 Hi, I'm Zor. Welcome to a new Zor education. Today we will continue solving certain problems. Well, sometimes I call these problems problems and sometimes exercises. In this case, I decided to call it exercises number three in this particular category of derivatives. Well, the reason is I was thinking that maybe if it's something which which needs some creativity, then it's a problem. If you just rely on the skills which have been already explained in the lecture, that's exercise, but obviously there is no clean separation between these two. So anyway, exercises about derivatives. So basically, we will just take some derivatives. This lecture is part of the website Unizor.com. It's dedicated to presentation of advanced course of mathematics for teenagers and high school students. I do recommend you to watch this lecture from the website because it contains very nice notes, very detailed notes for each lecture. And in addition, the website contains some other educational material. Also, people who can sign in, they actually can participate in the educational process, like taking exams, for instance. All right, so let's get back to business. Well, the first problem is actually kind of a joke. I was always fascinated if I can approach one particular problem in one way and get the result. And then I completely different way amusing and well, all of a sudden I got exactly the same result. Well, I think it's to be expected, but it's basically a confirmation that the world which we have basically created with all these mathematical objects is created correctly. I mean, if I can come with something one way and using basically another way, my result is different than something is wrong. If result is the same, it confirms that everything is actually glued together in the proper way. So my first problem is take derivative of the function sine square x plus cosine square x. Well, everybody knows that basically sine and cosine square is equal to one. So which means I'm asking to take a derivative from a constant function which takes the value of one. And as we know, the derivative of the constant function is zero. Well, question is, can I get zero from here? Well, let's see. Well, now the derivative of a sum is sum of derivative. Now, derivative of sine square, well, basically it's a chain rule. First, the outer function is square. So it's derivative of two sine x and then the inner function is sine and the derivative of this is cosine. So that's what it is. Now, derivative of this is again, outer function is this and derivative of the cosine is minus sine, right? So I have to put minus here and sine. And as we see, it's equal to zero. Two sine cosine minus two sine, cosine sine. Well, great. Everything is fine. The world is in harmony. Let's put it this way. And and obviously corresponds to whatever we expected. Now, somebody might ask can this be a confirmation that sine square plus cosine square is some kind of a constant? Maybe not necessarily one, but some kind of a constant. Well, absolutely not. Because all these rules about what is the derivative from cosine and sine, they are based on certain properties of sine and cosine and the fact that sine square plus cosine square is equal to one is much closer to the definition of these properties. And the definition of the sine and cosine much closer than taking derivatives. So it's definitely, it would be a logical loop if we will say that, okay, because of this, this is equal to constant. That's not the case, all right? So in any way. Now, and I have another again half joking, half serious problem which we will approach in exactly the same way. We will approach it from some kind of a classical way of taking a derivative. And then we will just realize that it's much simpler to do it in a different way. So my function is natural logarithm of e to the power of x. Again, by definition of the logarithm, this is supposed to be equal to x, right? Because what is natural logarithm is, what is the power I have to raise e to get e to the power of x? Well, obviously x. So we know this is x and the and the derivative of function equal to x is 1, right? Okay, fine. So let's just check if we will do it. If we will get it exactly the same way, if we will just use as a regular composition of two functions. So we will use the chain rule. First, the derivative of the logarithm is 1 over whatever is under logarithm, right? So it's 1 over e to the power of x. And then we have to multiply by derivative of the inner function. Derivative of e to the power of x is e to the power of x. And this is obviously is equal to 1. And again, the world is in harmony. Good. Next is something serious and boring, if you wish. But anyway, just for, well, this is exercises, right? So it's not a problem. So that's why we are entitled to have some boring exercises. All right. So our function is square root of x plus square root of x. And all we need to do is just to find its derivative. And we will definitely use, again, the chain rule for compound functions. So my outer function is square root. And we know that the derivative of square root, which is basically, remember, x to the power of 1 half, right? Whenever we have a power, it's 1 half x to the power of whatever it was minus 1, right? Which is minus 1 half, which is equal to which is equal to this, right? So that's my derivative. Okay, so my outer function is square root. So my first is 2 square root of whatever it is. That's derivative of the outer function. It should be multiplied by derivative of the inner function, right? What's derivative of the inner function? It's the sum. So derivative of x is 1, derivative of square root of x is, we just talked about it. And that's what we have. Now, you can maybe simplify this, but anyway, this is exactly what is, what the derivative actually is. Okay, basically, that's it. And all I wanted to make sure that we understand that we have a composition of functions. And whenever you have a composition of functions, we are just using the chain rule. The outer function, we take derivative of the outer function using whatever is under it as an argument. And then we multiply it by derivative of the inner function. But in this case, by the way, we again have another kind of outer function. All right, so that's kind of exercise on the skills which we have developed to take derivatives