 Hi, I'm Zor. Welcome to Indizor Education. I would like to talk about one particular example where implicit differentiation is very, very handy. A couple of examples were in the previous lecture where I explained what exactly is implicit differentiation. And this is just one of the examples of some classical and simple function where it's very useful, actually, to have this particular technique. Now, this lecture is part of the course of Advanced Mathematics presented on Unizor.com. That's for teenagers and high school students. I suggest you to watch this lecture from the website because it has explanation, like a textbook, basically. It's notes for each lecture, very detailed notes. And in addition, there is certain educational functionality in the website. If you are signed on, then you can, for instance, take exams, which is very important as a basically development of your logical thinking, creativity, etc. The site is free, so you can take as many times exams as you want. Alright, so let's concentrate on this particular problem. So, my problem right now is to take a derivative of the function arc sin of x. Sometimes notation is sin to the minus one of x. I prefer not to use this notation because sometimes it means arc sin, sometimes it might actually mean really sin to the power of minus one, which is reciprocal. Alright, so let's forget about this. I will use arc sin. Now, before going into the detailed calculation, let's just talk a little bit about the function itself. Now, the function is the reverse of the sin, obviously. Now, the sin has a graph, as we know, something like this. And obviously, there is no real function which is defined for any reverse function. First of all, what is the domain of this inverse function? Well, the sin can take values from minus one to one. So, basically, this function has domain from minus one to one. Now, what are the values of this function? You see, the function sin is periodical, which means the same value of the function corresponds to more than one values of arguments. Actually, infinite number of arguments have exactly the same value of sin. So, if we want to talk about reversing the sin, which means from the value we should really get the argument, we really should restrict what kind of argument we are willing to allow as values of this function. So, the function basically gives you one concrete value for the argument. So, in this case, what is the angle sin of which is equal to x? Well, as we have many different angles. Now, traditionally, what people do, they just cut from minus pi over two to pi over two. Only this interval of angles are allowed. And as we see, as angle is within this particular interval, the value of the sin goes all the way from minus one to one. And this monotonically increasing function, which means it has the reverse function, which is arc sin. So, the values of this function are actually in the interval of minus pi over two to pi over two. And this is the graph of the function arc sin from minus one to one is domain. And from minus pi over two to pi over two, it's the values of the function. And as we know, the graph of this inverse function is symmetrical to the graph of the original function sin relative to the bisector of the main angle. So, at bisector of this at 45 degrees and if you will invert, you will get the sin. Okay, now we are ready to talk about derivative. Now, traditionally, how we derive the derivative for function sin, for instance, or for any other simple function like y is equal to x square or something. Well, we basically follow the definition of the derivative. Now, the definition of the derivative if you have to take increment of the function at point x divided by increment of the argument and have the limit as increment of the argument is infinitesimal value. Well, and we actually calculated exactly what is the derivative of the sin. Now, you can actually, if you want, you can review this, it's one of the previous lectures. And one of the very important points was the limit which we actually proved before that this, as x goes to zero, this goes to one. So, this is the base of the formula which we have derived for derivative of sin. Now, here, unfortunately, we cannot really do something in this particular way because if I will do increment arc sin of x plus delta x, well, it's kind of difficult to deal with. Just think about, how would you deal with this? I have no idea how to deal with arc sin of x plus delta x. When it was sin of x plus delta x, I know what is the sin of the sum of two angles. There is some kind of a formula I was using, et cetera. There is no formula for this. So, the whole method of direct calculation of the derivative might not really work here. At least, I don't know how to do it. However, there is a method of implicit differentiation which does help in this particular case because what does it mean that we have this function arc sin x? It means that the sin of arc sin of x is equal to x. That's what actually means that the value of arc sin, what is it? It's an angle sin of which is equal to argument x. That's the definition of the arc sin. So, this is just the definition of the arc sin. I mean, it falls from the definition of the arc sin. Why is this better? Well, here is why. Now, I will use this principle of implicit differentiation because this is the function and this is the function of x. For any x which belongs to this interval from minus 1 to 1, this thing is true. So, we have two functions which means I can take their derivatives and derivatives must be the same. If functions are the same, derivatives must be the same. Now, derivative of the left spot, well, this is a compound function and you remember the chain rule. If you have a compound function, first you have to take the derivative of the outer function. Derivative of the sin is a cosine of the same argument and multiply by derivative of the inner function. So, the derivative of arc sin of x, I will use prime as a symbol of derivative. And what is it equal to? What's derivative of x? Well, derivative of x is 1. Now, look at this. We basically have unknown derivative which we can determine from here. So, arc sin of x derivative equals to 1 over cosine of arc sin of x. Well, number one, I can just leave it as it is. Number two, I know that this is supposed to be somehow simplified. Let's just think about this again. Arc sin is an angle sin of which is equal to x. So, I know that the sin of this is equal to x. Now, what is the cosine? Well, obviously we know that the sin of square of phi plus cosine square of phi is equal to 1. So, cosine of the angle is equal to square root of 1 minus sin square of phi. Actually, technically I have to put plus minus here. Now, let's think about it. What is this f? It's this. Phi, I should say phi, Greek letters. So, I know that I am from minus pi over 2 to pi over 2, right? So, phi is from minus pi over 2 to pi over 2. Remember the graph of this cosine? This is minus pi over 2 and this is pi over 2. So, it's positive here, which means I can just take only the plus sign. And now I can have it equal to... So, instead of cosine of this, I will just put this. Square root of 1 minus sin square of arc sin x. But we know that sin of arc sin is equal to x, right? So, it's 1 minus... So, the answer is 1 minus x squared. So, that's my final formula. So, this is a derivative of the function arc sin. Well, it might be a little unusual, right? You remember from sin, you have cosine. From cosine, you have minus sin as a derivative. Well, you might expect that the derivative from arc sin must be something like arc cosine. I don't know. But anyway, this is not arc sin. It's completely different formula and that's actually what it is. There is nothing to argue about this. If you remember, the derivative of the function logarithm x is 1 over x, which is also kind of unexpected. But again, it is where it is. So, that's the answer. I do suggest you to go to the Unisor.com website and just read whatever the proof of the same thing actually is. It will probably help you to better understand this principle of implicit differentiation. So, sometimes the functions like this, and you can obviously do it yourself for arc cosine or anything like that. So, sometimes this implicit differentiation is very useful and it looks like otherwise you cannot really derive this formula. At least not easily. At least I don't know how. That's it for today. Thank you very much and good luck.