Transcript for:
Natural Logarithms in Calculus 2

well let's get started then on your first day of calculus 2 and like I said we're going to start with Section 6.1 and we're going to talk about the natural log function you know I know you guys have seen the natural log function before but it's probably a good idea to go over some of the properties because you might have forgot them uh we haven't really worked with them a lot since maybe your like pre-calculus class maybe your math C Class when you get it for the first time so we're going to start with some of the properties what you can and what you can't do with natural logarithm and with log functions in general and how that's going to help us with our derivatives and some of our integrals so let's talk about the properties the first one first one was kind of a nice one anytime you see this you loved it because it meant if you had l n of one that was always equal to say it again Z yeah it's exactly right so what what this is if you don't remember how logarithms work logarithms say a base what's the base here d e is e is the base uh for any logarithm you don't need to write this but check this out if we had this for instance this would be our base that would be our power base to the power gives you the argument of the logarithm so this would have an exponential notation like that if you forgot all about logarithms that's like a really crash course in logarithms okay so in other words when we have a natural logorithm that's based on or the base is e the natural number e e is about 2.71 so on so on so on so when we have this this says Hey e to the 0o power gives you well how much is anything to the zero power that's why this works so anytime you see Ln of one we automatically know that's zero show up hands you feel okay with that so far good okay fantastic number two anytime you have and this is called the product property of logarithms in case you're wondering anytime you have a product within any logarithm including the natural logarithm you can always break up that logarithm into two different logarithms do you remember how to do it if I have log or Ln of x * y how do I split that up fantastic product plus plus product PP that's funny uh but anyway so product gives you a plus here so if we have Ln of x * y That's Ln of X Plus Ln of Y one thing to remember here these are not just oneway streets so we can either expand by a product or we can combine two logarithms by addition the only thing that you need to check on is that they have the same base now naturally our natural log always going to have the same same base so we're good to go as far as that goes you guys okay with number two yeah okay number three this one's really really similar and I'm imagining that even if you didn't know the property which you probably do you can probably figure it out if we expand multiplication by addition how are we going to SP expand division subtraction yeah fantastic which one's going to go first the Ln X or Ln y l that's right so Ln x minus Ln y again we can combine them we can combine two logarithms are being subtracted or we can expand a logarithm that has division inside of it we still okay so far y okay and last one this is my favorite one I love this one this one's called the power property says that anytime you have a logarithm with an exponent in the argument of that logarithm we can do something with that exponent what can we do with it yeah that's great so Ln of x to the r same thing as R * lnx now how we use these is to expand or combine logarithms and it depends on what situation we have but sometimes we'll be expanding sometimes we'll be combining depend on whether we're taking derivatives or integrals or doing something with that so let's practice just a couple times no calculus yet I want to make sure that you guys can expand some of these logarithms so let's start with that one Ln of 3x and you know what I want to make sure you guys know this the Ln goes in front of it Ln is it's not like multiplied by this argument it's Ln of the argument it's a it's a function of whatever your argument is here so Ln of 3x 4 all over 2 y^2 let's expand this as much as possible just to get the feel for how this this stuff works so um give me one idea that you might have um the minus yeah very good we we're noticing that right now we're kind of working from and you did this when you did derivatives a long time ago we're working from the outside in we don't really want to start with moving the exponents forward yet because well we got a couple of them right where would we put it doesn't really make a lot of sense so expand from the outside in the big part to the little Parts the biggest part we see the thing that combines all of this argument is division right now so let's take care of that Division and what's uh what's division changing again fantastic okay so we're going to have Ln 3x 4us Ln that's exactly right true or false right now it would be a great idea to move the exponents to the front no true or false false why because the exponent goes to X Y that's a great point so do you guys see what he was saying there he said because this exponent doesn't go to this whole thing it just it's just attributed that X I can't move forward yet now if it had been like this would I move it forward that'd be fine but if it's not you got to be real careful on that do you guys see the simple mistake that that can really really cause a problem right so watch the small stuff here all right well if we can't move that forward what's something that we can do with that logarithm what do you think how would I break them up to yeah you're exactly right because that's a product in there I can certainly do that let's do the same thing over here okay I want you think carefully on what I've just WR wrote on the board am I right or am I wrong wrong some over here why am I what did I do here negative need the go to fantastic so if we're subtracting this Ln remember if you expand this L this still means this right it still needs to be subtracted so either you're going to draw those brackets or you're going to distribute that negative that minus sign you guys with me on that one again be careful the small stuff so we're going to do a couple steps right now there's one thing I want you to do to expand this one more time what what can we do now with our logarithms Lou now you can do like the exponent to the front yeah that's absolutely right and then I'm going to distribute that minus so we'll have Ln 3 + 4 Ln xus ln2 nothing we can do there but here we'll have minus 2 Ln y this when we get to this point there's literally nothing else we can do to expand this make it any longer than than what we have right now by show of hands do you feel okay with expanding our logarithms good with it any questions at all what you're going to find out is that because we can do all this what's going to be easier to take a derivative of this argument or that argument that's way easier that's a constant forg good to sake derivative constants are hey that's fantastic this going to be a piece of cake this zero this going to be a piece of cake so we're we're getting there but by expanding logarithms we make our derivatives and our integrals a whole lot easier when we're dealing with Ln okie dokie okay tell you what we'll try um we'll try one more together and I want to see what you can do on one of them just keep something in mind too these properties they're the only properties that we have for logorithm as far as expanding combining one big thing people do in math is they like to make up their own math their own properties look if they weren't listed on the board you can't do them so be real careful about what you can expand and what you can combine so don't make up your own stuff just follow the things that they given you so let's see what we can do with this again we'll do uh one more together and one on your own so Ln x^2 + 1 all over the < TK of X tell me someone on the left hand side of the room you guys over here if I want to expand this all the way where might I start Division I would say change the X to the like theot to the 12 first I could do that sure uh I can do that now or I can do it later do you guys see what he was talking about the square root we know that's 1/2 power so we're going to do that ultimately either now or later doesn't really matter when as far as expanding goes where would I want to start would I want to start moving exponents forward would I want to start separating products or separating uh quotients which one you do the subtraction thing Okay cool so we know that this is the same thing as well since that's a quotient I know I'm going to have subtract square otk of x one more thing just a little side note for you guys whenever you're expanding sometimes you'll have to put some parentheses in the argument of your logarithm especially like right here this is really ambiguous it doesn't really say whether I have Ln of x^2 + 1 or Ln of x^2 + 1 you want to make sure you show that um don't leave the little stuff hanging out especially in calculus 2 because you get a l the the way that you write problems is important it's not meaningless like when you're doing math 80 you almost never get wrong right you just write anything you want uh but here at this level you really need to be careful about about how you're writing things so make sure you're you're accurate now I like the one2 we're going to change in just a moment what I want to know right here can I split that up anymore can I expand that any longer no that's great no we can't do it at all we know that if it was a product we could expand it but if it's plus there's nothing that says we can expand plus so if we can't expand the plus we can't move the two forward we're basically done here the only thing we can do and this was already said that is actually a power and that can move forward that's about as good we can get uh we can't split that up for sure we've already moved our power forward and that's expanded I want you to try one your I'm going give you a hard one uh if you can do this one you know what you're doing booah I think it deserves a booya let's go for it show me what you got remember that what we're doing work from the outside in go for the big Parts first and split off everything that you can follow only the properties that I've given you we can separate products we can separate quotients we can move powers to the front so do the best you can by the way when I have you do something like this I'll typically walk around the room if you have questions by all means please ask me that's is what this is for yeah to when we have a a cube root what's the power of that cube root cool so a square root because we had a square root we have 1 over two we got power over root basically the power of the inside over the root so when we have a cube root don't be putting crazy things like 2/3 or anything weird it's just 1/3 unless you have a power like this if we had a cube root of X2 that's when you would get a 2/3 power or something okay it's power over root so we have hey no problem the first thing I saw I've got division here this is containing this whole argument so I'm going to separate that division by a subtraction from the logarithm I'm also going to change that power to 1/3 power and that says I can move that out front next thing we're looking for is any remaining products or quotients in here um do you see any other products in here how many one or two good you know some some people who just don't like trigonometry they'll say oh yeah you got one right here is this a product no no this is actually the argument of a sign so this is the really the angle of that sign function so we are going to separate this product but we can't separate this it's not a product so we'll do Ln of x 5 we'll do a plus sign because we got product plus Ln of sin 2X and we'll do minus Ln of x to 4 + 2 to the 1/3 and lastly anywhere you have a exponent of an argument are we okay anytime we have an exponent of an argument we're going to move it up front so we have 5 Ln X the two can I move the two forward what do you think what does this mean when I have what does that mean X's to the 4 power is it the X is it the function function that's literally what that means so whenever you see a sign squared like this this is not being squared if it was I would have it here uh don't change your work please but watch if this was being squared I'd have it here and I'd have that as the AR that's how that would look if I don't want to say that if I want to say the functions being squared I know it's a little weird uh but people have gotten the habit of just writing it right there that's the power of the entire thing okay the entire function of sign that's being squared right now so we have this that's really a power that's an exponent we can move that to the front of course our 1/3 okay we've broken it down can we go any further do you see any products any quotients or any powers that we can move move no let's check no this well that's connected we can't separate addition we can't move that power forward we're completely done show up hands if you got that on your own that's really good you guys know what you're doing then you know just like we can expand our logarithms we can also combine them I just want to practice real quick doing one or two of those then we'll start talking about how we take derivative of these things you guys seem like really smart people so we'll cut right to the chase we'll do hardw so what I want to do here I want to combine these