 Hi, I'm Zor. Welcome to a new Zor education. I would like to present one more, well, formula or property, if you wish, of derivative, which is related to compound functions, functions of function. How can we deal with the right derivative of this compound function if we know derivatives of its components separately for F and G? This lecture is part of the course of advanced mathematics for teenagers and high school students. It's presented on Unizor.com and if you go to this website to watch this lecture, you will see that there are very detailed notes, which you can basically use as a textbook. So, we are talking about the derivative of this function. Well, first of all, obviously we have to talk about domain. So, G is supposed to be defined in certain domain and its values are supposed to be within the domain and function F, right? So, otherwise we will not be able to calculate what's the value of F for any particular X. So, we assume that with this particular domain, store everything is okay. The easiest way is if you are dealing with functions, which are defined, for instance, for entire set of real numbers and in this case, whatever the value of G is, it will always fall within the domain of function F in any case. So, we assume that the store with domain is settled and all we need right now to do is to calculate the derivative relatively straightforward from this compound function. So, how can we do it? Well, let's first of all transform this function into a slightly different way, slightly different look actually, it's exactly the same way. I can say that this function is F at Y, where Y is equal to G of X. So, I didn't really do anything new right now. I have specified this compound function in this language. It's basically changing up the language, nothing more than that, right? Okay. Now, what I would also like to use is these delta notation. So, we have delta X, right? That's the infinitesimal increment of argument X. Now, we also can talk about G of X plus delta X minus G of X as delta G of X, correct? Now, we also can talk about delta Y, which is actually, since Y is my G of X, is actually delta G of X, right? Now, on another hand, I can always express F of delta Y as function F. What am I saying? No, what I meant is that using this language, I can have delta F of Y as F of Y plus delta Y minus F of Y. That's what I meant. So, we defined delta function of function G in terms of X. We defined the delta function F in terms of Y. But at the same time, I can obviously say that my derivative is F of G of X plus delta X, right? Minus F of G of X divided by delta X, limit, obviously. So, that's the derivative, which I would like to find out somehow. Now, let me express it in this particular language. Instead of G of X plus delta X, I will use G of X plus delta. Can I do it? Obviously. Right? From here, G of X plus delta X is equal to G of X plus delta G of X. So, this is F of G of X plus delta G of X minus F of G of X. And I missed the limit, but obviously you understand that this is the limit as delta X goes to zero. Equals. Now, G of X and delta G of X, I would actually like to express in this particular way. So, it's F of Y plus delta Y minus F of Y divided by delta X, right? Now, this is basically, I didn't do anything. I just expressed exactly the same thing in slightly different notation. I didn't do anything. The last thing which I would like to do actually is, instead of having this delta X, I will put delta Y and multiply by delta Y divided by delta X. Right? There's nothing wrong with that. And delta Y, again, I just use a different notation, is delta G of X. Now, why did I do it? For a very simple reason. Because this resembles the derivative of F as a simple function from some simple argument Y. Yes, later on, we will substitute instead of Y, we will substitute G of X. But right now, to use the rules for derivatives, I can just completely disregard what particular Y actually is. I can take a derivative of function F of some abstract argument Y, and then instead of argument Y, I will substitute G of X. Now, this obviously is a derivative of function G. This is the difference between this divided by delta X. And one more thing, I have this limit in front of it. So I have limit in front of this. Now, if delta X goes to zero, delta Y also goes to zero because G of X is a continuous function. So whenever argument is, but whenever two points in the argument are close, the values are also close. So I can split this limit of this product as limit of this thing where delta Y goes to zero, and limit of this thing where delta X goes to zero. So the first represents the derivative with argument equals to Y. And the second is derivative with argument equals to X. So what's my final verdict? My final verdict is that d dx of F of G of X equals to this, which is, well, I'll use this notation. d F of Y by dy times d G of X by dx, where Y is equal to G of X. That's all. And basically right now, the best thing is just to exemplify it with certain concrete examples. So here I will just put this formula here, and now I will use a different notation. Y is equal to G of X. All right, now, example number one, cosine of X squared. This is a compound function. G of X inner function is equal to X squared. F of Y is equal to cosine of Y, right? These are my two functions. Now, F of G of X is basically cosine of X squared. So now my derivative is equal to cosine Y derivative, where Y is equal to X squared times X squared derivative. Now, the derivative of a cosine is minus sine of Y. The derivative of X squared is 2X. Now, Y is equal to X squared. We know that, right? So it's 2 minus 2 sine of X squared X. So that's my final derivative using this compound formula. All right, next. Okay, now this is my formula. Second is, basically you can consider it as a compound formula because this is 1 over cosine of X, right? By definition. Second by definition is 1 over cosine. Now, before, when I was talking about examples of trigonometry, and then I was talking about examples of the function 1 over F of X, if you remember. It's, I think it was just the previous lecture. I separately derived