 Hi, I'm Zor. Welcome to Unizor Education. Today we will solve a couple of problems related to derivatives. This lecture is part of the whole course of Advanced Mathematics for Teenagers and High School Students. It's presented on Unizor.com. So if you are watching this lecture from any other website or YouTube, I do suggest you to go to the website to which this lecture from, because every lecture including this one has very detailed notes and for people who signed in to the website there are some different aspects of educational functionality, like for instance you can take exams and the site is completely free, no advertising. So please use it. Okay. A few problems on derivatives. Okay, problem number one. Now, we used to think about functions extremums, maximum and minimum, as the points associated with the first derivative equal to zero. So here is a very simple question. So first, let's consider you have a function f of x and it's sufficiently smooth. So there is a derivative. It's a continuous second derivative, whatever you want. And there is a certain condition that at certain point its derivative is equal to zero. So my question right now is are these conditions has extremum at x zero. The condition of the function to have an extremum at point x zero and the function to have a derivative equal to zero at point zero, at point x zero. So question is which of them are necessary or sufficient or necessary and sufficient condition relative to another one? So should I point this way or this way or this way, both ways. So that's my question right now. Now, if this is a statement, then this statement and that's basically the final solution to this problem is a necessary condition, which means from this follows this. The other direction is not correct. And here is an example where it's not correct. At point x equals to zero, the first derivative is equal to three x square and it is equal to zero. However, this is not an extremum for x cube looks like this. This point is not an extremum. It's not maximum nor minimum. It's an inflection point actually in this particular case. But doesn't really matter as long as it's not extremum. So from this this does not necessarily follow. So this condition is necessary for this if function has an extremum, a smooth function has an extremum at point x zero extremum being this maximum or this minimum, local extremum. Then at this point derivative is equal to zero and we have proven that. The other the reverse direction is not true. That's why the condition is necessary but not sufficient. So there is only one way of logical implication. If function has extremum, then at that point its derivative is equal to zero. You cannot say anything otherwise. However, if you're looking for extremum, you obviously start from these points because if there is an extremum, then the derivative is equal to zero at this point. But then you have to investigate the problem further. Okay. For instance, we have found the point where the derivative is equal to zero. Next, what should we do next? How can we determine whether it's maximum or minimum or none of them? Well, there are many different ways. One of them is to check the second derivative. But again, that's not sufficient. The best way is actually to step to the left a little bit and step to the right a little bit from x zero from x zero and see if in both cases you are either increasing the value of the function in which case x zero is minimum or you're decreasing the value of the function in which case x zero is maximum. So that is probably a much better choice. Now as an example, for instance, if you have f of x is equal to x to x square, now obviously derivative is two x and it's equal to zero at point zero. So let's step to the left from zero. Let's say x is equal to minus epsilon, where epsilon is some positive small value. Then the function is epsilon square. Now if you go to the positive, the value of the function also would be epsilon square. Now epsilon is small, but it's not equal to zero. Now these two values, no matter what epsilon is, as long as it's not equal to zero, are positive, which means at point zero you have zero, but at point minus epsilon and plus epsilon we always have positive, which is greater than zero obviously. So this is a clear indication of the point of minimum. All right, got that. Next problem. Next problem is about inflection points. You remember that if point has inflection at x zero, then its second derivative is equal to zero at that particular point. And by the way, I have just shown you an example of x to the third power of three that has an inflection point is zero. Now question again, is this necessary, sufficient or necessary and sufficient condition? Well, the answer is this is not a sufficient condition. This is necessary, same thing as with the first derivative and extremums. Now again, it was proven that in case of inflection point, that's where basically you have changed the direction of the first derivative. First it's supposed to be positive and then it's negative. For first it's negative and then it's positive. In this particular case you do have this particular condition, but in some cases this is not sufficient for function to take an inflection point. An example is very simple, x to the fourth degree. Here is the graph. It looks like a parabola, but it's much closer to the x axis in the vicinity of zero. So the first derivative is 4x cube. The second derivative is 12x square and it is equal to zero at point x equal to zero, right? But it's definitely not an inflection