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chapters three and four are about derivatives and applications that's what exam 2 for you is about pretty much everything you should learn about derivatives and applications in a calculus one course the heart of the course at the beginning of chapter 3 we started with a bunch of derivative rules the first rule really is linearity that the derivative of a sum is the sum of derivatives and also the derivative of a constant times a function is that constant times the derivative of
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the function those rules can be combined into the derivative of an arbitrary linear combination you could call it say C1 time f ofx plus C2 * G ofx and I could even extend this to more than two functions any finite number of functions we apply this rule all the time when we take derivatives of pols that maybe have five or six or seven terms to them even we say the derivative is the first constant C1 times the derivative of the first function
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plus the second constant C2 times the derivative of the second constant of the second function G that's called linearity another name for it is a superposition principle it it's got analoges in physics when you use the word superposition it's very important though we don't think about it very much if this were not true calculus would not be nearly so useful it's very very very very very good that this is true and again we use it all the time we also had well technically
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in chapter 2 we had the power rule but I'll include it here in chapter 3 as well that the derivative of x to the N power is n * X the nus1 power and that works no matter what n is including if n is something strange like Pi or even e you know the derivative of x to the E power I'll just use a prime notation here is not x to the E power this is not an exponential function this is not the same as e to the X it's a power function the variable is in the base X
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is the variable e is a constant it fits this pattern with n equal to e the answer is e * x to the E minus one of course the derivative of e to the x that exponential function is just e to the x it's a different kind of function okay A different kind of rule applies yes n can be anything here and that power rule applies it can also be fractions which also means you should be able to apply it to take derivatives of say x^2 time the fifth root of x
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cubed how write these two functions both as power functions the fifth root of x cubed is the same as x to the 35s power you could use the product rule here but you don't have to because you can just add those exponents to getting a common denominator of five is the same as uh 10 fifths 10 fths plus 3 fifths is 13 fths or what would that be 2.6 if you prefer though you should feel comfortable with fractions
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bring that power down 13 fths x subtract one 13 fths minus 5 fths is 8 fifths there's the answer there's one way to write the answer you also could write the answer as 13 fifths uh say x to the first Power Times the fifth root of x to the 8th uh there's there's various ways of writing it okay that's one way of writing it but this is good enough just to get an answer question is decimal form AC decimal form is okay yes 13 fths could be written as 2.6 and then this would be 1.6 yes
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that is okay another question um does this hold up for imaginary numbers as well it does turn out to hold up for imaginary numbers n could be the imaginary unit I or any other complex number in fact and it does work though understanding what that means is less clear another question there uh that was a mistake thank you for noticing that this should be X cubed here instead of x to the 8 okay so that's the power rule um we also apply this when
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we are taking derivatives of polinomial we combine linearity and the power rule when we're taking derivatives of polinomial for example 4X cubed - 5x^2 + 6 x + 2 that's 4 * X cubed Prime - 5 * x^ 2 Prime + 6 * X Prime + 2 * what x to the zero Prime you could think of it that way if you wanted to you don't have to X Cub Prime is 3x^ 2ar x^2 Prime is 2X x Prime Prime is 1 and X the 0 power Prime is 0o this simplifies to 12 x^2 - 10 x + 6
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of course You' got enough practice now that you can just jump from here to here and that's fine get the answer quickly it's fine to do that but I wanted to illustrate that the rules technically do apply we have the rule for derivatives of exponential functions I already mentioned one of them the derivative of e to X is e to the X you might say that's the most important Special Rule in calculus for the derivative of a special function and fortunately it's the easiest to
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remember the derivative of e to the x is itself the function outputs at any value of x give you the slope of the curve at that same value of x amazing if you've got a general base base B like two for example then you also have to multiply by Ln of B when b equals e we do still get the right answer because Ln of e is one natural log of e is one the log base e of e is one because e to the 1st equals e all right so you should
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know these then we moved on to our more general rules the product rule quotient Rule and chain rule product rule says in abstract form I'm going to write this out of course you should be able to apply the product rule whenever you need to but you also should know its abstract form and even be able to apply its abstract form even when you don't know formulas for f and g but maybe function values and derivative values like given in a table or maybe with a graph you should still be
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able to apply the product rule there's your left function f there's your right function G left d right plus right d left there you go there's the product rule you because plus sign there because addition is commutative you could switch the order of these things multiplication is also commutative you could switch the order of these things that works for the product rule it does not work for the quoti rule because subtraction is not commutative the derivative of f ofx ided
