 Hi, I'm Zor. Welcome to Unizor Education. We will continue to take derivatives, basically, from some basic functions, and today would be trigonometric functions, only basic one, sine and cosine. Now, this lecture is part of the Advanced Mathematics course presented on Unizor.com. It's for teenagers and high school students. I do suggest you to watch this lecture from the website, because it has nice notes for each lecture. All right, so trigonometric functions and how to take derivatives from it. Well, let's just recall the definition of the derivative where delta x goes to zero. Now we will consider the function sine. Okay, so let's just use this definition and take direct limit of whatever is necessary, which is sine of x plus delta x minus sine of x divided by delta x. Now, that's equal to, we have sine of the sum of two angles, right? So it's sine x cosine delta x plus cosine x sine delta x minus sine of x divided by delta x. All right, so what do we have here? Equals two. Well, we have two things here. We have sine x times cosine x minus one, sorry, delta x, divided by delta x plus cosine x times sine delta x divided by delta x. Now, as delta x goes to zero, again, this is one of the amazing limits which we know what happens. This goes to one, right? Oops, I made delta x goes to zero and this ratio goes to one. Now, this is something a little bit more complex, if you wish. Let's just talk about this separately. I'll do this way. Cosine of delta x is equal to cosine of delta x over two plus delta x over two, right? Which is, this is cosine of the sum. So it's cosine cosine minus sine sine. So it's a cosine square delta x over two minus sine square delta x over two. Or, in other words, if sine is one minus, if sine square is one minus cosine square, right? So we will have two cosine square delta x over two minus one, right? Or, if we replace it with sine, it would be minus one minus sine square. That probably would be better for us. So it's one minus sine square and another sine square plus sine square delta x over two. Yes, that's what we actually need. Now, this is the cosine x. Now, and if I will subtract this one, what happens? The result would be sine x times, it's minus two. So it's minus two sine square delta x over two divided by delta x. And here we see a very interesting plus cosine x and I will not really write this anymore because I know it will go to one anyway. Now, here is what's interesting point is. I know that sine of delta x over two divided by delta x over two, which I can actually put it here, right? That's the same thing. So sine of delta x over two divided by delta x over two goes to one if delta x goes to zero. But I don't have sine. I have sine square, which means sine times sine. So one sine and delta x, sine of delta x over two and delta x over two will go to one. But another sine will drag it to zero because delta x goes to zero. So the whole thing actually will go to zero as delta x goes to zero. Because again, one of these two sines is a sine times sine. One of them divided by delta x will go to one and another one would be zero in limit. So the only thing which remains is cosine. So the f prime of x is equal to cosine x. The derivative of sine is a cosine. Again, something which is quite interesting actually that trigonometric functions are related to each other in this particular way. It's kind of unusual, I would say, at least for me. All right. And what is the derivative of cosine? Well, if you are guessing it's a sine, it's almost correct. But let's check it out. So what do we have? We have cosine of x plus delta x minus cosine of x divided by delta x. Again, we will use the formula for the sum of two angles. So that's cosine x cosine delta x minus sine x sine delta x. And minus cosine x divided by delta x equals to cosine x times cosine delta x minus one. This plus this and this divided by delta x minus sine x and sine delta x divided by delta x. Now, as delta x goes to zero, this goes to one as we know. And from the previous problem actually, we did exactly the same thing. This thing goes to zero because this thing is actually with a minus sign. It's two sine square of delta x divided by delta x. So this square with a sine of delta x actually makes the whole thing going to zero. And the result is minus sine x. So for function equals to cosine x, derivative is equal to minus sine x. Okay, so these are basically two very important trigonometric functions. We will discuss some other like tangent, etc. But right now I think it's enough just to have this feeling of how exactly we derive with derivatives of certain basic functions. Well, that's how we just go and directly calculate whatever we can. I do suggest you to take a look at the nodes because the nodes contain basically exactly the same proof, very short proof of whatever I was just talking about. It's just good to do it. And I also suggest you to try to repeat yourself everything for x to the power of n, square root of x, e to the power of x, a to the power of x, sine of x and cosine of x. These are my last three lectures of these derivatives of elementary functions, which I was just trying to explain. Do it yourself. I think it's a very good exercise. And it would help you actually to understand how the derivatives behave better. Okay, that's it for today. Thank you very much and good luck.