logarithms to see if we can get as as few logarithms as possible in our case we're trying to get one so one logarithm so let's take a look at it you know what you might have noticed is that we typically deal with our exponents last when we are expanding so here we're going to probably have to deal with some of these guys first otherwise we can't combine them does that make sense in order to find our logarithms when you when you notice those properties earlier it was log X Plus log y gives you log of x y or Ln X Plus Ln y there's no numbers up front in other words we have to move our numbers up front our coefficients to exponents first so tell me one thing you're going to do here should I deal with the 2/3 first that one we can't okay that's that's out there this is going to be the last thing we do should I deal with the five first yeah that's got to move up and the 1 half has to move up and the four has to move up so we'll leave that 2/3 it says order operations anyway really doesn't it it says 2/3 is outside deal with this stuff first put this all together then do the 2/3 so inside of our our bracket here we got an Ln of x to the 5th we've got a Ln x^2 + 1 to the 12 we got a minus Ln x - 2 all 4 power and I'll finish the FR okay let's keep on on working uh right side people you guys what's the next thing you might want to do say what M sure you can actually do both at the same time if you really know what you're doing if you know this is going to be combined by multiplication and this one's going to be combined by division we're just going to have this times this all over that just do me one favor if you're going to combine these things please don't write different logarithms it becomes one logarithm it never becomes Ln over Ln that's not the idea we're getting Ln of this argument times this argument divided by that argument all in one little piece so the 2/3 is still hanging on we got Ln well we got X to 5th we know plus is going to give us a product tell me one other thing you could do with that one half power what could you change it to if you wanted to all over you were like my little math robots there it was awesome square root there we go okay as if it's not ugly enough we can do one more thing what can we do now now that we look at that that's a coefficient right in fact I don't even really need these brackets if I didn't want them the 2/3 could go right there if that's a coefficient of a logarithm I can move that up and becomes a power now so this our very last step is Ln x to the 5th whole bunch of nasty crap and then the whole thing now with the bracket inside you do need the bracket inside to the 2/3 power okay that's about as inp as I want to get do you guys feel comfortable on expanding and combining logarithms probably some good practice right because you haven't seen it in a while could we have and that first step it would have been okay to just distribute the 23 you just would ended up with smaller exponents at the end absolutely uh you could do here and here and here and then move those up uh sorry uh here here and here and here and move those Powers up you'll notice that you get the same exact thing because well what's interesting is that that when you distribute you multiply don't you you multiply and you multiply and you multiply what do you do when you take a power to a power you how about that it's the same stuff so even if you take a power to a power you're still multiplying you get 10/3 same thing would happen here you get 10/3 okay so either way you go you'd be fine I like this way because this is all to one power that's just uh preferential though good question thank you for that one any other questions at all are you ready to move on to finally get dis calculus okay now that we've covered that we will use it trust me we'll use it a lot I want to refresh your memory on one more thing and then we're going to start talking about some integrals and get on to what in the world we're going to be doing with uh all this junk so the last thing we got to cover do you remember the graph of this function okay do you remember the graph of this function I see a lot of already let's talk about it when you take a positive number which this is uh e is approximately 2.7 you take a positive number to any power what are you gonna get out you're gonna get out a positive number that's right can you ever get out zero can you ever get out zero no so what automatically know that this function is not going to cross the x-axis it's always going to be above the xaxis does that make sense to you okay so uh how about this what if you were to plug in zero here how much would you get so we know it's going to across the y AIS at precisely 01 you said that it's always going to be above the x-axis because it's a positive number being raised to a power it has to be above here you're never going to cross this line does that make sense uh now when I plug in big numbers does this thing get really big or really small small big numbers Take 2 to 4th 2 to the e8th two this is about two this going to biger or smaller bigger bigger way bigger way faster what's going to happen over here because when you have e to the -2 which is what the these are that's the same thing as 1 over e s does that make sense when you have a negative exponent you put on the dener so as that power gets smaller and smaller more and more negative that denominator gets bigger and bigger and bigger and bigger one over a really big number tends to be close to zero is it never going to touch zero no but it's going to look like this that's the graph of e to the X tell me something somebody out there what's the relationship between this function and that function inverse very good they're inverses what inverses are graphically are a reflection across the yal X line so if we want to find out that graph it's really stemming from this graph and other words if I take this this diagonal right here and I just flip it right over that's going to give me Ln X so tell me if 01 is on my e to the X function if 01 is on my e to X function what has to be on my Ln X function good because inverses just flip the X and the Y did you know that about inverses that inverses really just flip the X and the Y it's a reflection across yal X so it's just flipping those coordinates really so if we have 0 1 and inverse would have 1 0 if this is astoic tox this is Asm prodct to Y which means our graph looks about like that that's the graph of Ln X you kind of need to know the graph of what we're working with because the properties are going to make a whole lot more sense to you uh for instance the very first property said hey when you plug in one you're going to get zero hello when you plug in one you get zero remember that Ln of one is always zero it's on the graph it's right there can you plug in zero to Ln do you have a calculator have you ever tried to plug in zero to Ln press Ln Zer if you have a calculator right now try it try do you have an error I got did you get if you tried it if you press ln0 hopefully if your calculator's not broken it's going to give you error what are you trying to kill me that's what it's saying you're breaking its heart really is what you're doing uh because you're supposed to know this about n can you plug in zero it's undefined it's undefined can you plug in a negative to Ln do you think it's undefined there's nothing there so when you realize what the graph looks like you kind of understand you understand a little bit better about why you can't plug in why you can't have Ln of three it doesn't make any sense you can't have Ln of Zer it doesn't make any sense Ln of one gives you zero it's on the graph so when you get a better picture of what the graph looks like Len makes a little bit more sense she H feel okay with the graphs of these things did I do it well enough for you understand where it came from fantastic now let's see how good you are how much you remember about one of our favorite things in the [Applause] world you ready R the bat man I'm going for it holy cow oh dear you didn't think it was going to be this fast did you don't wor you should know this stuff right you should know the integrals integral oh my gosh integral is the same thing as what we call an anti-derivative basically the not basically it is the inverse of a dtive just going backwards do you remember how to take the integral of something like that where we have x to the 3 DX which says in with respect to X changing add add something add one very good so the power rule backwards said well hey if you take a derivative and you subtract one you move forward and you subtract one here we're going to add one and then divide so if we did x to the 5th as a derivative and that said move this forward then subtract one we're going to do complet completely the opposite thing now we're going to add one and divide by that new exponent so this said multiply by the old exponent then subtract one this one says add one to the exponent then divide by that new exponent that's exactly how you take an integral and I'm missing one thing what am I missing why do we have a plus c what we're doing here is we're finding a family of curves that work you see that plus C when you take a derivative of any constant is going to go to zero so we have a whole family of Curves at work the only time you don't have a family of Curves with integral is if you have either a definite integral or you have an initial condition uh which we'll talk about later you okay with that one basic basic okay let's move on just a little bit what would happen if I had a constant inside of my integral always move out a constant provided it's being multiplied not added not subtracted but multiplied by whatever my function is so sure we can move that outside we can go four times the integral just like derivatives we could do this x 5th DX uh tell me left side people only just you guys tell me the integral of x to the 5th please cool okay however you want to write that's fine so we'd have four just don't forget the four x 6 / 6 + C and what we going to do with that simplify absolutely however you want to write it if you want to do the 2/3 x 6 + C or 2 x 6 over 3 I don't care whatever you want to do there fine still okay so okay stretch your brain just a little bit think back to this one oh dear now I want you to really think about what happens with derivatives before you answer this unless it's just burning in your memory you know it but please think before you answer this question integral of sin x DX integral cosine or negative cosine why isn't it positive cosine netive sign good the derivative of cosine is negative sign so if I'm trying to get a positive sign well the appropriate anti-derivative here is negative cine X watching too much scrubs Dr Cox bless bless you thank you yeah plus see exactly last one let's try that one up for size what are you going to do with that problem uh the qu quo rule works for derivatives but like where he heads that substitution oh I love substitution I love substitution if oh my gosh your teacher should have taught you this it wasn't me uh if the derivative of what you're trying to substitute is actually in the integral so find a piece of there take a derivative do you see that piece in the integral you do I hope not that gives you 4X to the 3 right I don't care about the four but do you see the X to the 3 I don't either plus is on the denominator so that would be a problem too think it easier SE top so what do you mean separated like x 4 x 2us that let's do that so this is kind of a nice trick anytime you have a single term denominator you can always separate it you always make different fractions out of that so we could do x 4x^2 - 1x^2 yeah absolutely that's fantastic tell me the next thing we're going to do someone over here on the left hand side what now simplify simplify one thing for me what do you get would be X2 on top m- One X x s but you can get X to the yeah perfect go ahead and do the the integral forming on your own see if you can get that remember that while you can split this up into integral minus integral you don't have to you can just take it piece by piece so take the integral of this piece minus the integral of this piece of course you can do negatives you can change the minus to a plus you can do it just piece by piece when you have multiple terms in there you don't have to separate it all uh by the way just a little bit of of notation for you when you do the integral and you get X 3r over 3 or 1/3 X3 you guys got that right I know this because I I saw a lot of you in the math lab some of you when you do integrals by the way what's this become one good you add one to it and you divide by whatever the new exponent is did you all get that far okay some of you like to do this some of you like to take integrals and go oh yeah I just took an integral what that symbol means is that you're supposed to take the integral whatever the argument is next if you've already done the integral this does not go here okay you've already done that so erase that you you want to stand about the notation here this is telling you what you're supposed to do this says take an integral evaluate it from 1 to two this says take an integral haven't you already done that if you want to show the evaluation the appropriate way to show it is to put it over here from one to two that's where you put that you go okay I've done the integral it's a definite integral now I'm going to evaluate from 1 to two of course we want to make it a little bit nicer looking before we do that we'll do X 3r over 