Start  if f(x)=xlnx to the 8, some of you had a different number remember there are three versions of the exam calculate f prime of x okay so f prime of x thats the product rule so how do we do lnx to the 8, you put lnx to the 8 on the bottom put 8x to the 7 on top 
0:32  plus derivative of x=1 so log x to the 8 alrighty, 8x to the 7th is 8x to the 8 divided by x to the 8, this is just 8 plus lnx to the 8 oh wait i can put log of 8 in front of that and can get this and if you had a 6 it would just be 6 +6lnx if you had 11, i think the other number was 
1:02  you can leave it in this form if you want
sure okay
but the problem is you have to do part b
okay yes thats a 5
in theory you have 5 out of 5 on part a
so far so good?
now we have to find where x is 0 put the 8 on the other side 
1:31  you get 8lnx is 8 you get negative 1 so thats x e to the negative 1 also known as 1/e and where is it increasing or decreasing you could just say lets take a guess look at that derivative 
2:01  and plug in a number bigger than e like 1 natural log of 1 is 0 that would be 8, so it would be positive over here and then you should guess it should be positive over there or you could try and choose something greater than e like 1/2e or 1 over/e squared so this is decreasing less than 1/e increase in greater 1/e thats a hard question no doubt but this is not so hard 
2:30  and even getting to the 1/e gets you most of the way there its possible to pick up some credit for that feel better about that no im sorry, i am im sorry 
3:10  find the slope of the line tangent to the curve sin xy x squared + y squared minus pi i think i did this in a review or i did something really close to this every once in a while i get a little to close 
3:31  theres no rule against that, i try not to the derivative of sin xy is cosxy times derivative of xlog times the derivative of y plus y times the derivative of x equals 2x plus 2y derivative of y 
4:00  plus 0 or minus 0 because the derivative of pi is 0 you put the pi in there and a whole bunch of people start to freak out its amazing though now we plug in 0 and the square root of pi we didnt ask you for the equation we just asked you for the slope so dont do any algebra just plus in so this becomes cos of 0 cause 0 times the square root of pi is 0 and then this is 0 dy/dx 
4:31  plus the square root of pi
equals, okay so now
because you all know the unit circle
cos0 is 1
and you should make a guess
its a coin toss if you get 0 or 1
if its 0 it kind of wipes out the whole right side
makes it hard to solve the problem so its probably 1
did you get one?
are you sure, theres a curve at 0 
5:05  alright so this is 1 times 0 thats just the square root of pi okay other side 2 times 0 is 0 so you get 2 square root of pi dy/dx what happens at the square root of pi they cancel you have 1 is 2 dy/dx dy/dx is a half 
5:40  we feel better about that?
okay some of you can still go to medical school 
6:15  alright lets see thats all im going to go over for now now we learn something knew 
6:39  this is a 10 point question my guess is finding the derivative is 5 plugging in was 3 or 4 getting the actuall right answer was 1 or 2 thats my guess but its up to my tas different professors have different systems of grading some professors tell you exactly what everything is 
7:00  and some professors say i leave it up to the ta to decide how much the partical credit is and that stuff but thats kind of how i would of graded it, thats 5 that probably gets you 8 and successfully getting to here would get you to 10 kinda want to see if you know what the s=cs of 0 is anyone have plans to take take 126 you still need to know the unit circle for those of you who are planning to finish after this course 
7:30  you just want to pass you do it in minimal very soon you wont have to memorize this stuff which will be great news okay but youre not there yet okay so we are going to do one more new thing today local linear approximation what does that phrase mean this is the last part of chapter 3 in the book 
8:06  you have a curve if you look at a curve a very narrow part of the curve it kind of looks straight it doesnt really look like a curver it kind of looks like a line if you zoom in close enough thats what we call local linear approximation 
8:35  further more lets say thats the curve, you want to know whats going on there but the curve is complicated if you draw the tangent line youll kind of get an idea whats going on in the curve as long as youre close to where the tangent line hits the curve if youre far away here your y values are off 
9:00  not that intended the curve how much thats off, it could just be a billion or it could be a billion depending on what the scale is right but when you get close your y value is very close when you go up you go up the curve you go up the x and youre hitting the same spot here when you go up the x value you get the curve and you get the ling youre off by more so you can approximate things by using the tangent line instead of using the curve itself 
9:31  you get the tangent line by taking the derivative well how would you do that, well its calculus remember you have to do the derivative hang on i have to erase that its hard to write it 
10:02  okay thats our definition of our derivative you know that, this is the first example lets think of what that says for a minute that says to get the derivative of the function of x you sole for the tangent line you take the difference in the ys then you divide it by the difference in the x's you want to shrink that to a very small number take f of x and you take it f plus a little bit 
10:30  and you divide by that little bit
and then you shrink that little bit to 0
and that gives you the derivative
and thats the process, so if we got rid of this
now we can say the deirvative
is kind of like
f(x+h)+fx
over h but not exactly
but as long as h is small
thats still going to get you close
so what do we mean by small?