of the compound functions. Now, what's next? Next might require a little bit more So function is arc tangent of x. No, arc tangent of 1 over x. All right, so we have again one function which is arc tangent. And then we have as an outer function and inner function 1 over x. But to tell you the truth, I don't remember by heart what is derivative of arc tangent. So I have to derive it somehow first. So let me just spend some time and derive the arc tangent derivative. And then I will use whatever I have. So let's just consider function y is equal to arc tangent x. And we will find its derivative as the function of x because that's what I maybe I have to remember it, but I don't. So if I don't remember something, I'm trying to derive it, right? So how to take derivative of this thing? Well, this is an implicit definition, obviously. So what is arc tangent? It's an angle tangent of which is equal to x, right? So if I will take tangent of y is equal to x, right? Now I can take derivative of both. Now derivative of tangent is, oops, I don't remember derivative of tangent. That's a shame, right? Well, but I do remember, the only thing I actually remember with trigonometric function is sine and cosine. Derivative of sine is a cosine. Derivative of cosine is minus sine. All right, so how can I derive derivative of tangent? All right, so let's just consider another function. So tangent x, I have to take derivative. That's actually sine x divided by cosine x. Now, I do remember the derivative of the product. It's first time the derivative of second plus second time derivative of first, but I don't remember the ratio. I don't remember anything. I have to derive everything, right? So actually, I'm trying to kind of convey the message that if you don't remember something, don't despair. You can always derive something. If you know something else, derive the one which you don't know. So how can I derive this? Well, it's simple because it's actually sine x times 1 over cosine x. Okay, that's closer. That's actually something which I do remember. I remember what is the derivative of this, and this is a combination of two functions. 1 over, which I do remember, it's minus 1 over x squared because it's just the power, it's x to the power of minus 1, right? And the cosine, which I also know. So basically, I have reduced my problem to known problems. Okay, so the derivative of this, which is tangent of x is equal to first times derivative of the second, okay? Sine x times derivative of this. Now derivative of this is, again, it's a compound function. So the outer function is 1 over something, which is minus 1 over cosine square x, right? Because derivative of 1x is minus 1 over x square. This is minus 1, this is x to the minus 1. Right? So minus 1, that's minus, and then minus 1 from the power that gives me 2. Okay, so I've done that. Times derivative of the inner function. This is cosine, so derivative is minus sine plus. So I've done what? The first times derivative of the second. Now we have to have the derivative of the first and the second, which is cosine and cosine. So what's the result? Now this is 1, right? And this is minus and minus would be plus, so it's 1 plus tangent square x. One is from here. Now this is sine square over cosine square, which is tangent square. So the derivative of tangent is 1 plus tangent square. Okay, fine. I was supposed to remember it, but I didn't, which is fine. All right, so I've got that. So now I can use it here. So now I know that the tangent derivative is 1 plus tangent square, and that's why I can put this. Now if I derive, if I have a derivative from both sides, what happens? Derivative of the tangent is 1 plus tangent square of y times inner function, right? Y derivative. And that's equal to derivative of the x is 1. Okay. All right. Now what is tangent of y? Well, y is arc tangent, right? It's the angle tangent of which is equal to x. So tangent of y is equal to x, tangent of square, tangent y, tangent square of y is obviously x square. So what I have is 1 plus x square times y equals to 1, from which y prime is equal to 1 over 1 plus x square. So that's exactly my answer. No, that's not. This is just our tangent. Sorry about that. This is just our tangent. Now I have 1 over x. So my first my outer function, which is arc tangent, has derivative of 1 over 1 plus x square. But instead of x square, I have to put 1 over x, right? Times derivative of the inner function, which is minus 1 over x square, right? So that's my final combination. Now, let me just simplify it just a little bit because it's kind of obvious. It's equal to 1 plus x square and x square goes to the top, right? x square plus 1 and x square goes to the top. And now I have minus and x square here, so that's minus 1 over 1 plus x square. That's it. That's my final answer, right? Right. Okay. Next problem or exercise, if you wish. Well, incidentally, I have demonstrated that you don't really have to memorize everything because there is always some way to derive the formula which you don't remember. Like, I didn't remember the arc tangent or tangent derivative. I just derived it from something which I remember. And the only thing which I remember is the power function x to some power, including negative powers like 1 over x and sine and cosine, right? Everything else is derivable from this. All right. Next is... Okay, let me just talk about some weird function. x to the power of 1 over x. Well, I don't know how to derive it directly quite frankly. It's not easy because I cannot use x to the power of n where n is a constant. I mean, I know the rules about the power function. This is not a power function because the x is here as well as here. So we cannot apply our knowledge that the derivative of this is n x to the power of minus 1. This is absolutely not applicable to this because here n is a constant. Here x is the variable which is also moving, changing its variable. So what do I do? Well, let's see. If y is equal to x to the power of 1 over x, I can logarithm y is equal to logarithm of y x to the power of 1 over x which is equal to... this is just a power and logarithm with a power we know how it works. That's how it works. Okay, so what can we do now? Well, one of the things which we can do actually is to take derivative of both sides. All right, so let's take derivative. Derivative