3 plus one do you guys see where I'm getting the plus one overx from good cool and then we would evaluate it go ahead and evaluate it remember that when you evaluate we plug in this number first we figure out what that answer is and we subtract after we plug in that number so plug in the two figure out what you get subtract after you plug in one did you get it what' you get 116 116 double check 11 116 116 we're not going to spend a whole lot of time in here waiting for this I know you guys can do this I just want to make sure you see some finer points of doing integrals that way you don't get stuck on some of these easier ones okay don't don't make it harder than it is that's a lot with this classes about don't make it hard it is so fans be okay with our integrals okay I have one more for you give me ideas the top okay that looks pretty painless let's try that so every other time we had this we did uh let's see where were we add one divide by the new exponent are you listening add oneide by add one divide by the new exponent let's let's do that let's add one to this uh what do you get x to the well that's fine we' get one divid by that's not fine that's not fine at all that that's pretty nasty so what in the world are we going to do oh my goodness let's give up calculus 2 is over let's just go well this isn't going to work this is not good this is not good at all well let's think a little bit more about this um first things first do you know what the graph of 1x actually looks like think about it for a second do you remember what the gra of one/ X actually looks like approaches zero it does approach zero and it approaches Infinity it's like this little actually got the wrong sound effect my bad sort of thing what I'm going to do and you'll see Y in a minute I'm going to graph 1 over T you'll notice that 1 over T has a dummy variable in place of x uh you'll see Y in a second this graph if this is T and this is y looks about like that the graph looks like as I get closer to zero my number gets really really big positively as I get closer to Infinity my number gets really really small but still positive you with me okay true or false that function is continuous anything bigger than zero absolutely there's no holes there's no gaps um true or false it's differentiable anywhere past zero are there any sharp points no so this is continuous and differentiable well what that means by the fundamental theorem of calculus I'll write this out for you because 1 / T is [Applause] continuous on the interval that we're trying to talk about here by the I'm abbreviate by the fundamental theorem of calculus part one faki I made that up by faki by fundament sounds kind of cool though I think I think it'll catch on my YouTube people make sure it catches on the talking by the fundamental theor of calculus part one this thing oh sorry we must be able to find some differentiable function that gives us this thing for because this functions continuous on our domain we're looking at by the fundamental theorem of calculus we absolutely have to find some differentiable function we're going to call it capital F ofx big F ofx such this is true such that F ofx big f ofx equals the integral of our function from some defined point to X that's got to be the case if you don't believe me look up the fundamental theorem of calculus part one that's exactly what it states so this is just by definition I'm not making anything up here okay uh this says that this integral has to exist it's got to exist because it's continuous and fundamental theorem calculus part one which is proven says it exists it's got to be there also says this is just the the corollary that if I take a derivative of it that would be again since these things have to be equal it would be a derivative of this thing this you have seen before this is what you were taught in calculus one about the fundamental theorem of calculus it says when you take a derivative of an integral and you go from a defined point to x what this is is just 1 /x do you remember that yeah which is the function in terms of x these are the easy ones you had in the test where it's like oh I got a whole bunch of T's D I just put an X everywhere I see T do you remember that one it's basically what it said and now you know a little bit further what this is talking about is that if your function is continuous on this interal you have to be able to do this you have to be able to take the integral of that function uh and it it's defined it's a differentiable function and when you take of that function it gives you back your original function so uh basically it says there's got to be some function that gives me little F ofx when I take a derivative of it and the integral of little F ofx actually exists that's the basic idea here now what we do to make it just a little bit easier um we typically Define a to be one it really doesn't matter as long as it's bigger than zero and this thing works out pretty nicely so here's a couple things I want you I want to point out to you in the last five minutes you know what I'll write your too it turns out that the following things are true sorry my bad that should be a t there change your notes that needs to be a t I'm sorry you get the X when you evaluate from a to X I bet so first thing it turns out that lnx is actually equal to the integral of 1/ T from 1 to X that's a true statement it's by definition so that's a definition that we have also check this out this is where the fundamental theorem of calculus part one's going to come in watch very carefully okay did you know that you can do anything you want to to an equation as long as you do it to both sides yeah including take derivatives which is cool let's check this out if I do you believe me that this is true you have to darn you because it's a definition it's true if I take a der because it said so if I take a derivative of both sides check out what happens over here here if I take a derivative of Ln X I have to take a derivative of this nasty crap that's a little cramed sorry a little better derivative l x equals derivative do you see this piece anywhere else on the board that we' talked about what's it equal okay check it out by fundamental theorem of calculus part one we know that this is true by definition we know that this is true so by the transitive property if this we can take a derivative of both sides this equals this this part equals this oh you know what I'm going to do it I'll write over here so you have more room did I lose you do you guys have any questions on that I don't want to lose you on that one it's kind of a cool little proof um I hope it makes sense so here's what you need to know this is a really long way to say here's what it is so here's what it is is here's what you got to know the first thing yes we can now take integrals of 1 /x check it out if the derivative of Ln X = 1X that means the integral of 1x = Ln X the only thing that we have to be careful of is that we have to have an absolute value around the argument of Ln naturally we can't have an argument that's negative remember the picture remember the graph we can't have negatives in there uh so here's what we say we say this equals the Ln of the absolute value of x of course plus C that's number one thing so when you're taking the integral and you come across 1/x and you go holy crap if I add if I add one to the exponent I get zero over Z I get one over zero we can't do that well now you have something for that we've proven it over here when you take the integral of 1/x DX you get Ln absolute value of x plus C are you guys okay with that one did you get the the reason behind it if the derivative of l xal x almost the RSE is true the integral One X giv you Ln of absolute value of x if you have a definite enru the same thing is true we get Ln X from A to B if you're wondering well wait a minute Mr Leonard uh wait a second I thought you said that you had to have absolute value here check out what would have to happen for this to be integrable function you'd have to have something that doesn't go across zero do you see what I'm talking about you'd have to either have both positive both negative we're talking about both positive here so if you have both positive you're never going to have that negative argument inside the logarithm do you see what I'm talking about so we're fine as far as that goes the second part if we take a derivative of Ln absolute value of x we are going to get 1X say right here derivative of Ln X = 1X we have the absolute value so it looks a little bit more like this um really you're not going to see a whole lot of the absolute value unless you can have a function which goes negative like cosine or something okay you're going see that little bit so really this or this it doesn't really matter it's kind of all this is saying is that hey your arguments got to be positive so you make it positive should fans feel okay with with that one okay tell you what I know we haven't done a whole lot of examples we are going to get to that next time so today we're going to do a whole lot of examples with derivative of Ln last time what we learned was the expansion and the combination of natural logs we also learned a little bit of calculus behind it we learned that if we take the derivative of Ln X it's 1 /x we're going to do the integrals in just a bit but I want you to practice some of these derivatives right now before we get to it though uh there's one thing I want you to recall from your Calculus one class it was this it's if you take a derivative of some function with another function of X inside of it we called that do you remember what that was called chain rule yeah that's right this chain rule so if you remember the chain rule what the chain rule said to do was take the derivative of the outside function in this case F I call it the outside function it's just like the big function what's containing everything take the derivative of f leave the inside alone and then multiply by the derivative of the inside do you remember doing that so in in this case it would look like this it would be F Prime of G ofx youd leave that alone and then you'd multiply by the derivative of G ofx or you just say G Prime of x if you want I'm kind of mixing notation here uh one quick example if you want a really quick like little sub example here if you did like the derivative of x^2 + 1 to the thir power if I asked you for that derivative what we would do here I'm not going to ask you to to help me out on this one because you really you should know it I'm just going to go really quick through it this would be a chain rule uh some people call it the general power rule it says the big function is something a function to the third power that's the big function here do you see how it contains everything says you bring down the three like you normally would you leave the inside alone whatever's here goes here you subtract one from the power just like you normally would notice three subtract one becomes two but then you multiply by the derivative of whatever the inside function is our G of X in this case would be x^2 + 1 you guys recall that show up hands if you remember that okay and then of course you do the derivative you get 2x you pull it out front and you'd be done with our derivative I'm not going to finish that off for you but that's the idea we can do the same thing with lnx I want you to consider what's going on here if we have a derivative of Ln with some function inside of it in our case right now this F the function f is just Ln so the function Ln it takes the place of the function f right here does that make sense to you L natural log is a function so it's taking the place of f if we have any sort of function of X inside of that what we're going to do here we're going to do the chain rule the chain rule says all right well the first thing you need to consider is take the derivative of Ln what's the Der of Ln what's the Der of lnx say to do you put one over so you put one over whatever the function is in this case this is really easy because we had a function of X it was just well the Der of X is one really what you're doing here is the chain rule it says you're putting one over whatever's here so 1/x and if you think about it it's kind of weird but if you think about it if you multiply by the derivative of the inside here which is X what's a derivative of x y so you it's not hopefully it's not zero because the whole thing would go zero it is is one so we don't show that because a derivative of x is just one but that's really what you're doing here okay you're doing one over the inside times the derivative of the inside the same thing works here we're going to do one over the inside times the derivative of the inside what I want to be sure is that you guys see both Ln at work and that you see both the chain rule at work do you guys see both those things that work right now the 1 over G of X that's the same thing we're doing here this is 1 /x it just so happens we're calling the function inside G of X so one over the inside one over the inside times the derivative of the inside times the derivative of the inside that's the chain rule in respect with respect to Ln show hands you feel okay with that one okay we're going to practice that in a minute for those of you who had a different type of uh notation for chain Rule and give it to you this way as well it means exactly the same thing it's just some people like to see the U here where U is a function of X so if you have Ln of U where you is a function of X it looks just a little