the following content is provided under a Creative Commons license your support will help MIT open courseware continue to offer highquality educational resources for free to make a donation or to view additional materials from hundreds of MIT courses visit MIT open courseware at ocw.mit.edu so again welcome to 1801 we're getting started today with what we're calling unit one highly imaginative topic uh highly imaginative title and it's differentiation so let me first tell you briefly what's in store in the next couple of weeks the main topic today is what is a derivative and we're going to look at this from several different points of view and the first one is a the geometric interpretation and that's what we'll spend most of today on and then we'll also talk about a physical interpretation of what a derivative is and then there's going to be something else which I guess is maybe the reason why calculus is so fundamental and why we always start with it uh at in most science and engineering schools which is the importance of derivatives of of this to all measurements so that means pretty much every place that means in science and engineering in economics in uh political science Etc uh polling uh lots of commercial applications just just about everything now so that's what we'll be getting started with and then there's another thing that we're going to do in this unit which is we're going to explain how to differentiate anything so how to differentiate any function you know and that's kind of a tall order but let me just give you an example if you want to take the derivative this we'll see today as the notation for the derivative of something of some messy function like e to the X AR tan of X we'll work this out by the end of this unit all right so anything you can think of anything you can write down we can differentiated all right so that's what we're going to do and today as I said we're going to spend most of our time on this geometric interpretation so let's let's begin with that so here we go with the geometric interpretation of uh derivatives and what we're going to do is just ask the geometric problem of finding the tangent line to some graph of some function at some point which is say x0 y0 so that's the problem that we're addressing here um guess I should probably turn this off all right so here's our problem and now let me show you the solution so well let's graph the function so let's say here's its graph and here's some point all right maybe I should draw it just a bit lower so that I don't all right so here's a point P maybe it's above the point x0 x0 by the way this was supposed to be an x0 that was the some fixed place on the x axis and now in order to perform this this Mighty feat I will um use another color of chalk how about red okay so so here it is there's the tangent line well not quite straight close enough right I did it all right that's the end that's the geometric problem I achieved what I wanted to do and uh it's kind of an interesting question which unfortunately I can't solve for you in this class which is how did I do that that is how physically did I manage to know what to do to draw this tangent line but that's what geometric problems are like um we visualize it we can figure it out somewhere in our brains it happens and the task that we have now is to figure out how to do it analytically to do it in a way that uh a machine could do just as well as I did in drawing this tangent line so so what do we learn in high school about what a tangent line is well a tangent line has an equation and any line through Point has the equation y - y0 is equal to M the slope time x - x0 so so here's the the equation for that line and now there are two pieces of information that we're going to need to work out uh what the line is the first one is the point that's that point P there and to specify P given given X we need to know the uh the the level of Y which is of course just f ofx z now that's that's not a calculus problem but anyway that's a very important part of the process so that's the first thing we need to know and the second thing we need to know is the slope and that's this number M and in calculus we have another name for it we call it frime of x0 namely the derivative of f so that's the calculus part that's the tricky part and that's the part that we have to discuss now so just to make that uh explicit here I'm going to make a definition which is that frime of x0 which is known as the derivative of F at x0 all right is the slope of the tangent line to Y = FX at the point uh uh let's just call it P all right so so that's what it is but still I haven't made any progress in figuring out any better how I drew that line so I have to say something that's more concrete because I want to be able to cook up what these numbers are I have to figure out what this number m is uh and one way of thinking about that let me just uh try it is so I certainly am taking for granted the sort of non-calculus part that I know what a line through a point is so I know this equation but another possibility might be you know this line here how do I know well fortunately I didn't draw it quite straight but there it is how do I know that this orange line is not a tangent line but this other line is a tangent line well it's it's actually not so obvious and but I'm going to describe it a little bit it's it's not really the fact this thing crosses at some other place which is this point q but it's not really the fact that the thing crosses at two place because the line could be Wiggly the curve could be Wiggly and it could cross back and forth a number of times that's not what distinguishes the tangent line so I'm going to have to somehow grasp this and I first do it in language and it it's the following idea