11:00  well if f of x is a million then h is going to be a small number like 10 or 20 thats still a very small part of a million then the next is 2 then h is going to say, .1 or .01 so its a small part of the whole okay what do you say when h is small its an irrelevant term, its small compared to the whole thing so if youre doing the distance from here to the sun time 3 million miles then the distance from here to new york 
11:30  very small approximation with that but then if youre doing the distance fromhere to ny then to something, huntington its a pretty big fraction of that when you do the approximation you want to the small rather than the whole thing now we just do a little re arranging, multiply by plus h and youre going to add the fx to the other side 
12:00  and im gonna get f of x +h of f of x +f prime of x times h, i did a little rearranging okay this equation is the same as this eqution with crossing multiplying and adding and all sorts of stuff like that and that is your linear approximation some books write it this way 
12:44  i want to match what the book says, right, okay this is the same thing, theres two ways to right the same equation so this says to find f as a specific number 
13:03  you take x of some other number and add the derivative and the difference of those two numbers this says to find f at a number plus the difference you start with the number and then you multiply it by the derivative times the differece its the same thing theres just two ways to write the same information i like the top one stewart likes the bottom one, it doesnt really matter 
13:31  but suppose, lets approximate the square root of 16 .1 now you say to me why would i approximate the square root of 16.1 i can take out my calculator and find the square root of 16.1 youre right but your calculator is still in approximation the reason you want to do this is 
14:00  want to give you something where the numbers are less obvious you get the concept of you can get very close to the answer by using the derivative than the function and of course before calculators were invented you didnt have these fall backs you say yes we do have calculators and computers so why bother the answer is i want you to get the idea on whats going on so what would our function be, our function 
14:30  would be square root x and what value would we do it it at, well if x=16 but i would be off by .1 so my h would equal .1 or if you were using the second method your a would be 16 and your x minus a would be .1 same thing 
15:01  now im going to approximate the square root of 16.1 so first i need to find the derivative and of course you memorize its 1/2 square root of x told you it would be handy this would be a perfectly good question to ask later on in the course now we use our formula this says if i want to find 
15:33  f at 16.1 its about f of 16 + f prime of 16 times .1 f at 16.1 is about f of 16 plus the derivative of f of 16 times the error, times .1 by how far im off 
16:00  so what is f of 16, well f of 16 is the square root of 16, thats 4 square root of 16.1 about 4 what is f prime of 16, f prime of 16 equals 1/ 2square root of 16 1/ two times 4 equals 8 so thats 1/8 by the way turn point 1 into a fraction 
16:30  its 1/10 it about 4 and 1/80 so lets see how we did 4+1/80 is 4.0125 
17:02  and now i do the square root of 16.1 and i get 4.01248 okay not bad thats a pretty good approximation were only off by .0002 so you did pretty good at estimate 
17:31  now what happens if you dont do a very good approximation so lets say were not that close to 16 lets say we want to use this for the square root of 18 i want to do the square root of 18 id be using the same formula id say thats about the square root of 16 plus 1/2 radical 16 so 1/8 how far am i off by? im off by 2 
18:03  my 18 is 2 because if my x is 16 its x plus h is 18 this 4.25 now what is the square root of 18 4.24 26 still good but im not nearly as good as i was there 
18:34  imagine do it with big numbers you do the square root of 1000 you do the square root of thousand whatever, you could start to be off so the tangent line is a pretty good approximation now some curves is excellent approximation some curves is not as good lets do another one 
19:20  say i want to approximate 2.1 to the 5 you can try 2.1 to the 5th with out a calculator, you can just sit there for a bit 
19:30  work it out what would f of x be, f of x would be x to the 5 
20:01  the derivative would be
im sorry, and x would be 2
h=.1
that make sense? see where the /1 is coming from?