bit different I hope you understand it means the same thing you put one over U you can see it's the same exact thing times the derivative of U with respect to X DX same same idea just a little bit of different notation what do you say we practice some examples right now would you like that okay let's go for it uh I promise you these really aren't that hard if you keep in mind this is a chain rule what a lot of people do here is they'll they'll forget to do this part just like a lot of you in your Calculus one class when you first learned it some well maybe not a lot of you but some of you forgot to do this a lot you probably did especially with like s of 5x you know where 5x is inside I know when I taught calcul this one a lot of people forget that that's what's not cosine of 5x it's 5 cosine 5x because you have to take the derivative of the inside same thing happens here don't forget that when we're doing logarithms you still have to take the derivative of this piece it's kind of like two derivatives derivative of Ln is here derivative inside is here let's practice it [Music] [Music] okay so we got a derivative we got Ln you guys seeing what I'm talking about a function within within a function here we got a function of X inside the function Ln so the first thing that our chain rule says to do is all right take the derivative of the outside function in this this case the Ln function contains everything so if we take the derivative of Ln it says you're going to put one over I don't even care what's here you put one over whatever is on the inside so we're going to have 1 over 3x^2 - 1 looks a whole lot like this idea right now you're taking a derivative of the power but you're leaving the inside alone taking a derivative of Ln you're leaving the inside alone now what do we do leave it keep going keep going by multiplying by what sure I would like to see this work right now you guys get all advanced and you're in forc or whatever do whatever you want okay but right now I want to see you doing this that way if you make a mistake I can see where it's coming from so I want to see your work on these problems I want to see the one over this I want to see times a derivative some form you can use prime if you want I prefer the DDX because when I teach calculus I tell people hey follow the DDX wherever you have a DDX that's where you're doing derivatives that's where you're doing calculus all right so if you show that it's going to let you know what you do next so in our case we'd have derivative of Ln no problem one over the inside times the derivative of the inside this lets you know where you're doing the calculus next this you're good right there's no DDX so you're fine this you need to keep going take a derivative and then we'll put it together and be done what is the derivative of 3x^ 2us 1 what do we have there fantastic so we're going to have 1 over 3x^2 - 1 * 6 x where does the one go turns zero yeah sure the derivative of any constant always zero so we're good on that put it together make one fraction out of it I pray that you get the right answer here can I simplify anything can I start Crossing out sixes and threes it still happens sometimes I swear I still catch people this level yeah don't do that please don't do that unless you can factor out and simplify we're good uh do you all understand the first example yeah you okay with it not so bad right it's very much like chain wheel it's just you have a new function now it's not even a hard function either it's just you put one over the inside and and you're basically done let's see how much we can uh step this up okay that looks nasty enough I think we'll start there give me some ideas on what you could do with this uh true true or false let me ask you actually true or false can I move this cube root from the outside to the inside as a 1/3 power can I do that no no no no if it was 13 lnx absolutely but remember I told you about not making up your own mathematics don't make stuff up it's if it's not on one of those properties I gave you yesterday you can't do it so now I can't move move it here that's not what this is saying what this is saying is that this is supposed to be the derivative of lnx all to the 1/3 power do you guys see the difference there it's not here we can't we can't move it forward it's not an exponent of the inside of our logarithm of the argument of our logarithm it's that whole log X or lnx raised to the 1/3 power okay now you tell me what to do what do you think chain Ru chain rule I love the chain rule uh chain rule says to do you guys over here what what would you do here take the derivative outside take derivative outside do you guys see the similarity between the problem I just gave you and this problem right here or no I know there's an Ln over here but what we had here we had something to some power check it out if I cover this with my hand do I have something to some power and do the same thing what do you do with the 1/3 to the front sure I don't even care what's inside I know for a fact whatever's here if it's differentiable this 1/3 goes forward and something goes right here what goes right there 23 how do you get -2/3 what's the operation 13us on the inside what's going to go on the inside do we take the derivative of Ln now or do we take it later later okay so this we leave alone just like we left this alone we leave this alone but we're not done what's the next thing we're going to do perfect so we're going to show that right now DDX lnx so we understand well a cube root yeah it's still a 1/3 power but it's not a 1/3 power on the X3 power of the entire function here what that means that we we have a chain rule we have a general power rule we bring the one thir down no problem we taking a derivative we subtract one from it but then we don't forget that this stays the same on the inside of our function and we multiply by the derivative later of the inside of our function now this should be an easy one what's the derivative of Ln X fantastic so we got 1/3 Ln x^ -23 * 1 /x you know we can clean this up just a little bit of course this is a -2/3 which means that you can take this whole thing and put it where yeah we can do that we can change it to a cube root of this to the second power if you want so this you might be done with the calculus but clean it up just a little bit don't leave it so nasty we could have one 3x no problem we can have a cube root of lnx to the 2 that's how I'd like you guys to write it make it clean uh typically sometimes I let you do it with negative exponents most of the time I don't want negative exponents I want them nice and neat uh the only time that that's not the case if it's just a super huge uh derivative and it would just waste your time to do it so that's an appropriate answer show hands feel okay with that one fun right right lie to me right okay all right let's keep going so we know how to take derivatives of lnx we now know that the chain Rule still applies to that let's keep on working with it this is kind of good practice because it gets you back in the habit of doing derivatives and you're learning something new what do you think we're going to do first what do you think is this like a general power rule like we had last time is it a chain Rule still yes anytime you got a function instead of a function it's a chain rule pretty much anytime you do a derivative it's a chain rule sometimes it's easy because you just have X but it's always some sort of chain rule so we have Len of some function uh tell me people in the middle what's the first thing that I should write here up the L okay let's talk about it can we break up the Ln can we break up that natural laog oh remember that we can break up multiplication division and powers of the entire function inside the Ln can we break this one up can you break up addition so then we can't touch this can't touch this anyway we can't break this up because of that nasty plus there uh Ln of something plus something does not equal l n of something plus Ln of something it doesn't work if that had been multiplication of course we could but we can't do it here gcy okay so now we got to do the derivative like it is right now what would I do first one over the infin yeah that's fantastic this is a chain rule right the derivative of Ln says you put one over whatever this is I don't care what it is that's how the chain rule works for Ln so the first thing absolutely is one over just that nasty stuff do you guys understand that that is the derivative of Ln that's what that says Ln just says one over the inside that's all it means okay now to deal with the inside of it you use the chain rule so what are we going to do next the the inside perfect I'm going to change that right now so I have to do it again instead of the square root let's just make that the 1/2 power right now can you guys do that for me once youall do that take care of that derivative I'll write it on the board I want you guys to try it on your own I want to see if you got it did you get it did you guys notice that there's a double chain rule in here it's awkward right but that happens a lot in calculus we deal with one chain rule boom we got another one in there derivative of 2x is of course two we bring down the 1/2 power bring down the 1/2 power leave the inside alone subtract one from our power and then multiply by the derivative of the inside um this sort of thing you can do without showing me the extra step I want to see the extra I want to see this stuff for Ln because that's what we're learning but those other the ones that you should know already feel free to just do them quickly okay uh I don't expect that we have any issues with this I hope do we well I hope I hope not do you do you okay if you do you let me know I can cover this a little more thoroughly uh do you guys see what I'm talking about though the chain rule at work we have bring down leave it inside uh subtract one and multiply by the derivative inside after this it's kind of your job to clean this up I'm not going to do any more work on this particular problem what I suggest this this gets really messy really fast when you have multiple chain rules especially with these negative exponents this right here of course you can move this to the denominator I'll show you one more step but the rest of it honestly is just some algebra you could do 3x^2 over 2 < TK x cubus one do you see how that works yeah we can bring this down that's a negative power that's a square root X - 1 this 2 is on the denominator 3x2 goes numerator of course you have to show me the parentheses there that's important but in order to make this one fraction you'd have some work to do you'd have to get common denominator here and then put those two things together that's kind of up to you um I would say that this would be fine for for one of my answers in class uh in the back of the book like when you're checking your work or whatever it might not look like this most of the time they do the extra work uh they make the comma denominator they distribute they combine like terms they make one fraction out of it so if you want to get yourself really good at algebra try doing stuff like that you guys okay with this one so far so calculus is done the only thing you'd have to do now just a little bit of algebra y'all have any questions or can I move on move on WE you good yeah am I going too fast for you you okay all right tell you what we'll do about U three more of these just to get your heads wrapped around some of these other Concepts I want you to know uh we'll talk about a new technique that uh you probably haven't ever seen before we'll talk about implicit differentiation then we'll do log LS oops oh dear here we go first things first here let's look at our problem uh when you're doing derivatives or integrals it's really good to spend some time and diagnose what's going on sometimes these are going to be pretty easy but you still want to think about them what's the big idea on this derivative does Ln contain everything in this problem then it's not a good idea to start with Ln first it's a first good idea to start with say it again product yeah yeah a product contains all of this you guys see it we have X Cub anytime you have something in front of Ln works just like something in front of parentheses is being multiplied so there's a product right there do you you remember you remember remember the product rule yeah show me the product rule here go for it write it out product has a p in it plus has a p in it so product plus so product will you're going to have to add something together so I'm going to show my work that way you see where things coming from product work works like this you do the derivative of the first times the second plus the first times the deriva of the second some people do it backwards some people do the first times der of the second plus the second times der of the first or whatever I prefer this way just that's the way I grew up learning it derivative of the first times the second plus the first times the derivative of the second did you remember it if you didn't that's okay we're just starting here but relearn it right now because you're going to use it a lot all right again say what now say that again the derivative of the first time the second derivative of the first * the second plus the derivative of the second Plus time first sure yeah something like that something like that it's one of those things derivative of the first times the second plus the first times the derivative of the second derivative of the