of logarithm is 1 over y times derivative of y, right, because it's a compound function. It's logarithm of y and y is some kind of a function, derivative of which we want to know and this is something which we want to know and 1 over y is because of the logarithm. That's a do-remember by the way. Derivative of natural logarithm is 1 over x. So this is 1 over y times this. Now this derivative of this is, well, again, that's simple. That's a product. I do remember how the product is working. So it's the first one times the derivative of the second plus derivative of this one which is minus 1 over x square and the first. Well, that's actually not bad because now we can say y y prime which is derivative of y is equal to y times this. We know what y is which is this and we have to multiply by 1 over x square. Uh-huh. We can take it out, factor out times 1 minus logarithm x. Am I right? Yeah, I'm right. So that's the final derivative of this function. Looks scary, right? But don't worry about it. It doesn't really matter how it looks. As long as we know how to derive it. Okay, and now another thing which looks even weirder than the previous one, our function is implicitly defined using the following and don't ask me how it looks. I don't know how it looks, but it's the function, right? For any pair of x, y, I can say whether it belongs to this function or it doesn't. I mean, if I'm just thinking about how to graphically put this, how to have a graph of this function, I know for sure, for instance, that the main diagonal where x is equal to y is always part of the graph. But there are some other pieces which might not actually, I don't even know how they look. In any case, so we have to derive, we have to find out how the derivative of this function looks like. Well, but here is a very interesting point. If we do not actually know whether this is a true function in a true sense of this world, which means for every x there is one and only one y, in this case it might not be the case actually. For instance, by the way, if you have, for instance, function which is given this way. Well, basically you can say that it's like two different functions. One is y is equal to square root of x and another is minus square root of x. This is not a true function, right? Because for each x you have two different values of y. If x is equal, for instance, four, then two and minus two works here, right? So that's why it's not really a true function. So I cannot expect the derivative of this function y to be expressed in real formula as a function of x. Most likely, I will have it also in this kind of a form, implicit form, right? So my function is implicitly defined, as well as in this, by the way. It's an implicit definition of the function. And that's why it leads to this type of not really functional property. So this is as well. So it's implicit definition of function. So I should expect implicit definition of the derivative. So let's see how I can approach this. Well, obviously, I have to apply logarithm to both cases, right? So logarithm of this is equal to logarithm of that. y times logarithm of x is equal to x times logarithm of y. Okay, this is easier. This I can deal with, okay? So this is exactly equivalent to this one. Now let's take the derivative of both sides. Okay, here, y derivative times 1x, 1 over x, that's derivative of this. Oh, no, I'm sorry. Derivative of this times this plus this times derivative of this. Okay, that's on the left. Here, derivative of this, which is 1 times logarithm of y plus x times 1 over y. Times derivative of y in the function, right? This is a three functions, basically. Outer is logarithm, inner is y, but it's also a function of something. And that's why when I'm talking about derivative of this thing, I have to use first the outermost function, which is logarithm, which is 1 over y. And then I have to have the inner function, right? Everything else seems to be fine. I hope. Now we can resolve it relative to y. Y prime, right? So what can we do? We multiply by y and I will get y prime times y times logarithm x. Well, actually, let's multiply by xy. So we'll have x as well here. Y times x times y times logarithm. Plus, this is y square, right? We multiply by xy. So x is reduced and y will be square. Is equal to xy, logarithm y, plus x square y prime. From which y prime is equal to this and this? Okay, let me just do it in two steps. So y prime times x times y times logarithm x minus x square, right? Equals to x times y times logarithm y minus y square. And now y prime is equal to xy logarithm y minus y square divided by xy logarithm x minus x square. I think something like this. Yeah, that's the answer. Now, obviously, it's not without problems, this type of a solution. For instance, I'm taking logarithm of both sides. Well, I can take logarithm only if both sides are positive, right? Now, are they positive? Well, that's a big question. Now let's take a look at this. We know that this type of exponential functions are defined only for positive arguments. So I have to really analyze what is my argument. And since I have two exponents here and here, I must have x and y both positive. Which is a good thing. Now, if they are positive, then both sides are positive. So logarithm is fine, no problem. Now here, I have 1 over y, and y is not equal to 0. Okay, so that works too, since both x and y are positive. So it's not like it's without problems. I just have to analyze that there is no problems. Okay, that's what I'm doing right now. So we have to be careful with this. So considering x and y and both positive, that seems to be okay. Now, obviously under certain circumstances, this thing probably might be equal to 0, in which case function will have an infinite derivative. I don't know. It's difficult to say, because I have absolutely no idea how this function in the denominator behaves. Because I don't know the y, which is defined by God knows how. But in any case, this is some kind of a parametric expression. I cannot substitute instead of y something, because y is not explicitly defined. So y is defined implicitly, and that's why the whole thing becomes an implicit definition of the derivative. And that was my last problem. I do recommend you to read notes for this lecture on unizor.com. Other than that, that's it. Thank you very much, and good luck.