay just say it say again okay so there are two steps to the proof and the croup the step that you're asking about is the first step right and what I'm saying is if you have a number and you multiply it by 10 over 10 it's the same number if you multiply it by 3 over 3 it's the same number 2 over 2 1 over 1 and so on so so it's okay to change this to this it's exactly the same thing that's that's the first step yes the question was how does the proof how does this line yeah where the question mark is so what I checked was that this number which is on the left hand side is equal to this note this very long complicated number which is equal to this number which is equal to this number and so I've checked that this number is equal to 0 because the last thing is 0 this is equal to that is equal to that is equal to 0 and that's that's the proof yes yeah so you're that's it that's a different question okay so the the hypothesis of differentiability I use because this limit is equal to this number that that limit exists that's how I use the hypothesis of the theorem the conclusion of the theorem is the same as this because being continuous is the same as limit as X goes to X zero of f of X is equal to f of X zero that's the definition of continuity and I subtracted f of X zero from both sides to get this as being the same thing so this claim is continuity and it's the same as this this question here this issue here last question how do we get the zero from this so the claim that's being made so the claim is why is this tending to that so for example I'm going to have to erase something to explain that so the claim is that the limit as X goes to X 0 of X minus X 0 is equal to 0 that's that's what I'm claiming ok does that answer your question ok all right ask me else other stuff after lecture