first one times the second one plus the first one leave it alone times a derivative of the second one so you take a derivative of one of them you leave one alone add take a derivative of the other one leave the other one alone oh the quotient is the quotient first happens when you have a right uh low D High minus high D low Square the bottom away we're done wait that's the quo rule that's the quo rule I remember it works for quotients product rule for products but product rule you're your derivative of the first times the second plus the the first times the Der of the second add them together that's the idea I don't really care the order in which you go U make sure you're taking the derivative of one of these factors multiply by the other one one add take a Dera of the other Factor multiply by the first one that's the idea now the derivatives themselves should be pretty straightforward uh you let me know X Cub what's the Dera of X Cub 3 sure so 3x2 you'll notice we leave Ln 5x alone we don't do anything with that then we have this add we have X cubed time how about the derivative of Ln 5x what do we have to use there chain rule so chain rule here says what perfect times sure I'm going to show it right now just so we get used to doing that times the derivative of 5x some people forget to do this if you forget to do this you can change your problem a lot you have a a one over five instead of 1x okay you you have something you have something different here so three x^2 Ln 5x no problem plus how much is that what happens to our fives that's what I mean about you have a 1 when you're not supposed to how about our X's simplify the things that you can so if we have x to the 3 here and we have an X down here we can simplify that sorry I know I I did a whole lot of simplification all at once uh but if you need this one more time the derivative of 5x is 5 simplify the fives we still have an X we have x to the 3r that makes x to the 2 show F be okay with that one what's one more thing you could do if you felt like doing some extra math here what could you do take out the X squ good you could Factor it that's right you could Factor this out and get 3 Ln 5x + 1 x^2 on the outside uh do I care not really if you get a different answer in the back of the book maybe that's what they do uh so just be aware of that can we move on do you guys have any questions at all are you getting your brains a little bit un cobwebbed with some of this stuff it's going to take some time it is uh but don't worry about it right now you will you'll get there okay derivative Ln absolute value cosine x one thing I want you to notice is that sometimes we have absolute value around our function and sometimes we don't um really when it comes down to it don't worry about it because when I gave you the definition the very first time you don't need to write this you already have in your notes when I gave you the definition the first time I really gave you the definition of Ln absolute value of x is 1 /x so we don't really need to worry all that much about those absolute values it's just there to make sure that you it's telling you hey don't worry you don't have any negatives inside of your natural log function because we can't have those of course cosine has positive and negative correct so when they give this to you they're saying we're just focusing on positive cosine that's what we're talking about all right so don't worry too much about that but let's worry more about doing this derivative first thing Ln is the overarching function it contains everything here so let's take the derivative of the Ln you got y'all should be pretty good at this by now we've done a few examples what's the first thing I'm going to write one over cosine X is absolutely true yeah for sure 1 / cosine X you'll notice there's no more absolute value the derivative of absolute value X is 1x so that goes away don't no need to worry about that one are we done no this is what again sure so we're going to do the derivative we should be pros of this derivative of cosine X come on come on oh I heard some people say the nasty one and some people say the right one I know it's it's it's sign somehow right is it positive or negative negative very good so this is going to be 1 over oops that's not right 1 cosin x * sinx let's put that together we got sinx over cosine X say it again perfect that's tangent this this is tangent so this isga tan X cool that was interesting wasn't it so sometimes these logarithm functions of sign functions gives us other sorry logarithmic functions of U trigonometric functions gives us other triat functions it's kind of weird but a lot of those relationships show up with Ln it's really interesting can we do one more is this good practice for you are you getting something out of it I hope all right good I don't want to waste your time or my time should I stop there no don't worry I haven't gone crazy I know right it's crazy now I think you'll agree that looks pretty bad to start with right does to me at least go my goodness gracious that's not fun well I want you to think about what we can do and why we did the whole yesterday's lesson on uh learning how to expand and combine logarithms in every other example that we had before we can't expand it can't we could have but it really wouldn't have done anything here because that's pretty easy we couldn't really expand it for anything reasonable definitely can't expand that can we expand this one the more that you do the easier that your derivative becomes so expand these as much as you can then do derivative of each little piece man these become way easier uh I'm I'm going to kind of cheat here I'm going to do the whole expansion in my head right now hopefully you can follow me along we're going to talk about it though so stick with me notice the LNS over the entire thing you need that what do you do with this division how do you separate that so we're going to subtract off another Ln of this piece are you with me yeah okay what are we going to do about this product right here so we're going to subtract or we're going to add another Ln on there so we're going to have Ln of this plus Ln of this minus Ln of this okie dokie maybe I'll do two steps for you notice one thing um when you're breaking up your Ln or your logarithms you're not doing derivatives you're not doing any calculus at all so you still need that DDX we're not doing anything right now except using the properties of logarithm so please maintain this until you actually take the derivative once you take the derivative of course get rid of it so we got Ln of X2 no problem plus + Ln 2x^2 + 1 all to the 3 powerus Ln of 5 - x^2 under root we'll make sure that's in the bracket because we want to take the d entire thing let's make sure I'm right am I right this plus this one because we had a product minus this one because we had a quotient okay uh should I stop here or should I keep going keep going the more you do the easier this becomes so let's keep on going what should I do square square fantastic what else going to square root up square root okay so that's a 1/2 power I'm going to move that to the front Cube Cube to the front let's do this so every one of these arguments has an exponent let's move to the front let's look and see what we did um you should know that when you take derives you can take derivative of each individual piece as you go is this derivative easy or hard easy that's a piece of cake this one easy pretty easy involves a little chain rule but not a big deal this Oney also easy we just made something that looks really nasty into three really easy pieces really easy pieces that's nice try that when you get to those heavy duty ones in a home you go oh my gosh what in the world see if you can split them up now don't split up things that you're not supposed to for instance can I split this up no it's got a plus this one no it's got a minus I can't touch those but we definitely want to be putting our expon for it I definitely want to be separating products and quotients still okay so far okay question you I don't think so when you bring down that okay remember that right now we still have this DDX out front it's telling us we're not doing derivatives yet we're not touching calculus all this was was logarithm properties so all this was was hey remember how we brought the three down we're not subtracting it because that's where exponents go remember how that's a 1/2 power we're bringing down the 1/2 power we're not subtracting CU we're not doing a derivative right now now is the time where we take a d question I think what you subraction for the to it and it was exponent so would it be an addition like at the bottom no it doesn't matter um like when we did something like this Ln X Y we do Ln xus Ln y this takes care of the negative exponent for us do you see why you could actually write this Ln x * y y1 correct you can separate this by Ln X Ln y1 that's an exponent that goes down in front lnx minus Ln y That's why that works okay that's actually a little proof for you didn't want to have to show that but there you go does that make sense okay now we you to do derivatives what I want you to do is practice this do the derivative of this piece this piece and this piece it's nothing fancy it's just little chain rolles so show that to me and I'm going to do it over here for you okay do yall get it already yeah felt good all right you I got space over oh let me think so derivative means that because I can split up derivatives by addition I can do that for for derivatives uh I get to do derivative this and then this and then that so I know I've got 2 * the derivative remember with with constants you don't need a product rule uh if you did the derivative of two would be zero so that would kind of eliminate half of your product rule anyway you don't need it for constants only functions of x times functions of X so no product rules here on any of this 2 * 1X no problem 3 * again that's a constant we have 1 2x^2 + one one over the inside times the derivative of the inside three stays there derivative of Ln of this nasty junk we put one over the inside time derivative of the inside minus again we leave the2 there derivative Ln 5 - x^2 well we do one over the inside time derivative of the inside let's see what we got I know I can put that together that's 2x I know that this what's the Der of that one so I got 4X * 3 that's going to give me 12x over this stuff none of it simplifies so that's going to be part of my answer you guys two for two did you get both of them Dam how much is this derivative what's that give you so this is let's write that this would be 4X that's how we got the 12x this you said this is what okay so if we put this together hopefully I'm not going too fast for you -2 pos2 minus what's the sign going to be for this fraction posi yeah sure minus and negative gives us a plus 2 over 2 those simplify out so we're not going to have a a two in here we're going to have one over this that's it X say what now oh yeah X X's aren't really that important are they I mean yeah okay you're right I forgot so X over 5 - x^2 did I miss anything else let's let's double check 4X * 3 that gives us our 12x this stuff doesn't change twos are gone minus with the negative gets to plus X over 5 - x^2 we got that that's as good as I want how many people got that one that's good if you did that you know what you're doing okay um if you didn't if you struggled a little bit this is one of those times when before you start your homework you should go back through your notes do the problems that we've done in class like don't just follow them actually do them like different sheet of paper put your notes aside see if you can do them do them until you can get it right and then start your homework I promise it'll be a whole lot easier if you can do that okay if you jump right into your homework you go I'm a little overwhelmed because you never really learned the material you need to learn the stuff before you try the homework okay you with me the homework's there to kind of beat it into your head so you get the repetitive nature of this okay are you ready for something a little bit new this thing this is cool I promise you're going to like this this is this is nice you you really will I do at least I think you like you're going to like it um what this is called this next little piece we're going to talk about is called logarithmic differentiation so when they ask you for that and you go what in the world is logarithmic differentiation that sounds horrible it's really not horrible it's actually not fun uh this is what they're talking about and it stems from this idea let's suppose that we have some sort of a a function like this one question is this still 6.1 sure is I'll let you know when we change we got a function like this one that well we could probably take the derivative of that couldn't we just like it is I think I could you guys had it earlier what you could do what what what could you do here quo rule would have to be done then then you'd have a chain Rule and you'd have a chain rule but you can do it uh it's not going to look very pretty but you could you could definitely do it but there's another option if you have some of these derivatives you go man that's it's kind of nasty and they get way this is kind of an easier example to do this with but then get really nasty okay especially if I had like another function of X up here you'd have a quotient rule then a then a product rule inside the quotient rule then some chain rules it' get nasty really