that this derivative is equal to minus F divided by F squared. I independently derived this formula. However, now, what we can do, we can use the compound formula for compound functions because this is a combination of two functions. Now, inner function is cosine X, and outer function of Y is equal to 1 over Y. And I can, I know what's the derivative of each one of them is. So my formula based on this would be F derivative of Y times G derivative of X where Y is equal to cosine X. So that's derivative of the cosine is minus sine Y. Derivative of this cosine is, oh, I'm sorry, F of Y, that's 1 over Y. So this is minus 1 over Y squared. The derivative from 1 over Y is 1 over Y squared. And derivative of the cosine is minus sine of X, right? Now, Y is cosine, so minus and minus, so I have cosine squared X and sine X. So that's my answer. Now, if I'm using this thing, it probably would be the same thing, right? Let's check it out. So F is cosine, so minus F derivative is sine, and the minus would be minus sine, and another minus would be sine X. And F squared is cosine squared, the same thing. Yeah, the same thing. So this formula actually gives exactly the same thing. I would be surprised if it would not, but obviously it gives the same answer and it's supposed to, and this is the answer. So we can actually use many different approaches to take derivative, right? But in some cases, using this combined formula style approach is probably a little bit easier than to derive the derivatives of this type directly with going to a limit, etc., etc. Okay, next example. Next example is e to the power of sine X. Now, inner function is sine, and outer function of Y is equal to e to the power of Y. So my derivative is derivative of the outer function, which is e to the power of Y, derivative from e to the power of Y by I, but by Y is e to the power of Y, right? Now, my inner function derivative is cosine X, and Y is g of X, right? So it's e to the power of sine X times cosine X. That's the result of the derivative. Next one. It's really very, very simple exercises. Sine square of X. What kind of functions do we have here? Well, g of X is equal to sine X. f of Y, now this is Y, and f of Y is equal to Y square, right? So first we do sine, and then the result of this we square, separately doing the derivatives. So derivative of the f would be 2Y, right? Derivative of the Y square is 2Y. Derivative of sine is cosine. So it's 2 times Y, which is sine X times cosine X, which, by the way, happens to be sine of 2X, right? Okay. Next. Next I will use the same example as I was using before, but we'll do it slightly differently. Sine of 2X. Now, you remember to take the derivative of this function before, what I was doing was I converted it into this, and then I was using the formula for derivative of the product, remember? f times g is equal to fg plus fg. Okay? First derivative of first times second, derivative of second times first. So the derivative of this is, well, 2 is just a factor out. So it's sine by derivative of cosine, which is minus sine. So it's minus 2 sine square X, right? Plus derivative of this, which is cosine X, times this one. So it's square. So it's 2 cosine square minus sine square, which is 2 cosine of 2X, right? Remember cosine square minus sine square is cosine of 2X. So this is the way how I did it using the formula of the derivative of the product of two functions. Now, on another hand, I can use this as a composition of two functions. So the first function in the function is 2X, and the outer function of Y is equal to sine Y. And this is Y. Now, let's take the derivative using the derivative of the compound function. So first, I do this. The derivative of sine is a cosine of Y times derivative of the inner function, which is 2X. So derivative is equal to Y, where Y is equal to this, right? So it's equal to 2 cosine of 2X. You see? Exactly the same answer, but completely different way of deriving this answer, all right? And the last example which I wanted to present, also the one which we were using before to illustrate a different approach. Now, e to the power of minus X. Now, you remember that the way how I did it in one of the previous lecture, I expressed this first as 1 over e to the power of X. And then I was using this formula for the derivative of the reciprocal of the function. Now, using this formula, the f at X is e to the power of X, right? So I have 1 over e to the power of X square, which is e to the power of 2X, and minus. And the derivative of e to the power of X is e to the power of X, right? Now, I can reduce it, and I will get minus 1 over e to the power of X, which is equal to minus e to the power of minus X, right? Now, these are relatively lengthy calculations in remembering this formula, right? Instead, what we can do is consider this. I can put G is equal to G of X is equal to minus X, and this is my Y function. And my f at Y is equal to e to the power of Y, right? e to the power of Y, and Y is minus X gives me e to the power of minus X. Now, the derivative of the outer function is from e to the power of Y is e to the power of Y, and derivative of the inner function from minus X is minus 1. So, my result is this, where Y is equal to this, so it's e to the power of minus X times this, which is minus 1, which is this. And as you see, exactly the same thing, but derived in a faster way. So, in many cases, using this particular approach to take the derivative is really very, very beneficial, and it might actually help you a lot. I suggest you to read the notes for this lecture on Unisor.com just to make sure that you understand this concept completely. It's a very powerful, actually, technique to use the compound function, and it helps a lot. But what's important is to understand how I derived to this particular function all these delta of function G, delta of function F, then using Y instead of GFX, etc. So, I would suggest you to read again the notes for this lecture, they are quite detailed, and the proof is actually written there as well. So, read it and you will understand it better. That's it for tonight, thank you very much and good luck.