point. This is the minimum point. So incidentally we have proven that even if we have a minimum we still can have the second derivative equal to zero. So if you are analyzing the extremums, you check the first derivative, you found for instance point x is equal to zero. You have the second derivative and you're thinking that if the second derivative is positive it's minimum and if it's negative it's maximum, but in this case it's zero and it is minimum. So again our second derivative also is not sufficient. So all these properties of derivatives are necessary for certain behavior of the function. For the extremum the first derivative equal to zero is a necessity but not sufficient condition. For a function having an inflection point the second derivative equals to zero is necessary but not sufficient. This is the end of my second problem. Next. Next is a little bit more involved I would say but faster. We are talking about derivative of the inverse function. Let me start from the example first. For instance you have a function let's say it's e to the power of x. Now its inverse is logarithm x as we know, right? Derivative of this guy is e to the power of x. Derivative of this guy is 1 over x. Now what does it mean that this is inverse of this? Well it basically means that if composition is retaining the value x. We start from x, we go to e to the power of x and then you go to logarithms. So logarithm of e to the power of x is x. Now this is we start from x, we do logarithm of x first and then we e to this power and this is also x, right? So this is basically the property of the inverse function. Now let's talk about their derivatives. Now I would like to say actually that this derivative is equal to 1 over derivative of another function at point g to the power g of x. f derivative is e to the power of x of logarithm x that seems to be and that's x so it's 1 over x. Now I basically am stating that this is a general rule. Now how can we prove it? Well there is a very easy proof something like this. If you would like to let's say take a derivative d for dx of f of g of x what is this? It's a composition of two functions, right? Now the composition of two functions if you would like to differentiate it it's a differentiation of the first function at the same point g of x times derivative of the inner function, right? But from the other hand this is x, right? So derivative of x is 1 so it's equal to 1 from which you get this. So from this you get this. So this is the general proof basically. So we have proven this particular thing. Now in my notes to this lecture actually I have another example not only this one but this one works as well. Now this one works in this direction let me just do it in another direction. What if I would like to have f1 of x which is e to the power of x. Now if I will do this manipulation I'm supposed to get 1 over g of f of x, right? Now g is logarithm so g derivative is 1 over x so 1 over would be would be x but instead of argument I put f at x so I will get again this is equal to e to the power of x, right? So it's e let me just do it step by step. It's 1 over f at x, right? That's what it is in this case because this is g is logarithm the derivative of the logarithm is 1 over x so that's why 1 over and then instead of x we should substitute f at x which is e to the power of x. So it's 1 over 1 over what is this e to the power of x which is equal to e to the power of x, right? Same thing is here. So g is logarithm f is exponential so derivative of logarithm is this 1 over and exponential is this and that's why I have this, okay? So that's my third problem and finally I have something which is more of a you know practical aspect if you wish my first problem is related to doing something physical. Let's assume that you would like to make a metal box in the form of a regular quadratic prism. So the base is square A by A and some kind of a high altitude. Now you have certain material to make this box. Now your material is basically what you have been given which means you basically have the total area of this box. That's the amount of material which you have. Let's say you want to make a box of aluminum so you have certain amount of aluminum in square meters or whatever you want some kind of area, okay? And you would like to make your aluminum box from this but you would like to make it in such a way that it would maximize the volume. So this is given. I don't know exactly A or H but I would like to make this particular box in such a way with these parameters A and H such that the volume will be maximized from the same amount of aluminum which I have, all right? Sheet aluminum. All right, so how can we do it? Let's do it step by step. First of all, S is equal to, you have two bases each A square so it's two A square. So I would like to basically use A and H and then come out with an S. Plus I have four sides. Each of them are A times H, right? Plus four sides A times H. That's what I have. Now my volume is equal to area of the base times height, right? So it's A square times H. So how can I maximize the volume? Well, first of all, there are two variables but I know that S is given. So I can find one variable from here and substitute here to get only one variable from which I will depend, from which volume will depend and then I will try to maximize volume as a function of that variable, right? Well, it's obviously easier to find H. So H is equal to S minus two A square divided by four A. Now my D, my volume as a function of A now would be A square times S minus two A square