really fast sometimes we can get around that and here's how you do it the way that you do is called logarithm differentiation what you need to know is that you can do anything you want as long as you do it to both sides of an equation practically anything you want well one of those things that we could do looks a little weird but one of those things we can do we can actually take an Ln of both sides of our equation can someone out there explain to me why this would actually help us here aha if I taking Lon on both sides this doesn't help us very much but this does because I can separate it I can expand it all out are you with me yeah okay so what we're going to do before you start showing me derivative before you start showing DDX or doing anything like that you're going to expand this thing so no DDX is right now we expand it first so what that means is Ln y equals well let's see what we can do we got a quotient here what's a quotient change to everybody real quick subtraction very good do we have any products here no but we have some exponents we have a four and we have a cool so I'm going to do all that right now so we got Ln 4 * Ln of x -1 -3 Ln 2x -1 can you guys double check me make sure I did that right okay cool is that as expanded as we can get yes can we break this up please please goodness say say no can you break that up how about this one no we can't we can't separate subtraction so this four came forward no problem this we have subtraction 1/3 came forward no problem now we're ready to take a derivative do you remember doing implicit differentiation y Prime yeah we're going to have y primes everywhere you touch a function of Y you get that y Prime I like the dydx because that's how I grew up with the liven this notation but here's the idea check this out with me we're going to do a derivative of both sides so here and here you'll notice that our derivative is with respect to X which means that we treat y as a function of X what that means is that we do our derivative just like normal so the derivative of Ln Y what would the Der of Ln y be very good but check this out because this is a function of X we have to show the chain rule it is a chain rule here so what this looks like is 1 over y for sure but check this out why you get the Y Prime why and I don't really really like the Y Prime because notation wise it kind of fails you here but notice you would do a derivative of the inside does that make sense you'd have hey one over y this is the the Der of Ln times the derivative of y look at this this is dy DX that's what that is that gives you your derivative so I'm going to go down here and give you Dy DX that's this side that that's all you got to do so we have how do you take a Dera l and y no big deal it's one over y just like always but then you do a chain rule derivative of y hey that's it that's your derivative this is what we're going to solve for in just a second now let's do the derivative of this side this should be pretty easy I want you guys to do it go for it we'll do one more and we'll call it good uh for today's lesson oh goodness uh oh did you were you able to do this derivative yeah should get 4 -1 did you get 4 -1 we do the chain rule but the derivative of x -1 is simply one so this is 4X -1 minus well we got a 1/3 here we've got a 1 over 2x - 1 because we're taking a derivative Ln but now we do have a chain rule don't forget about the chain rule my goodness derivative of the inside gives you what how many people made it that far that's cool that's great it's fantastic let's clean this up a little bit a couple things need to happen firstly we have a 4X -1 we've got a 2 over three don't forget parenthesis here we get a 2x- one can you distribute that just make it you can you can distribute it U generally people don't typically distribute denominators because if you have to create a common uh denominator it's easier to do it when it's Factor okay CU here you you distribute but then to find a common denom you have to Factory so typically people don't do that it's not a big deal if you do or not okay now the last thing if you'll notice this we haven't solve for our derivative yet we have 1 y * dydx what you need to do when you're doing this process solve it for dydx this stands for your derivative so if you get this by itself you're done tell me how can I get this piece by itself fantastic so if I multiply by y and multiply this whole thing by y well these are gone that's cool we get dydx equals I'm going to leave all this stuff the same 4X -1 2 over 3 2x - 1 time y okay I want to show of hands real quick if you're okay with what we've done so far in the problem good that's fantastic now question for you is it appropriate if I can change the Y should I leave the y the way it is or would you rather see function of X right now this is implicit if I can solve for an explicit derivative that's a little bit bit better look back at our problem do you know how much Y is equal to yes do you you should how much is y that that bunch of crap hey look at that we got it so all you got to do I know it's fun told you it was fun day sure we took L on both sides that's so that we can split all this junk up okay we split all the junk up we took a derivative after that notice here's our DDX this is where you do calculus calculus this side is implicit calculus this side is pretty easy put it together as much as you can solve for your derivative whatever that is if you used y Prime no problem solve for y Prime solve for dydx here by multi applying y on both sides now just make a substitution this should be explicitly defined somewhere in fact it's right here so this piece goes here if we do that then we're done um I really don't care how nice you make this you're going to find out that your derivatives can look quite different uh from the back of the book um right now I'm not trying to get you to be super excellent at algebra I'm putting this stuff together you really should know that so you should know how to find common denominator and put that together okay I'm trying to get you to be good at the calculus part of it um and that's that's really our our goal here so I'm going to leave it with these two fractions you need to know that you can manipulate it a little bit so we'd have 4x - 1 - 2 over 3 2x - 1 big old bracket and then we multiply by however much the Y is equal to since the Y is this stuff that's what we put on the back end my test reasons will you mark us down for not doing the Y what it equals two or like but y absolutely I do want the Y Prime is is perfectly fine but I do need you to notice that this is a function of X this should be substituted back for I don't want you to leave it why I want you because this is what logarithmic differentiation is it's kind of going around the derivative but still getting a proper derivative in terms of X okie dokie are you guys okay with this idea question so if we leave it like that and don't do the algebra we're fine this is fine for me I need to tell you that in the back of the book uh typically what they'll do is they will find a common genom they'll put this together it's going to look a little bit different because they want one fraction if possible do you see what I'm talking about if they get one common denominator put this together then you could make one fraction out of it I'm not going to spend the time to do that because we could spend all day doing things like that so F be okay with it okay now the last thing we're going to talk about we got about two minutes here I want to show you something like this I don't know how much we're going to get through um I'll probably just go pretty pretty quick at it but I want you to think through this uh firstly can you think about this as an implicit problem do you guys see that we have x's and y's all tied together and we can't really get them separated okay there is one thing that we can do to make a problem easier what is it you have it it's in your head I can see it the Ln separate yeah the Ln absolutely let's separate that thing that makes our problem easier do you guys see it so we would do Ln 2x - Ln y - sin y + x^2 = 0 that's as much that's as easy as we can get it okay we can't do anything really substantial to make this problem any easier we could separate these but it wouldn't make a difference because that's a constant who cares derivative that thing going be pretty py so we would take a derivative of both sides now would it be easier to separate the x's and y's on their respective sides you can it doesn't really matter you're going to do it at one point or another okay so here I'm going to be kind of quick about this I I really don't want to keep you over so uh derivative of Ln 2x can you tell me what that is 2 2x sure it's 2 2x it's 1 2x * 2 because we have the chain rule here we have -1 over y * y Prime if you like the prime notation that's y Prime minus derivative of sign everybody der of sign real quick so we would have cosine y but check this out please please because you took a der of Y you're going to have a y prime plus this one der of x^2 is perfect D of zero cool 2x cosine y y Prime 1 y y Prime 2's are gone we get 1/x all you have to do now get all your x's on one side Factor your y Prime just like implicit differentiation does and you're done so let's see if we do that we're going to have a 1 y * y Prime cosine y * y Prime on the right hand side we have a -2X - 1X how' you get um what I do you put 1X Plus what are you talking about just checking as soon as you have your y Prime terms on one side and everything else on the other side this is when you want to factor out your y Prime and then just a so break I guess the long story short break down the LNS as much as you can break them down then start taking derivatives whether you need logarithmic differentiation okay whether you need implicit differentiation okay if it's implicit you're not going to get many of these but if it's implicit make sure you take and do a y prime or dydx everywhere you take a derivative of y get all those terms to one side everything else on the other side Factor y Prime divide typically and you're done um this this answer by the way it can look different you need to know something about this you can multiply and get a common denominator you multiply get a common denominator do a complex fraction on that and it's going to look a little different what I want to know is is everyone okay getting down to at least this far so you are yes is it making sense to you yes okay so we don't all right so now it's time to keep going with our our natural logarithms we've learned all about the derivatives we did a lot of examples from that now we're going to start talking about integrals so if you remember the first thing I actually proved other than just the properties of logarithms was the integral and derivative of that so what we should have down I think you guys already have this but we'll talk a lot about it today our goal was to find out what in the world happened when I take an integral of something like 1 / U we tried that right right we did it at first and like man this didn't work like the normal way we we were to do this we can't make this U to the1 and then add one to it because we get U to the 0o we get one over Z that's a big problem so we said well well how else how are we going to do that so we did the uh the whole Ln thing we said well this really this really is Ln of the absolute value of U plus C do you remember the absolute value that we had okay so that's the integral that we're we're talking about integral of one over U gives you Ln absolute value of U plus C that's what we're working with today what I want to do is just practice a few very basic examples with a little bit of substitution for right now and we'll gradually gradually work our way up to some trig functions um and see how that works for us you guys ready yeah so first thing let's consider in integral of 1 5x - 2 DX now if you remember back to your integrals your idea was and we haven't really learned any integration techniques yet you're still in the same boat the same the boat being you either have to have something that's directly on your table remember the integration table you guys have I know some of you had the little sheet out right now uh either if it's something directly on your table like X 3r or just secant squar or something like like that or you have to substitute to make it fit your table well here's one more thing that we have added to our table so if we can make it look like this then we can do the integral does this look exactly like this no it doesn't so if it doesn't look like that we've got to make it look like that what's a good way to make it look like that subti substitution is fantastic remember how substitutions work substitutions work when you look at a part that you don't like in your problem like this I don't like that part and the derivative of this part has to be somewhere in your function disregarding the constants so what's the derivative of this five but there's no X's right so that's cool we can pull fives out in front that's the idea we're going to make a substitution to make this guy look like this guy so our substitution here has to be U = 5x - 2 let's see well oh you know what I want you to practice that too what do we do when we do a substitution what's the next step yeah okay cuz we got to we have to not only substitute for this but we've got to change this as well this has to match up with whatever variable you are substituting for so yeah we need a du equals mhm 5 DX um I