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g ofx top function bottom or high and low is low D High minus high D L over the square of what's below common mistake is to get the Ator in the opposite order the wrong order don't make that mistake you might wonder maybe I talked about this question a few weeks ago if subtraction's commutative and subtraction is also addition how does that work subtraction is not commutative you know 5 minus 2 does not equal 2 minus 5 right three does not equal
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negative3 but wait a minute isn't subtraction can't that be defined in terms of addition 5 minus 2 is the same as 5 + -2 well 2 minus 5 is the same as 2+ neg5 it's not the same addition that's the problem so subtraction is not communative you cannot switch the order of things in the top and then we have the chain rule for the derivative of a composition F of G ofx think in terms of outside and inside functions G gets applied first when you
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plug a number in for X that's the inside function f gets applied second that's the outside function the chain rule says take the derivative of the outside but then plug in the inside function then extra step multiply times the derivative of the inside function
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you should feel comfortable with livets notation for this as well if U equal G ofx and yal F of U so that y also equals F of G ofx by PL by replacing you with G ofx you can write the chain rule as dydx equals Dy du times du DX and it's as if you're canceling the duus to get dydx not literally but you could pretend however if you use this livess notation for the chain rule you need to realize in the end you got to get everything in terms of X so even though you see a dyd there
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which would typically involve you in the end you want to replace any use with G ofx and simplify to have an answer that only depends on X okay but it's kind of a nice way to remember the chain rule that's why we teach it those are our general rules in section 3.5 we went back to more specific rules the derivative of trig functions in particular the derivative of s of X is cosine of x the derivative
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of cosine of x is negative s of X the derivative of tangent of X yes it is one over co^ squ X but more typically people write it as secant squ x and the derivative of secant x is secant of X time tangent of X which is also the same as Sin of X divid cos s x what about the derivative of coent and cosecant do you need to know those they can be found but you don't need to know them okay you can use the fact that coent is cosine divided by S or another way to think if it is one over
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tangent and use the quotient rule to derive the derivative of cotangent and cosecant of X is one/ s of X and you can derive its derivative with the quotient rule as well but even though that can be done we don't typically bother these are good enough for us and it's at this point well let me remind you of two things let me remind you in order to use these equations it is assumed that X
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is measured in radians if you're thinking of it as an angle it's assumed that you're thinking about the ordinary graphs of s cosine tangent and secant that we defined by thinking about the unit circle definition of those fun functions thinking of X as being an angle measured in radians but then we make a graph of it then X becomes the horizontal axis and we don't have to think of it as an angle anymore okay so X has to be in radians your calculator needs to be in radiant modes for
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this to make sense second thing I want to mention here to bring up right now is start relating what I've just been talking about the last few minutes two indefinite integrals as well we have talked about indefinite integrals as an extra subject in chapter 3 it was technically from chapter six in the textbook but I like to introduce it early indefinite integrals are easier than definite
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integrals effectively when you write the integral sign and you write your function in this case I'm going to write a cosine of x and then put a DX to emphasize the variables X just just an indefinite integral sign with no limits of integration all this means is write down the most general anti-derivative of this function since the derivative of derivative of s is cosine the most general anti-derivative of cosine is s plus C you don't want to forget your plus C we want to
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allow for arbitrary constants here because well any arbitrary constant still works for one thing because the derivative of a constant is zero it also turns out by adding constants here it helps you solve problems especially I'm thinking of differential equations if you take the class called differential equations making sure you put the plus C there helps you not that's not the end
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of the problem you got to finish it from there you got to find What's called the initial value of the function based on some initial condition like maybe y of 0 is seven or something well since s of0 is zero in order for this to equal seven when X is zero C would have to be seven you have to be able to allow yourself arbitrary constants to be able to solve those kinds of equations so it's a good thing to get in the habit of and I do require it it is assumed that you will put
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it there when you write indefinite integral signs for the anti-derivative of s of X since the derivative of cosine is negative s of X the anti-derivative of s of X is negative cosine of x plus C so it kind of switches around right from what's going on with the derivative it has to the derivative of s is cosine therefore the integral of cosine is s plus C derivative of cosine is negative s so integral of s is got to be negative cosine because if you differentiate
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this the negative sign stays there initially but then you get an extra negative sign from this fact and then the two negative signs cancel to give you a positive sign which is what you see there these facts about the derivative of tangent and secant to tell you the truth the thing they're most useful for is doing integrals and that that applies to some derivatives here as well that I'll talk about in a minute for example the integral of secant s x because the