it's that if you take this orange line which is uh called a secant line and you think of the Q the point Q is getting closer and closer to P then the slope of that line will get closer and closer to the slope of the red line and if we draw it close enough then that's going to be the correct line so that's really what I did sort of in my brain when I drew that first line and so that's the way I'm going to articulate it first now so the tangent line is equal to the limit of what so-called secant lines PQ as Q TS to p and here we're thinking of p is being fixed and Q is varying all right so so that's the the G again this is still a geometric discussion but now uh we're going to be able to put symbols and formulas to this computation and we'll be able to um to work out uh formulas in any example so so let's do that so so first of all I'm going to write out these points p and Q again so maybe we'll put P here and Q here and I'm thinking of this line through them I guess it was orange so we'll leave it as orange all right and now I want to compute its slope and so this is gradually we'll do this in two steps and these steps will introduce to the basic notations which are used throughout calculus including multivariable calculus across the board so the first notation that's used is you imagine here's the x-axis underneath and here's the x0 the location directly below the point p and we're traveling here a horizontal distance which is denoted by Delta X so that's Delta X socaled and we could also call it the change in X all right so that's one thing we want to measure in order to get the slope of this line PQ and the other thing is this height so that's this distance here which we denote Delta F which is the change in F and then the slope is just the ratio Delta F over Delta X so this is the slope of the of the secant and the process I just described over here with this limit applies not just to the whole line itself but also in particular to its slope and the way we write that is the limit as Delta X goes to zero and that's going to be our slope so this is the slope of the tangent line okay now this is still a little a little general and I'm going to I want to work out a more usable form here I want to work out a better formula for this and in order to do that I'm going to write Delta F the numerator more explicitly here the change in F so remember that the point p is the point x0 F of x0 all right that's what we got from our formula for the point and in order to compute these distances and in particular the vertical distance here I'm going to have to get a formula for Q as well so if this horizontal distance is Delta X then this location is x0 plus Delta X and so the point above that point has a formula which is x0 plus uh sorry plus Delta x f of and this is a mouthful x0 + Delta X all right so there's the formula for the point Q here's the formula for the point p and now I can write a different formula for the derivative which is the following so this frime of x0 which is the same as m is going to be the limit as Delta X goes to zero of the change in F Well the change in F is the value of F at the upper Point here which is x0 + Delta X and minus its value at the lower Point P which is f of x0 divided by Delta X all right so this is the formula I'm going to put this in a little box because this is by far the most important formula today which we use to derive pretty much everything else and this is the way that we're going to be able to compute these numbers so let's let's do an example this example so we'll call this example one uh we'll take the function f ofx which is 1/x that's sufficiently complicated to have an interesting answer and uh sufficiently straightforward that we can compute the derivative fairly quickly so so what is it that we're going to do here all we're going to do is we're going to plug in this this formula here for for that function that's that's all we're going to do and Visually what we're accomplishing is somehow to take the hyperbola and take a point on the hyperbola and figure out some tangent line all right that's what we're accomplishing when we do that so we're accomplishing this geometrically but we'll be doing it algebraically so first we consider this difference Delta F over Delta X and write out its formula so I have to have a place so I'm going to make it again above this point x0 which is a general point we'll make the general calculation so the value of F at the top when we move to the right by F ofx so I just read off from this read off from here the uh the formula the first thing I get here is 1 over x0 + Delta X that's the leftand term minus 1x0 that's the right hand term and then I have to divide that by Delta X okay so here's our expression and by by the way this has a name this thing is called a difference quotient it's pretty complicated because there's always a difference in the numerator and in Disguise the denominator is a difference because it's the difference between the value on the right side and the value on the left side here okay so now we're going to simplify it by some algebra so let's just take a look so this is equal to let's continue on the next level here this is equal to 1 / Delta x * now all I'm going to do is put it over a common denominator so the common denominator is x0 + Delta x * x0 and so in the numerator for the first expression I have x0 and for the second expression I have x0 + Delta X so this is a the same thing as I had in the numerator before factoring out this denominator and here I put that numerator into this more amable form and now there are two basic cancellations the first one is that x0 and x0 cancel so we have this and then the second step is that these two expressions cancel right the numerator and denominator now we have um