learned it this way you guys can do your substitution however you want I learned to solve for DX so you'll always see me doing du over 5al DX some people have done it other way it's fine I don't care what way you do it okay uh so we'll do a substitution we still have a one uh what does the 5x - 2 become good and instead of DX I put perfect what do you typically do with that five move it out to the yeah that way it doesn't affect any of your makes it more difficult or anything makes it a little easier so 1/ integral of 1 U du we want to make sure this matches this your variables have to match did we make it look good now does it fit part of our table y that's fantastic that's the whole idea here so uh don't trick yourself and thinking these things are ridiculously hard right now pretty much all you have all you know is it either fits your table or you substitute to make it fit your table that's about it so we did a little substitution I know the 1 stays there that doesn't affect anything everybody integral of one to U du is Ln good absolute value is important plus C sure you put that now or you can put at the very end it really doesn't matter there is one thing that is important though what variable did we start with an x what variable do we have now that's a problem so what are you going to do put X back in yeah so if we have this substitution U = 5x - 2 that U becomes 5x - 2 so our integral is 1 Ln 5x - 2 in absolute value plus C okay that's our first example it's a really simple one but I want to make sure that you guys are okay with it show fans if you are you feel okay with that one perfect fantastic let's step it up just a little bit I want you to know that we're not always substituting just to make an Ln sometimes we're going to have a sorry I one over you to get a integral that equals Ln sometimes you'll have Ln in our integral so let's practice one of those to all right looks a little scary but let's look at it the first thing you should do whenever you do an integral is to see if it fits your table is this anywhere in an integration table that you know of no it looks pretty nasty actually there's a square root there's an lnx it's got a fraction that's all the bad stuff at once right that's no good but then we think if it doesn't fit the integration table we should look to make uh sub that's right let's make a substitution here the appropriate substitution is one that's typically inside another function it's usually like a denominator or it's inside parenthesis or it's inside a root it's usually inside something and the derivative has to be there disregarding constants so let's look at the appropriate substitution is the appropriate substitution 3x no no the derivative of that is three I don't that's not going to help us with our Ln is the appropriate substitution the square root of lnx with the square root yeah it's typically not the whole function guys it's typically the inside of a function you you take the ugly part out now here's what I mean by that look if you had a u in here if you had a u could you take the integral of U to the 1/2 power yeah so don't worry about that square root also the derivative of lnx I'll think back to yesterday the Der of Ln X is here's a one do you see it here's a 1x that's something that we're looking at so you want the derivative to be in there and typically they use the inside of some other function so let's try it = lnx and you already told me that when we take our derivative we're going to get du = 1 /x DX again I always solve for DX you don't have to but you can If I multiply both sides by X I get X dual DX do you guys see what I'm talking about about your derivative has to be there because right now when we subtitute this in for DX our X's are going to go by by and what you have to have is all one type of variable and that's got to match up your du or DX or whatever you have you guys ready to do the rest so let's see uh you tell me what I'm going to have do I still have a square root yes sare root of you inside that square root is M denominator I got a 3X still I haven't changed changed that yet but I have an X du yeah that's right the x is due so X's are gone that's fantastic we had to have that happen if that doesn't happen you've made the wrong substitution okay you have to get rid of all the other variable or be able to substitute for it in a different way you have to have these things match up and have no other extraneous variables going on tell me one other thing two other thing that things that we're going to do on this problem 1/3 goes out what else perfect square root to 12 mhm so 1/3 no problem we've already got rid of our X's that was a big deal We've Got U to the 12 power du be careful when you're doing your integrals remember that we're always adding one we're not subtracting that's a derivative so when we do our integral here we got 1/3 no problem we've got U to [Music] the 2 23 Goodtimes 2/3 where we getting the 2/3 from I'm going to show it here this would be over 3es but then what you know is you got a complex fraction the two just moves to the numerator so really we've got 1/3 * 23 U to the 3es nothing simplifies here we've got hopefully I did it right 2 n u to 3 plus C yes and we're done right you got to put the we're not done we're close to done we're almost done the last thing we got to do is make sure that if we've done a substitution you got to substitute back for it so since we did all the way back over up here we're going to put this back for this that we'll be done by the way is my math all right yes okay good so we've got 2 9ths yeah lnx we'll just move that three halves up like that around the whole thing uh one one quick question can can I move this three Hales to the front yes or no yes or no what do you think can okay so question is this the same as this no no okay this one you can this one you can't this is the whole function raised to the power does that make sense to you yeah okay so be careful on that again don't invent your own mathematics please plus C that's done show be okay with that one boom fantastic that's really I'm only going to give you two of these ones they're pretty basic you're typically going to use a substitution you're either going to have something like one over U uh to substitute or sorry after substitution or you can substitute lnx see if you can work around with that of course I can't give you every type of example you're going to have but these two cover a lot of bases here so look for substitutions try to make it fit something that's in your integration table now the next one next one the now now gets fun this is the good stuff I know you've been waiting for it I know you've been waiting for the trigonometry right right am I right yeah I know well so far here's what here's a couple of them that you know we know this one and this one and this one and we know I going just do a few just these ones for right now you know this one let's practice see what you guys know um let's start with this one what's that how about that one good oh you guys are good at this some of you in the math lab took my advice and actually learned these when I told you to that's fantastic how about good D okay thanks one of us there's a couple other ones uh what's another give me another one I think you got cosecant cotangent yeah that's the one and then squ what's this one that's right and then the last one should be uh cosecant squ right yeah let's see if secant s = Tan x cose s should equal Co something Coan negative negative that's right that right now is your integration table for trigonometric functions I believe that's about all that's on there right we're going to create some more right now so this is what you're kind of expected to know uh coming in this class everyone from your verbal responses knew these ones yay uh but these ones are a little bit harder to memorize because we don't use them as often we're going to use these okay we're going to use them so get used to them uh write this down have that in your your memory banks you okay okay now what we don't see up here is stuff like this one we don't see the integral of tan X if you've wondered why in the world don't we ever see the integral of tan X why couldn't we deal with that well we're going to find out right now why you couldn't deal with that in calculus one so uh give me some ideas on how we might go go about now that we know some of the stuff that we do know how we we might go about solving this particular intergal what would you want to do first good we use an identity uh you'll find out that with trig you use a lot of identities I wasn't lying to some of you who came to me earlier and said what do I need to know learn your identities learn your identities because we're going to use a lot of them so yeah let's bring it up tangent is sinx over cosine X go through the steps I'm talking about here now we're going to look at this does this fit doeses this fit over here exactly no does it everybody does it fit over here exactly no no it doesn't look exactly the same if it doesn't look exactly the same then you're going to use substition yeah you're going to make a substitution to make it look the same okay well let's do a substitution here now it's true that the derivative of sign is here but if I solved for it it wouldn't cross it it wouldn't simplify that out of the problem so you being sign's probably not a good choice We want to pick the worst part of this and call it u in our case right here we're going to pick U equals the coine cosine X do you guys see what I'm talking about about the worst part of this being uh being cosine is on the denominator also it's a great part for us because if I do the derivative the derivative of cosine is and besides the negative the sign parts right there that's something we're going to be able to simplify out so that's a good choice so dual s x DX you know me I do du over sinx = DX let's see what this does uh do we still have a sign well it depends on how you do the integral for me yeah I still do because I I like to do this do we still have a cosine what's our cosine do we still have a DX no du for me or sorry DX for me is Du over negative sinx our signs are gone tell me a couple things that we're going to do that so this is negative integral 1 / U man that made it a lot nicer can we do the integral no you can we can now because we learned about what the um we learned about that thing what's that thing we learn about Ln that's right we learned about natural law so notice we do have the negative Ln of you're forgetting something absolute value absolute value is absolutely important okay that's that's cool we're almost done let's go one more step I think someone else said that already you can't just leave it in terms of you what are we going to put in place of our U so this equals I'm going to go right over here so it looks better this equals negative Ln absolute value cosine X plus C show be okay with that one isn't that interesting did I I think I told you I think I prefaced this last time and said it's weird but a lot of trig functions get related to Ln somehow don't you it's interesting isn't it that when you take the integral of tangent you somehow get Ln of absolute value negative Ln absolute value of cosine you guys okay with that one it's pretty straightforward right now there's one more thing that we're going to do here what does this count as what does that mean an exponent that is an exponent so if this is a negative one then really what we can do we can move that as an exponent so this is Ln absolute value cosine X to the 1 tell me something how much is that equal to you should know it one over cosine very good this is one over cosine one / cosine is also known asant that's right so we get another identity for us what we know now is that the integral of tan X DX is equal to Ln absolute value see X plus C some people out there might be wondering wait a second um isn't this a problem because is if I plug in if I plug in Pi / 2 that's zero are you with me you should know your trig right if I plug in pi over two you get zero so wouldn't this be one over Z isn't that undefined and answer is yeah would be until you consider where this is coming from tangent is not even defined at pi/ 2 it's continuous and it's defined between not including pi2 and pi2 remember those S curves that you get so we don't really have to worry about the pi/2 because you would even be able to integrate It Anyway does that make sense to you okay you might not have thought of that but there it is so we got another one so we got these ones now we got integral of tangent equals Ln absolute value see X plus C here's the the deal in this class once you prove something you don't have to keep redoing it so next time you see the integral of tangent or you see something that has a tangent in there don't reinvent the wheel don't you don't have to break it up into sign over cosine we just did it anything that we prove in class like this one you can use so now use that keep that in mind next time you do your homework you go oh okay how about the uh you know we're not going to do this one but how about the integral of x tan x^2 DX well you're thinking uh substitution for the X now that's in your table you're good to go does that make sense makes a whole lot easier than having to go through this all the way over again and again you guys want to see one more did you like that one it's kind of interesting right it's kind of fun uh tell you what we'll do run out of room here I think we can do this this one's tricky it's it's really tricky it It's