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derivative of tangent of X is secant squ is tangent of X plus C if you didn't know that the derivative of tangent of X was secant squ x you'd have no idea what to do there it's because you know this fact that you can say that fact very quickly does the integral secant s x come up in calculus 2 yes it does in applications even and in calculus 3 which is multivariable calculus the integral of secant squ comes up in applications
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it is relevant for Applied problems kind of amazingly this is an aside you don't need to know this here's an aside the integral of secant of x while it can be done is very complicated and the integral of secant cubed x while it can be done is also very complicated secant squared is a special case that turns out to be easy by memorization I happen to know the answer to this off the top of my head it's because I've done it
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so many times the answer to the integral of secant of x is natural log of the absolute value of secant of X X Plus tangent of X plus C what log yes natural log I won't show you why okay likewise this integral this derivative fact leads to a fact about integrals that the integral of secant of X time tangent of X would just be secant of X plus c
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the aside comment again was that these are complicated you don't need to know those for your Calculus one any problems in calculus one but they do come up in calculus 2 integrals in general are much harder sometimes they very very complicated sometimes they are even impossible oh here's another aside that will be very relevant for us the last couple weeks of class integrals like this say the integral of cos squ x it's
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more than just complicated it's impossible and you can actually prove it's impossible uh well is it really impossible with a caveat and the caveat is this impossibility is with respect to ordinary functions that you're you are used to from pre-calculus integral of Co squ cosine of X2 is does exist it is a function it can be defined but it's a function you've never heard of yes there are there are hundreds if not thousands of functions that you have never heard
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of to this point in your life there are hundreds if not thousands that I've never heard of I've heard of more than you but there are hundreds if not thousands that I have never heard of are they useful yeah they they wouldn't be out there if they weren't useful they are specialized kinds of functions they're called non-elementary functions because they're not the kind you see in elementary school no I I meant to say not the kind you see in pre-calculus and calculus you
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don't see these functions in elementary school that's a joke right well yeah you don't you don't you're right just as all these functions were new to you when you learned pre-calculus in trigonometry there are other functions that are new to you that you've never heard of before from that point we went on in section 3.6 to derivatives of inverse functions examples of which include the derivative of natural log of x being one /x the course this is
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if x is positive because the domain of Ln of X is only positive numbers we also did show that the derivative of natural log of the absolute value of x is also one overx and that's if X is not zero it could be negative there and that does mean that the integral of 1 /x should in general be written as natural log of the absolute value of x plus C based on this fact down here actually that's not even the most General anti-derivative this is a bit of a
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fib and even more general anti-derivative of 1/x is a pie wise defined function you don't need to know this for the test but Ln of X Plus say C1 if x is positive and Ln ofx + C2 if if x is negative and I could have written an absolute value of x there instead of a Negative X because if x is negative then Negative X is positive don't worry about that that's another aside more inverse functions include the derivative of inverse tangent of X called
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arct tangent of X and you should be okay with either not is 1 1 + x^2 which does then mean the integral of 1 1 + x^2 is something to memorize inverse tangent of X plus C as with those trig function ones with Tangent and secant the main application of this fact is this fact you want to know the derivative of inverse tangent to be this so that you can integrate that that's more important for applications plenty of applications of integrals like this in calculus
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2 corresponding fact for inverse sign of X also called AR sign of X is 1 over theare < TK of 1 - x^2
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so that yes the corresponding factor for integrals which is still pretty important but maybe a little less important than the previous one is that the integral of 1 over < TK 1 - x^2 is inverse sin of X plus C I'm about to write down an equation you you don't technically know need to know for the test but I think I mentioned it once this is another aside the derivative of inverse cosine of x also called Arc cosine of x amazingly is just the opposite of the derivative of inverse
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s of X which means you also can write this integral as negative inverse cosine of x plus C isn't that interesting why would that happen I actually encouraged you one day about three weeks ago to explore why maybe four weeks ago by adding these two functions something interesting happens and that if you do that that adding and you graph the sum it will be more clear why their derivatives are opposites of each other give it a try after class I'm not going
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to do it now okay just kind of an interesting fact but you don't need to know it for the test are there any functions that I'm missing uh I don't think so though we also did have the abstract formula for the derivative of a general inverse function that you should know and be able to apply the derivative of f inverse of X is one over the derivative of f where you plug in F inverse of X into fime that's a general fact you remember der