divided by four A. Well, one of them is going out. So it's one quarter which is this one times minus two A cube plus SA, okay? Now this is the function of A and I can maximize it. Well, obviously, A has certain boundaries, obviously, right? One boundary is that A is greater than zero and another boundary is obviously A cannot be greater than total amount of aluminum used to increase A. Now, how big can A be in theory? Well, it can be as big as H would be equal to zero and A would be so large that these two bases together will make up the entire S. So maximum is square root of S2, S over two. So these are minimum and maximum values for A which are actually allowed. Now out of curiosity, I do suggest to have a graph of this function. Well, first of all, we obviously know that function is equal to zero. Well, it's a cubic polynomial, right? So the general shape would be like this or like this. Now in this case, since this is minimum, since this is negative, the shape would be like this, right? Now, it equals to zero at A is equal to zero. And as A goes to negative infinity, it will go to positive because A to the third degree would be negative and minus so it will be positive. So it goes this way. Am I right? No. S is equal to, if A goes to negative infinity, it would be negative. If it goes to right infinity, it would be positive. Now something is wrong. Another root would be A square is equal to, so it's square root of S over two. This is zero as well. And it goes to plus. No, this goes to minus because A is very large, of course. This goes to there. This goes there. Okay. And what happens in between? Well, in between, it should go something like this. Am I right? Let's just think about it. If A goes to plus infinity, this is very fast because it's cube, so it goes to minus since this is minus. Right. That seems to be the right thing. Let me just straighten it out. Okay. And probably, it will go slightly differently here. Probably, it will go this way. That's what it goes. That's what it is. But we don't really care about negative side because A cannot be negative anyway, right? So A is supposed to be greater than zero. So it looks like it's supposed to be some kind of a maximum, right? At this point. Okay. That sounds reasonable. Now, as far as roots, that should be another zero somewhere, right? So, well, actually, we can solve this. A is equal to zero is one root. Another is A is equal to square root of S over two. And then there is a negative S over two. So that's basically what it is. Yeah. I have to put it a little bit further. So it's negative square root of S over two. So it has three zeros, but we don't care about the left part. Okay. So all we have to do right now, which is a very simple thing is to find this point of local maximum. Now, obviously, we start from first derivative equal to zero, right? So first derivative is, well, we don't care about one fourth, right? Because it will be just a factor. And whenever you're equal to zero, it will cancel out. So it's minus six A square plus S equals to zero, right? That's my equation for the first derivative of this equal to zero, which means A is equal to, and again, we will take only the positive solution. It's square root of S over six. So that's where my local maximum probably is. Well, since from our graphical analysis, the shape is like this, it's most likely the only one. We can obviously take the second derivative and the second derivative is minus 12 A, right? Minus six A square. So the second derivative is minus 12 A. And for a positive A, it's negative. And negative means we have local maximum. Negative of the second derivative I meant. All right. So we have actually determined an exact value of the base of our prism. And all we need to do is to find out what's the height, what's the altitude. Let's just go back to our original equation about S, which is 2A square plus 4AH. Now we know what's the value of A is. It's this one. All right. So let's just substitute it. So S is equal to 2A square, which is S over six plus 4A. Well, A is square root of S over six, H. So H is equal to S minus 26 is S over three divided by four square root of S over six, right? So this is two-thirds, two-thirds S. This is four square root of S square root of six equals two. Now this is two. Now square root of S and S is square root of S in the numerator. Square root of six and six is square root of six in the denominator, which is square root of S over six. Notice it's exactly the same as A. What does it mean? Well, it means that we are dealing with a cube. H is equal to the side of the square in the base, right? So if you have certain amount of sheet aluminum and you have to make a rectangular prism with the highest volume, you just make a cube out of it. You calculate what exactly is your side and then you cut whatever is necessary from this aluminum piece and make it the cube. So cube has the highest volume among all the regular square prisms with the same surface area. All right. Well, these are four relatively simple problems which I wanted to present today. I do suggest you to go to the website Unisor.com and try to go through these problems just by yourself. And well, there are answers and in case of the forest problem, there is a complete solution, but I do suggest you to do it yourself manually before you go any further. I think it's a very good exercise and probably I will offer a few more problems of the same kind of type on maximum, minimum, inflection point or whatever. All right, that's it for today. Thank you very much and good luck.