a cancellation that we can make use of so we'll write that under here and this is uh equals Min -1 over x0 + Delta x * x0 and then the very last step is to take the limit as Delta X tends to zero and now we can do it before we couldn't do it why because the numerator and the denominator gave us 0 over Z but now that I've made this cancellation I can pass to the Limit and all that happens is I set this Delta x equal to Z and I get minus1 /x 0^ 2ar all right so that's the answer right so in other words what I've shown let me put it up here is that fime of x0 is -1 / x0 2 now uh let's let's look at the graph just a little bit to check this for plausibility all right uh what's Happening Here is first of all it's negative right it's less than zero which is a good thing you see that slope there is negative that's the simplest check that you could make and the second thing that I would just like to point out is that as X goes to Infinity that that is if as we go farther to the right it gets less and less steep so uh less and whoops as X go x0 goes to Infinity not not zero as x0 goes to Infinity less and less steep so that's also consistent here is when x0 is very large this is a smaller and smaller number in in magnitude although it's always negative it's always sloping down all right uh so I've managed to fill the boards so maybe I should stop for a question or two yes so the question is to explain again this uh limiting process so the formula here is We have basically two numbers so in other words why is it that this expression when Delta x ts to 0 is equal to - 1x0 2 let me let me illustrate it by sticking in a number for x0 to make it more explicit all right so for instance let me stick in here for x0 the number 3 then it's -1 over 3 + Delta x * 3 that's the situation that we've got and now the question is what happens as this number gets smaller and smaller and smaller and gets to be practically zero well literally what we can do is just plug in zero there then you get 3 + 0 * 3 in the denominator minus one in the numerator so this tends to tends to -1 over 9 over 3^ SAR and that's what I'm saying in general with this with this extra number here other other questions yes how did you simplify Del Delta X how you simplify from the origal equation to so the question is how what happened between this step and this step right explain this this step here all right so there were two parts to that the first is this Delta X which was sitting in the denominator I factored all the way out front and so what's in the parentheses is supposed to be the same as what's in the numerator of this other expression and then at the same time as doing that I put that expression which is a difference of two fractions I expressed it with a common denominator so in the denominator here you see the product of the denominators of the two fractions and then I just figured out what the numerator had to be without really yeah other questions okay okay so now uh so I I claim that on the whole calculus is uh gets a bad wrap that it's um actually easier than than most things um but it has there's a perception that it's that it's that it's harder and so I really have a duty to to give you the calculus made harder a story here so we we we we have to make things harder because that's that's our job and this is actually what most people do in calculus and it's the reason why calculus has a bad reputation so the the the secret is that when people ask problems in calculus they generally ask them in context and there are many many other things going on and so the little piece of the problem which is calculus is actually fairly routine and has to be isolated and gotten through but all the rest of it relies on everything else you learned in mathematics up to this stage from grade school to through high school so so that's the complication so now we're going to do a little bit of calculus made hard by uh uh talking about a word problem now we we only have one sort of word problem that we can pose because all we've talked about is this geometry uh uh point of view so so far those are the only kinds of word problems we can pose so what we're going to do is just pose such a problem so find the areas of triangles enclosed by the axes and the tangent to um y = 1X okay so that's a geometry problem and let me draw a picture of it it's practically the the same as the picture for example one of course so here's we only consider the first quadrant here's our shape all right it's the hyperbola and here's maybe one of our tangent lines which is coming in like this and then we're trying to find this area here all right so there's our problem so why does it have to do with Calculus it has to do with Calculus because there's a tangent line in it and so we're going to need to do some calculus to to answer this question but as you'll see the calculus is the easy part is and once I figured out what the tangent line is the rest of the problem is no longer calculus it's just that slope that we need so what's the formula for the tangent line put that over here it's going to be Yus y0 is equal to and here's the magic number we already calculated it it's in the box over there it's -1 /x 0^ 2ar x - x0 so this is the only bit of calculus in this problem but now we're not done we have to finish it we have to figure out all the rest of these quantities so we can figure out the area all right so how do we do that well to find this point this has a name we're going to find the um so-called x intercept that's the first thing we're going to do so to do that what we need to do is to find where this horizontal line meets that diagonal line and the equation for the x intercept is y = 0 all right so we plug in y equals 0 that's this horizontal line and we find this point so let's do