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these facts we did a bunch you should be able to do those derivations on the exam for example to derive the derivative of Ln of X it's best to use the fact that e to the Ln of x equals x if x is positive and then differentiate both sides with the chain rule assuming Ln of X is differentiable when you do so remember doing that maybe you don't we did it we also did it for example to find the derivative of inverse tangent using this equation that's actually true for all
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X inverse tangent of X has a domain equal to all X its range is not all X it's the graph of inverse tangents looks like this it's got two different horizontal ASM tootes at negative pi over 2 and positive pi over 2 so it's defined for all X but it's range is not all possible values of Y anyway this is true for all X then you differentiate both sides including the left side with a chain rule assuming inverse tangent is
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differentiable do you need to be able to prove other derivatives you should be able to do basic limits for derivatives of simple functions say like X2 or X Cub that was UN even in exam one I'm not saying it couldn't be on exam two you should be able to do that because in the beginning of chapter 3 there were various limit calculations we did sort of derive the derivative of e thex as well and S and cosine using trig
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identities it was a while ago you may want to look back in your notes for those things might have even been before exam one late September early October it might have been
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we did these derivations we talked about how to derive the product rule and quotient rule product rule we made the picture of the rectangle probably wouldn't put a full proof of the product rule in the test but you should study the proof anyway and understand it maybe I'd asked you to explain part of it maybe I don't know quotient rule we derived using the product rule and the chain rule we derived that and that's simple enough you should be able to do it right F ofx / G ofx as F
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ofx times G ofx to the1 power use the product rule and the chain rule to derive the quotient rule at this point we moved on to implicit functions and multivariable calculus and local [Music] linearization trying to put all those ideas in the context of multivariable calculus oh take a simple example here like x^2 * y cubed oh maybe I should do something a little else here uh plus X let's say that's an example of a multivariable calculus function certainly you should be able to plug
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numbers in to find outputs you know what is f of 21 replace x with two and Y with Nega one and simplify 2^2 is 4- 1 cubed is 1 so we get -4 + 2 the output is -2 so you should be able to plug numbers in should you be able to graph multivariable functions that's probably too complicated for us I mostly even myself relying technology for that but you should understand the idea of the graph the concept is you got a three-dimensional
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coordinate system you imagine this to be three-dimensional space and x-axis coming out of the paper towards you A y- axis going to the right and a z-axis going upward and the concept is you let X and Y vary over the entire plane where Z is zero and you find the outputs and you plot the corresponding points and what you in general get is some sort of surface in general it could be some wacky looking surface that could be the graph of your multivariable
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Calculus function though we did special cases where one looked like kind of like a bow shape that was x2+ y^2 what is the graph of this look like I have no idea off hand I'm sure it does not look like that and you should also understand the concept of level curves where we imagine cutting the surface with horizontal planes at constant values of z z equals c for example Z equals -2 I drew it so C looks like it's positive here but it could have been negative when you slice
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a surface with a horizontal plane you typically get a curve it's called a level curve and you can make a graph of the level curve in an ordinary XY plane like what is the level curve where this function equals -2 one point I know for sure it includes is the point toga1 21 is one point on this curve where this equals that all the points where this expression equals -2 what are some other points I don't know could I estimate what the curve looks like near this point I can
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try I can try using implicit differentiation to try to estim try to find the slope of the tangent line there to the curve assuming it exists now it might not be a differentiable curve that's possible though rare in these kinds of circumstances it's also possible though rare that we might be at a point where the tangent line is vertical and has infinite slope and if that is true then implicit differentiation is going to give us a dividing by zero and
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we would see that but probably probably the slope does exist and is not infinite so let's just go ahead and assume that's the case and use implicit differentiation to try to find the slope there take this equation I'm going to rewrite it again x^2 y Cub + x = -2 and assume
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assume Y is a function of X near the given point assume that is it a function of X actually it's it's a function of X we actually could solve for sometimes it's complicated enough that you can't solve for it but here we actually could solve it for it we could subtract X from both sides and write x^2 y Cub is -2 - x then we could divide both sides by X2 and write this and then we could take the cube root to both sides and even with minus signs on the right that's okay to do because
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we can take cube roots of negative numbers that's all okay okay so there there's what it equals but you don't need to know that instead you can assume it and differentiate implicitly based on that assumption and in the first term you would need the product rule for differentiating this product and you would also need the chain rule for differentiating y cubed left dite which I'll initially write just DDX of Y cubed plus WR D left differentiate with respect
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to X then plus the derivative of X which is one equals the derivative of -2 which is a constant so it's derivative is zero but don't stop we're not done with this derivative we need the chain rule differentiate with respect to X Y Cub the answer is not zero because y depends on X it's not 3 y^2 even though we can start writing that because we need the chain rule to multiply times the derivative of the inside function with respect to X which is