tricky because we use a trick here that I I wouldn't expect you to know I didn't know it until I saw it so if you're like well where's that coming from so was I the first time okay what we're going to do here is we're going to multiply by one in a really really unique way we're going to change the look of this thing and we're going to change the look so that we can use a substitution that actually makes sense because right now doing the one over coine thing we're going stuck in a right because if you do one over Co are you familiar with the one over cosine here that I'm talking about you do one over cosine you do a substitution for cosine to get the one over you it doesn't the sign doesn't eliminate so you'd be stuck in a right back and forth and back and forth It's like Catch 22 okay so we're going to do a nice little trick here instead of just having secant x what I'm going to do is I'm going to keep the secant x and I'm going to multiply by see X Plus tangent X on both the numerator and denominator if you're like Leonard what in the world are you doing it's a trick that we're going to use to create a substitution all right I know it looks weird it's like where how would you even come with that I didn't come with it myself honestly I read it out of the book I'm like oh man these tricky guys you son of a gun you did it well here's the deal and you're going to see it right away as soon as we distribute this numerator notice numerator gets distributed denominator does not so we have here see 2 X Plus SEC tangent uh sorry seing X tangent X you guys okay with the distribution yes no se Square see tangent here's why we did that I want you to look at what's the worst part of this what's making this harder than it needs to be the denominator is the denominator is the bad part here if we do a substitution for this what's the derivative of secant squ oh yeah what's derivative of tangent oh sorry I had these backwards but what's the derivative of secant that one what's the derivative of tangent so if we do it together we're going to get that whole piece out of it does that make sense we're inventing a substitution here it's a little awkward cuz yeah it's hard to see it it's wow how in the world but someone figured it out and we're just going to use it so if we do U equals the denominator SEC X Plus tangent X and we do a derivative again let's try this derivative of see X everybody what is that very good plus so we did this this is that part plus that's right C squ I'm going to do the substitution the different way some of you are familiar with this uh you look here you go okay DX I try to look for this stuff too do you guys see that this is this whole part right here yeah that's our substitution that's awesome I love this game so we do a substitution we have tell me what the integral is now what's on the numerator one yeah one one's or du sure over what what did we call this U and then this whole thing became that du so you either have du over U or you have one over U du either way is fine can we do the integral now can now it's a piece of cake Ln absolute value of U plus C no problem the only that we're going to do is what put the U Back yeah put it right back in there so what we know now is that the integral of secant x DX got it from there is Ln absolute value remember this was our U see X Plus tangent X plus C that's the reason why you didn't get a whole bunch of these things in your Calculus one class is because you can't do them you can't do them until you learn about what the integral of one over you is once you do that the whole world opens up for you as far as trigonometric integrals uh did I explain this well enough for you show hands if I did you're okay with that one what I'm going to do now I'm not going to prove all of them I'm just going to complete this list for you that way you have the entire thing okay so we just did tangent we just did secant there's a couple of those I want to give you so as add on to whatever list you have the following things first one we already did we know that the integral of tangent U du is Ln absolute value see U plus C is that's not the same thing as saying x uh you know it is the same the reason why they have it as you is to let you know that you can make a substitution change it into that so you can put it X or a u um actually all of these should probably be used I think it's that way in the integration table in your book or whatever you're using um my bad I just did them on top of my head so I used X because I'm used to it um use X or youu interchangeably the reason again why they're using U here is to let you know that a substitution is possible in them so don't forget about that other words one we didn't do we didn't do coent U but I hope that you understand it would be really really similar to tangent all you'd have is cosine over s you do practically the same thing and you're going to get this this one we did do this one we did do did right over there so this would be Ln absolute value see U plus tangent U plus C and last one again the uh gu see what I'm talking about about the cotangent would look really the proof would look really really similar to this one well the cosecant proof would look really really similar to this it's it's a it's similar to that one so we'd have Ln absolute value cose U minus Cent U and there's some similarities here uh instead of secant we get cosecant instead of tangent we get Cent the plus we get minus it's kind of like the reverse of all of my suppos if you wanted to call it that show fans feel okay with with these ones I'm not going to prove them uh the proofs are really similar so you can either do them on your own look them up I think some of them are exercises for you what I do want to do is two examples to illustrate how you would go about doing these would you like to see some yeah okay so you guys like tattoos right just get an arm tattoo it'd be the coolest thing like ever no can you rip your arm off then yes do it I won't give you extra credit but you'll be awesome just tell them it's some sort of like I don't know Japanese writing most people w't know the difference some says honor and strength I did have a student once who came to school I got to tell the story came to school like he wore the same shirt like I think three times in the class and it was Japanese writing he was not Japanese so hey what's your shirt me he goes oh I really don't know and so I had two Japanese students I'm like do you know what this he left class uh to use restroom something I said do you know what that shirt means like yeah I said what's it mean I don't want to tell you like no come on tell me it means I want a girlfriend so funny he had no idea classic that that was a funny day I never told him I never told him what I meant any I don't want to make fun of them that would be mean f of okay so let's start with our problem does it look exactly like something in our integration table uh yeah no it's got X's it's got an X squ there it doesn't look like that so if it doesn't look exactly the same you're going to do what sub let's do a substitution uh what's the appropriate thing to substitute for here good yeah it's usually the inside of something the derivative is always in the integral you don't want to pick the whole thing because then you're going to get a trig function that's not in there again so that would be a problem so typically try the small stuff first like X squ in this case so if we substitute for x^2 of course du = 2x DX du over 2x = DX well then we have an X still here depending on how you substitute see U du over 2x is it going to work out for us yeah X's are gone it's fantastic tell me what else we do 12 in Fr yeah 12 see U du you look what our substitution did what it did was it matched this up with something that we just learned how to do you don't have to reinvent the wheel all the time that be ridiculous so as soon as you do your substitution now you start looking at your your table of integration you go oh well how much is that do you know how much that is you know what the integral of that is think I'm covering it up it's right here so we still have the 1/2 we've got Ln we've got absolute value sure see U plus tangent U plus C the reason why they use the U here is because you are probably going to use the use here that's a lot of use do use understand me joking uh anyway the last thing that you want to do though please please don't forget to substitute back in a lot of people when are doing this they either do us they go I'm done that was so hard oh finally done or they just put X's right off the bat you don't want to do that either okay so what goes here here and here yeah so our answer is going to be 12 Ln absolute value X2 + tangent X2 + C you guys are right with that one can you put that one2 of an exponent as well you can not such a big deal I don't really care about that it is it is an exponent it would be the square root over all this stuff all right uh not super necessary for you to do that good question though any other question questions before we go on you ready for one more does anybody else need this okay cool okay this one's going to bend your brain a little bit you going have to think about this one that's another wait a minute uh that hey that definitely doesn't look anything like anything in our integration table right away right so we're going to have to do a substitution let's check out the appropriate substitution should it be 2x no if I do that it's not going to get rid of any of this stuff should it be the numerator at all no because if I do this it's not going to touch this down here so probably it's going to be something down here now let's talk about the denominator if I pick the entire denominator let's try the derivative of that what's the derivative of one well that's fantastic so when we have that get rid of that one for sure can you do the derivative of s squared oh come on what rule is that rule it's a chain rule that's right do you remember that sin^2 X means sin x all of it squared we'll bring down the two we we leave the inside alone we'll multiply by the derivative of the inside are you with me yeah we'll bring down the two we'll leave the inside alone we'll subtract one from it and we'll multiply by the derivative of the inside I want to know if you're okay with 2 sinx cosine X do you not so much derivative of sin x^ 2 that's what this means you have to know that part okay if we do the Dera of this this is 2 sin x to the first Power time the derivative of sin x what's the D of sin x two this is meaningless X that's our appropriate derivative yes no I know I went fast on that but that's old stuff are we okay with it yeah okay now wait a second does that even help us no explain why well because s of 2x the trick identity for that is 2 X I love those identities don't you love those identities this identity right here this is two SX cosine X that's the identity it's double angle you got to know those identities if not you got to got to refresh your memory like now okay go home and start you guys like tattoos right just joking you have a lot of tattoos let's see what happens here okay I forgot the DX sorry see what happens um let's do the easy for what's this become mhm if this is you know what I'm going to do this in two steps I'm sorry I'll do it in two steps the first step that I've done here it's just the identity I've just changed this into this um if you don't know that identity by heart you can look it up later it is a true identity okay so this is true at this point I would do my substitution now check out what happens I have a 2 sinx cosin x DX that's exactly this piece so this is going to change into my du so I get du over U one over U du or du over U it's the same thing okay I don't want to have I don't want to go too fast on that one I need to make sure that you guys are with me uh do you are you comfortable with this identity now that I told you that it's true are you okay with picking this as your U if you are then the derivative is 2 sin x cosine X that's literally exactly what we have here so this piece becomes du this piece becomes U and that's a really really easy integral how much is that integral abute yep we got one more thing to do we're going to put the U so Final Answer Ln absolute value 1 + s^ 2 X plus C it's actually technically not the final answer we're going to do one more thing with this one more thing I want to make sure you're okay with this so far though are you the last thing we're going to do is we're going to drop the absolute value do you know why because it can never be negative that's right it can't ever be negative this is always positive if you add one to something that's always positive it's always positive so the absolute values are kind of redundant you can do that if you know your argument's always positive so we could write this instead as just Ln this stuff plus C and that's probably how you'll see it we good that's it that's 6.1 we're done so we just started off our Ln uh we know how to take derivatives and integrals do I have any questions whatsoever how many people feel okay with with the idea you guys have questions now so now you know how to do derivatives with Ln yes just do one over inside times derivative inside integrals a little bit more labor intensive you have to do some substitutions especially with trig stuff you're going to get a lot of trig stuff in this class get you used to it learn those identities and learn how to use them okay so right now what we're going to start we're going to start section 6.2 [Applause]