 Hi, I'm Zor. Welcome to Unizor Education. Among the main theorems which are presented here related to derivatives, I would like to stop on one particular very, very simple case when the function which we take derivative from is just a constant. Well, it actually seems to be a very simple thing because let me just start from a very trivial example. If I have a function let's say y is equal to c, where c is a constant and I would like to take a derivative from it. Now, it's obvious its derivative is equal to zero because what is a derivative? Well, it's a limit of function at some point with some increment minus the function value of that point divided by increment where this increment is infinitesimal. Now, but if my function is constant this is equal to this so it cancels out and I always have zero. So that's a trivial thing. But what's interesting is a converse theorem takes place which means if the derivative of the function is equal to zero is it true that the function is constant? Well, this is much less obviously a trivial fact and it requires certain proof and that's what I'm going to do today. So today I'm going to talk about how to prove that the function is constant if its derivative equals to zero. Okay, so this lecture is part of the course which is presented at unizor.com. It's a course of advanced mathematics for teenagers and high school students. I recommend you to actually watch this lecture from this website unizor.com because it contains very important notes for each lecture and also it has functionality which you might find useful like taking them for example. Alright, so back to this particular theorem. So let's forget about constant which is not actually known. What is known is that the derivative is equal to zero of the function which is defined let's say on some segment AB. Actually it can be an infinite interval as well, it doesn't really matter. But let's consider it's a segment and the function is smooth which means it's differentiable maybe once, maybe twice as needed basically. So whenever I'm staging something and it requires certain continuity or differentiability of the function it's assumed. So functions which I'm talking about are nice smooths without any kind of abnormalities which basically are all the functions which we are usually dealing with in algebra let's say. Alright, so the derivative is equal to zero. Now I would like to prove that the function is constant on this particular interval. Okay, now before I'm doing that let me recall the Lagrange theorem which I have just proven in the previous lecture. Okay, the Lagrange theorem is and let me just put the graph of it. So let's say your function is something like this and this is your segment AB. This is the chord, this is function at B, this is function of A. So if I do this, this particular piece is FB minus FA, this is B minus A. So if I divide this by this, FB minus FA divided by B minus A I have basically a tangent of that angle. So I have a chord and now the theorem Lagrange theorem says that there exists some point at zero in between A and B where the tangential line is parallel to this chord which means that basically the tangent of the chord is equal to derivative at point X0. Okay, so we have proven that theorem in the previous lecture so I'm going to use it as given. Now, what do I do now? Well, now I know that my function derivative is equal to zero at each point of this particular segment. Alright, fine. So let's take a particular point X0 and let's take point A and consider the function only from A to X0 on this particular segment where X0 is just taken in between A and B at will. What do we know? Now let's assume that we know the Lagrange theorem and we can say that the function of X0 minus function of A divided by X0 minus A is equal to derivative of the function at some point, I don't know, X1, let's say, where X1 is somewhere here in between A and X0. Now, I know that the function derivative is equal to zero at each point of the interval AB. Well, that includes the point X1, obviously, so this is zero. Now, X0 minus A is some positive number, right? X0 is to the right of the A, so this might be equal to zero, right? Which means F of X0 is equal to F of A. Now, let's go back to the beginning of my proof. Where did I get the X0? I said, let's just take any point X0 in between A and B. So it looks like for any point X0, the value of the function is equal to the function value at the beginning of this segment, which means it's constant. So at each point from A to B, function value is exactly the same as in the beginning. Now, obviously, we are talking about function differentiable at this particular point. I cannot say the function is differentiable at exactly edges of this, A and B. However, since the function is differentiable and I actually suggested that it's smooth to degree whatever is necessary, including continuity, so if function is equal to some specific value from A to B without the edges, then obviously at the edges it must be also equal to the same value, otherwise we would have no continuity. So the function is always equal to the same value from A to B, so it's constant. Now, if A or B are infinite values, which means that the function is defined everywhere, then I don't have any specific consideration, so any point X0 would be fine, and the value of the function would be basically exactly equal to this particular value. Now, if I have to really apply the Lagrange series, I need finite A and B, and my function is actually defined on infinity, and within the whole infinity interval it's equal to zero. Then I just take two points, any two points, call it A and B, do whatever this logic actually is, and then I can expand it to the left or right to any extent. All right, so that basically concludes a very simple proof that the function which has a derivative equal to zero is equal to constant. Now, the very simple corollary of this, very simple other theorem, if you wish, which is related to this is the following, which immediately follows from this one. Now, if you have two functions, and their derivative is the same on this particular segment, then they must differ only by constant, which means, let's say, f of X is equal to g of X plus some constant C, constant, which is the same for every X. Now, how can I prove this? Well, let's just think about it. What is f of X minus g of X? Well, it's some other function, h of X, right? What can I know about derivative of the h of X? Well, derivative of difference between two functions is equal to difference between derivatives, and since I said that they are the same, that's zero. It follows from this that the derivative of h of X is equal to zero, that this function is constant C, from which follows this. So, difference is the constant C. That's a very simple consequence of the first theorem. And now, the final point of this lecture is as follows. For some reason, you know the derivative of the function, and you are given the task to find the function, original function, which has this derivative. Let me just give an example. For instance, you know that derivative of the function is equal to X squared. And now, you are given the task to find all the functions in the world, which might have this particular derivative. All right, now, you know that this is X squared, and you probably remember that if you have a polynomial, then if you take derivative, the degree of polynomial is decreasing. X3, if you have function, and then you take a derivative from it, it would be 3X squared. Okay, now, you have just plain X squared. Now, you know this, which probably means that if you have this function, one-third of X cubed, well, you know that the factor can go outside of the derivative, so it will be one-third 3X squared, which is equal to X squared. So, you have guessed correctly that this is exactly the function, one particular function, which has this derivative. Great, but your task was to find all the functions which have this derivative. And obviously, you can use whatever couple of theorems I have just proven before, if there is any other function with the same derivative, some function g of X, such that g derivative is also equals to X squared. I know that g of X is only a constant different from f of X. So, you can say that all the functions which have this derivative can be described as one-third X cubed plus c, where c is any constant. So, it's like a family of functions. Well, obviously, infinite number of functions, but they all kind of look alike. They all contain some formula, some expression which contains an unknown X and plus the constant. And this is actually the basis for an operation which is opposite to differentiation. It's called integration, where we find, from a derivative, we find the original function. So, all I'm saying is that it's sufficient to find or to guess or whatever only one function. And then you have a plus constant to this function, and that describes all other functions with the same derivative. Okay, that's it for today. I do recommend you to go to the Unisor.com website and read the notes for this lecture. Quite a useful exercise, and don't hesitate to actually sign in on the website and take the whole course with all the exams, etc. So, that would be a very good exercise for your mind. Alright, thanks. That's it. Thank you very much, and good luck.