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dydx and now you want to solve this equation for dydx x 3x^2 y^2 dydx bring the these things to the other side equals -2X y Cub minus one now divide both sides by 3x^2 y^2
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and then to find the slope at the point in question just plug numbers in the point in question was x = 2 y = -1 yeah we're not going to divide by zero X is 2 and Y is1 be careful not to make a mistake hopefully I have not made a mistake to this point-1 Cub is1 * -4 is pos4 on the bottom negative 1 squ is positive 1 looks like we get a 12 3 12ths or 1/4 if I have not made a mistake is the slope of the tangent line to the curve at the point which then allows us
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to approximate the curve near that point with a linear approximation tangent line approximation
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how uh well we know a point on the curve 2 negative 1 we also know the slope of the tangent is 1/4 that's the tangent line there it's not the curve itself because it's straight the curve itself just looks like the tangent line near this point it's a good approximation when X is close to two so I can use point slope form
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y --1 equals the slope 1/4 * x - 2 this the same as y + 1 this is the same as 1/4x - 12 subtract 1 from both sides y will be 1/4x minus 3es that'll be the equation of the tangent line at that point and then its graph will approximate the curve near that point and you should also be able to use it to approximate values on the curve near the point for example if x is 2.1 what is y it's actually
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better to use this form of it with one subtracted say when X is 2.1 Y is approximately 1/4 * 2.1 - 2 and then Min -1 because I'm subtracting one from both sides that's 0.25 * 1 which is 025 so if I have not made a mistake this would be. n75 is your approximation for the y-coordinate of a point on the curve when X is 2.1 x is a little bit bigger than two I moved to the right a little bit that slope is slightly positive
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y should be a bit bigger than negative one so assuming I haven't made any mistakes which I'm not guaranteeing just the idea is what I'm trying to mostly get across here running low on time oh boy this is the further stuff further back stuff so oh we haven't had much time for parametric equations oh they just I I probably should not have done a CP but it was good to review anyway
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hang with me here just a few more minutes um we need to go quicker with what remains let me just very quickly remind you what remains there were the well there's applying linear approximations to specific functions as well like oh for example e to 2X or sine of 5x or or cube root of 1 - x you could do linear approximations for all those you did examples like that in your homework there's also the theorems about differentiable functions including the mean value theorem the
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increasing function theorem and the RAC trck principle were the main ones we focused on do you need to do proofs on the test possibly using the constant function theorem I forgot to mention a proof could show up on the test using that similar to the one we did in class after that point into chapter four it was mostly optimization at the beginning that included things like the first and second derivative tests for trying to decide whether a critical point was a local Min or a local
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Max so the first derivative test you focused on does the first derivative change sign like from positive to negative at a critical point that would mean the function itself changes from increasing to decreasing you have a local max if it's the opposite changing from negative to positive you got a local Min how you show work for the first derivative test make a number line well find your critical points first make a number line mark off the critical points and
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test the sign of the derivative in the different intervals between the critical points that's how you show your work the second derivative test you find the critical points you find the second derivative you plug the critical points into the second derivative if you get a positive second derivative the graph is concave up so it's a local men you get a negative second derivative the graph is concave down so you get a local Max some critical points are neither local Maxes nor
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local mins like xals Z for X cubed the first derivative does not change sign even though it is zero at x equals z the second derivative at zero is zero so it's neither a local Max nor a local Min that happens sometimes and you should be aware of that there was the global optimization where you had closed intervals and you also had to check the end points not just the critical points in the in the interior but also the end points the global
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Max or in could occur at the end points there was the Practical applied problems the word problems maximizing or minimizing surface area for a given volume for example you could also maximize volume for a given surface area for cylinders and that kind of thing uh there's the related rates where you are assuming all the quantities depend on time and you're differentiating with respect to T and you're relating those rates to solve
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for an unknown rate key thing is you only plug the numbers in at the very [Music] end
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then there was lol's rule for finding limits of indeterminant form zero over zero Infinity over infinity there were more complicated ones I think I'd give you just a more basic one in the test and then there's parametric equations and curves so we haven't had a ton of time for that um so so probably if I test you on parametric equations it would be on mostly what we talked about on Monday which was parameterizing a line on your homework you also thought about
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circles and ellipses and that kind of thing bit trickier all Focus if it's on your test on lines
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