thats how far i am off by the way i can let x be 3 and h would be .9 but why would i do that i could make and x be 2.09 and h be .01 but that wouldnt help me im trying to find a number that i can easily 
20:30  and find out how far off i am so the derivative 5x tot he fourth lets see what f(2), f of 2 is 2 to the 5 is 32 f prime of 2 is 5 times 2 to the 4 which is 80 
21:06  okay so this says f of x 2.1 to the 5th that would equal to 2 to the 5th plus 5 times 2 to the 4 times .1 
21:31  about 32 plus 8 about 40 sounds good doesnt it it should be 80, understand why it should be 80 2 to the 4th is 16, 16 times 5 is 80 oh no why did i write 8 
22:01  because its 5 times 2 to the 4th times .1 i did that in my head, i multiplied that by .1 pretty easy to multiply by .1, you multiply by 1 most of you can multiply by 1 you can also multiply by 10 multiply by 0 you can multiply by 0 
22:33  so we get 40, so whats 2.1 to the 5 2.1 to the 5 is actually 40.4 and thats okay you taken chemistry whats the error, you take the difference divide by the actual number 
23:03  ehh its .02, its a 2% error thats not bad got the idea, lets have you guys try one 
23:32  approximate the cubed root of 8.1 we spent long enough what would f of x be here, f of x also known as x to the 1/3 where x is 8 h=.1 
24:00  by the way rememvber you dont have calculators for the exams so we cant give you credit for difficult numbers to work with on webassign you can use caluculators so we can have messier numbers cause you can use calculators for webassign f prime of x 1/3 x to the negative 2/3 what does 2/3 mean means one over 
24:31  cubed root of x squared thats what negative 2/3 means lets see f is 8 is the cubed root of 8 is 2 f prime of 8 is 1/3 1/ cubed root of 8 squared cubed root is 2, 2 squared is 4 3rd times the fourth is 1/12 
25:02  so we plug this into our formula we get the cubed root of 8.1 would be about cube root of 8 plus 1/12 times .1 which is 1/10 about 2 and 1/120 so if you want to decimal that 
25:33  i heard someone ask this question
what happened to the 1/3?
1/3 cube root of 8 is 2, 2 squared is 4 1/4 times 1/3 is 1/12 this is 2. how many 0's 0083 
26:04  so we get 2.00833 so the actual answer is 2.00833 thats very good why is that such a good approximation remember what were actually doing cubed root of x looks something like that we want to find the cubed root of x when x is 8 
26:31  you look at the tangent line and you say well i dont know what it is right here but i can go to the tangent line its going to be very close to the curve theres no difference between the curve and the tangent line thats why youre so close 
27:07  so whats going on, youre using the tangent line very close to whats going on with the curve lets do another one of these to see if everyone gets the concept lot of things to learn i want to learn this together, and know the trick whats the approximate of sin44 degrees 
27:31  well f of x would be sin ofx
where x is 45 degrees
how far am i off?
1 my h would be minus 1 degrees f prime of x is cos x 
28:02  so lets use our formula its going to be sin of x plus the derivative times h right so thats going to be sin45 plus cos45 times 1 whats the sin of 45 
28:31  square root of 2 over 2
cos 45
same thing but wait a second
i have that as 0
0 is no where near, theres something wrong
it cant possibly be true it cant be 0
0 is no where near square root of 2 over 2
so what am i doing wrong?
why is that not right? what did i do? 
29:09  i heard it over here have to do it in rdians ah you havent seen degrees in the exams soo far have you you always see the pi stuff wso 45 degrees would be pi/4 and your error would be 1 degree should it would be pi/180 
29:41  so here is sin pi/4 plus cos pi/4 times minus pi/180 thats about 
30:00  square root 2/2 minus pi radical 2 over 360 and you work that out very close 
30:48  this comes out .682 and sin44 is actually .695 so you are pretty good 
31:01  not amazing still off a little but but pretty good so thats a favorite trap professor put that on the exam but i caught it all the stuff you do do it in radians dont do it in degrees derivative of sin is cos if you in radians theres one other thing you can do with this, another approximation 
31:35  this is whats called differential differentials have on the test we expect you to know what square root of 2/180 is oh sure 
32:01  oh 180, oh of course you wouldnt have to do that actually its more accurate yea much closer than i though .6947 thats actually very accurate sorry about that no of course we wouldnt expect you to do somthing that messy 
33:03  differentials, alright ive been doing this for a while if you think of the derivative of y is dy/dx and f prime of x remember what dy/dx means dy means the change in y dx means the change in x and we have f prime so if you cross multiply 
33:30  you get something that looks like that thats your differential formula and what that really says is a little change in y is equal to the derivative times the little change in x this is what you would use if you want to find out how something changes, so this actually shows up a lot in chemistry physics engineering and biology you make small changes and you want to know what kind of change in theother end so for example 
34:00  suppose you have 
35:04  approximate the change in volume in a cube when its side increase from 3 to 3.1 cm so what would we do well use this dy is f prime of x dx now f of x would be the bottom of the cube so f of x 
35:30  would be x cubed where x is 3 and what would dx be well dx would be the change in the side of the cube so dx would be .1 so we need f prime of x, f prime of x is 3x squared 
36:05  to get dy basically the change change in volume would be 3x squared so 3 times 2 squared times .1 so that would be about 2.7 cm and then you pull out a calculator and you say whats the actually change in volume well lets see 
36:34  i go from 3 cubed to 3.1 cubed which is 29.791 this says it will increase by 2.7 so 29.7 thats pretty good okay so differentials help you figureout the change the tangent line approximation helps you figure out the actual result 
37:00  differential is the change in result, both of them are useful well i know how much you love math but im going to stop early today because you had the exam yesterday, ill see you guys on monday 