that into star so we get 0 minus oh one other thing we need to know we know that y0 is f of x0 and F ofx is 1 /x so this thing is 1 / x0 right and that's equal to -1 /x 0^ 2ar and here's X and here's x0 all right so in order to find this x value I have to uh plug in one equation into the other so this simplifies a bit uh let's put let's see this is uh - xx0 2 and this is plus 1 over x0 because the x0 and x0 squar cancel somewhat and so if I put this on the other side I get x / x 0^ 2 is equal to 2 x0 and if I then multiply through so that's what this implies and If I multiply through by uh x0 2 I get X is equal to 2 x0 okay okay so I claim that this point we've just calculated it's 2x0 now I'm almost done I need to get the other one I need to get this one up here now I'm going to use a very big shortcut to do that so so the shortcut to the Y intercept sorry yeah the Y intercept um is to use symmetry right I claim I can stare at this and I can look at that and I know the formula for the Y intercept it's equal to 2 y0 all right that's what that one is so this one is 2 y0 and the reason I know this is the following so here's the symmetry of the situation which is not completely direct it's a kind of mirror symmetry around the diagonal it involves the ex change of XY with YX SO trading the roles of X and Y so the Symmetry that I'm using is that any formula I get that involves x's and y's if I trade all the x's and replace them by y's and trade all the Y's and replace them by X's then I'll have it a correct formula on the other way so everywhere I see a y I'm making an X and everywhere I see an X I make it a y the switch will take place so why is that that's because the that's just an accident of this equation that's because so the Symmetry explained is that the equation is y = 1 /x but that's the same thing as XY = 1 If I multiply through by X which is the same thing as x = 1 y so here's where the X and the Y get reversed okay now if you don't trust this explanation you can also get get the Y intercept by plugging xal 0 into the into the equation star okay we plugged y equals 0 in and we got the x value and you could do the same thing analogously the other way all right so I'm almost done with the with the geometry problem and uh let's uh let's finish it off now well let me hold off for one second before I finish it off what I'd like to say is just make one more tiny remark all right and this is the hardest part of calculus in my opinion so the hardest part of calculus is that we call it one variable calculus but we're perfectly happy to deal with four variables at a time or five or any number in this problem I had an x a y an x0 and a y z that's already four different things they have various interrelationships between them so of course the manipulations we do with them are algebraic and when we're doing the the the derivatives we just consider one what's known as one variable calculus but really there are millions of variables floating around potentially so that's what makes things complicated and that's something that you have to get used to now there's something else which is more subtle and that I think many people who teach the subject or use the subject aren't aware because they've already entered into the language and they're not uh they're so comfortable with it that they don't even notice this confusion there's something deliberately sloppy about the way we deal with these variables the reason is very simple there are already four variables here I don't want to create six names for variables or eight names for variables and but really in this problem there were about eight I just slipped them by you so why is that well notice that the first time that I got a formula for y z here it was this point and so the formula for y0 which I plugged in right here was from the the equation of the curve y0 = 1x0 the second time I did it I did not use y = 1X I used this equation here so this is not y = 1X that's the wrong thing to do that's an easy mistake to make if if the formulas are all a blur to you and you're not paying attention to where they are on the diagram you see that Y intercept that x intercept calculation there involved where this horizontal line met this diagonal line and Y equals z represented this line here so the liness is that y means two different things and we do this constantly because it's way way more complicated not to do it Con to do it it's much more convenient for us to allow ourselves a flexibility to change the role that this letter plays in the middle of the of the computation and similarly later on if I had done this by this more straightforward method for the uh Y intercept I would have set x equal to Z that would have been this vertical line which is x equals 0 but I didn't change the letter X when I did that because that would be a waste for us so this this is this is one of the main confusions that happens if you can uh keep yourself straight you're you're a lot better off and and as I say this is this is uh this is one of the complexities all right so now let's finish off the problem and let me finally get this area here so actually I'll just finish it off right here so the area of the triangle is well it's the base times the height the base is 2x0 the height is 2 y0 and a half of that so it's a half 2x0 * 2 y0 which is 2 x0 y0 which is lo and behold two so the amusing thing in this case is it actually didn't matter what x0 and Y Z are we get the same answer every time that's just an accident of the function 1 /x happens to be the function with that property all right so we have still have more business today serious business so let me continue so first of all I want to give you a few more notations and these are just other ways that people uh refer uh notations that people