### Stony Brook MAT 125 Spring 2015 Definition of the Derivative, pre-lecture

 Start what we want to do now is make sense of the idea of infinitesimal rate of change or rate of change on a vanishingly small scale so we have some function here it is and i have a point that I care about it doesn't really matter where the axes are I have some point here that I care about here's FX X there's its graph of course we can do this with formulas and I will 0:31 do this with formula to do with pictures for right now and I want to make sense of what is the line that most looks like the function at that point now that doesn't really make a whole lot of sense because it seems like i could wait around a little bit not very well-defined concept however if I take a little region around here and blow it up to a bigger scale 1:05 than this function looks something like that my point of interest is right here and well the line is a little easier to see but it doesn't quite match now I'm gonna blow up another tiny region here and on this scale my function looks 1:36 pretty much like a straight line that is I can't tell the green line from the graph the function let's say on this picture on this scale this is a point of interest let's call it a just to let it be a general point and let's say that this 2:00 distance is this point 1 and if I blow it up again in here is a but this is a plus . 01 if I blow it up again here's a this is f of a then this is a plus . 001 and I can 2:31 continue this infinitely along i can take the limit of this picture and in the limit if everything is nice the green line and the white line are exactly the same and so I can do my calculation now how would I do this calculation? well we get blue color so here on this scale your say a plus 3:00 here's a plus five i compute the slope of this line is purple line and what is the slope of this perfect well this with is . 5 and what is the difference in Heights here well this point here is always the point a F a and this point up here is . a plus 3:42 . 5 f of a plus 1/2 so the slope of this purple line is going to be the difference in the heights so this is the change in height divided by the change --that's a c-- 4:18 change in the width -- the rise over the run-- that's our definition of slope and so the change in height here is we go from 4:33 here to here that distance is whatever the change is there and the the winds here is only five between a and a + . 5 in this picture we can do the 5:00 same thing put it up there when you have the same kind of formula f(a+.01)-f(a) over this width, which is .01 we have f(a+.001)-f(a) over .001 and we can continue in 5:33 this way and in general we want to look at F of a plus some little distance we moved away minus a and / how far away we moved all of these are of that form .5 .1 .001 and so on so that gives us the slope at this point notice that this number H could be negative i could look on this side 6:04 instead but in all of these cases i want to compute some quantity like this and now really what I want to do just rewrite that same formula is I want to know what happens when this distance gets very very very small that 6:37 is what is the limit as H goes to 0 this and this we define to be the slope of F at x= a and we give it a name we call that F prime of a if it works 7:05 maybe this limit doesn't exist in which case there is no well-defined slope if the function is sufficiently nice then this limit will exist and we can make sense of this slope notice that this is just erase some of the pictures ok we can think of this is a another way as saying take some point B and some other point a and compute the slope heres a 7:40 that I care about here's b some other point compute the slope between them that's going to be enough f of over here so that's the change in height and this is the change width 8:04 we take the limit as two points crash together take b and squeeze it towards a so this is the limit as the wind goes to 0 width which is this which is the same as this here b-a=h b is a plus a little bit is itself 8:34 this gives us the slope ok so that's the basic idea let's do it with an actual function actual number one actual example and we'll choose a fairly straightforward one with the limit is not too bad but it's also not completely 9:02 simple so let's take our function just squaring function graph of it looks like this let's take the point x equals 2 is highest for because 2 squared is 4 and let's ask the question what is the slope of the tangent line to f of x at x 9:44 equals 2 and we can compute this directly with my point of interest is to its values 4 i want to know what the slope of this line that passes through two and just barely 10:01 touches the graph is and I'll compute it by taking some point over here which is two plus a little bit finding out where it is on the graph well if i plug that in that would be 2 plus h squared and then I'll take the line between them and then i will run these points together and I claim that 10:32 this green line will approach the purple line as I get closer and closer and closer so let's see what the slope of this Green Line is so Green Line is evaluate the function and 2 plus h and subtract off the value to this is the change in height and i divide that by the value over here 2 plus h minus 2 this 11:15 is the change in width that's the slope the green line noticed that h is 0 I get 0 over 0 so I'm going to have to do some 11:37 stuff have to take a limit let's plug in what this is two plus h is (x^2+h)^2 subtract off f(2) that's 2 squared call that 4. 2 plus h minus 2h so that I have to do this i want to be able to do this as H goes to 0 so let's clean 12:04 it up a little bit so this will be 2 plus h quantity squared be 4 plus the middle term will be 2 H Double which is 4 h plus h squared that's just that i'm going to subtract the four divided by H now really what I want to do is compute the limit of this quantity as h goes to zero notice that I can't plug in h yet 12:35 because if i plug in h equals 0 i get 4 plus 0 plus 0 1 is 4 is 0 on the top / 0 on the bottom so i can't just compute it when H equals 0 what I want to do is compute the limit as H goes to 0 f of 2 plus h minus 2 over age which will be all of 13:04 this stuff here i can clean up now this will be the limit as H goes to 0 of 4 minus 4 id 0 that leaves me for h plus h squared on the top / h on the bottom and can factor out an H that's four plus h on the top here just factored the h out and now as long as H is 13:37 not 0 H over H is one as long as H is not 0 this will be equal to the limit as H goes to 0 plus4 plus h that's just for so this slope here is 4 so we've done this calculation so we computed is that F prime 2 is 4 it's also f of 2 maybe that was 14:07 an unfortunate choice but there it is.