use to refer to derivatives and the first one is the following we already wrote Y is equal to F ofx and so when we write deltay that means the same thing as Delta F that's a typical notation and previously we wrote um f Prime for the derivative so this is this is so this is Newton's notation for the derivative okay but there are other notations and one of them is DF DX and another one is d ydx meaning exactly the same thing and sometimes we let the function slip down below so that becomes d by DX of f or D by DX of Y so these are all notations that are used for the derivative and these were initiated by livets and these notations are um used interchangeably sometimes uh practically together they both turn out to be extremely useful this one omits notice that this thing omits the uh underlying base Point x0 that's one of the nuisances it doesn't give you all the information but there are lots of uh situations like that where where uh people leave out some of the important information you have to fill it in from Context so that's another couple of notations so now I have one more calculation for you today uh I carried out this calculation of the derivative of the um of the um the the derivative of the function 1 /x I want to take care of some other powers so let's do that so example two is going to be the function F ofx is X to the n and = 1 2 3 one of these guys and now what we're trying to figure out is the derivative with respect to X of x to the n in our new new notation what this is equal to so again we're going to form this expression Delta F Delta X and we're going to make some algebraic simplification so what we plug in for Delta f is X+ Delta x to the N minus X the N / Delta X now before let me just stick this in and I'm going to erase it before I wrote x0 here and x0 there but now I'm going to get rid of it because in this particular calculation it's a nuisance I don't have an X floating around which means something different from the X zero and I just don't want to have to keep on writing all those symbols it's a waste of Blackboard uh energy uh there's a total amount of energy that I'm you know I've already filled up so many blackboards that it's just a limited amount of plus I'm trying to conserve chalk okay anyway no zeros so think of X is fixed again um in this case Delta X moves and X is fixed in this in this calculation all right now in order to simplify this in order to understand algebraically what's going on I need to understand what the nth power of a sum is and that's a famous formula we only need a little tiny bit of it called the binomial theorem so the binomial theorem binomial theorem which is in your text and uh explained in an in an exercise says uh in an appendex sorry says that if you take the sum of two guys and you take them to the nth power that of course is X Plus Delta X multiplied by itself n times and so the first term is X to the N that's when all of the N factors come in and then you could have this factor of Delta X and all the rest X's so at least one term of the form X the n minus1 * Delta X and how many times does that happen well it happens when there's a factor from here from the next factor and so on and so on and so on there's a total of n possible times that that happens and now the great thing is that with this alone all the rest of the terms are junk that we won't have to worry about so to be more specific the junk there's a very careful notation for the junk the junk is what's called Big O of Delta x s what that means is that these are terms of order uh so with Delta X2 Delta X cubed or higher right that's how very exciting higher order terms okay so this is the only algebra that we need to do and now we just need to combine it together to get our result so now I'm going to just carry out the cancellations that we need so here we go we have Delta F over Delta X which remember was 1/ Delta X time this which is this times now this is X the n + n x nus1 Delta X plus this junk term minus x to the N all right so that's what we have so far based on our previous calculations now I'm going to do the main C cancellation which is this all right so that's one over Delta x * n x nus1 Delta X plus this term here and now I can divide in by Delta X so I get N X nus1 Plus now it's O of Delta X there's at least one factor of Delta X not two factors of Delta X because I have to cancel one of them and now I can just take the limit in the limit this term is going to be zero that's why I called it junk originally because it disapp appears and in math junk is something that goes away so this tends to as Delta X goes to zero n x n minus1 and so what I've shown you is that d by DX of x n minus sorry n is equal to n x nus one so now this is going to be super important to you right on your problem set in every possible way and I want to tell you one thing one way in which it's very important and one way that extends it immediately so this thing extends to pols we get quite a lot out of this one calculation namely if I take d by DX of something like X cubed + 5 x 10th power that's going to be equal to 3x^2 that's applying this rule to X cubed and then here I'll get 5 * 10 so 50 x 9th so this is the type of thing that we get out of it and we're going to make more hay with that next time question yes I turn myself off yes Del the question is the question was the binomial theorem only works when x uh Delta X goes to zero no the the binomial theorem is a general formula which also specifies exactly what the junk is it's very much more detailed but we only needed this part we didn't care what all these crazy terms were it's it's it's junk for our purposes now because we don't happen to need any more than those first two terms yes because the death Lex goes to zero okay see you next time