 Hi, I'm Zor. Welcome to Unizor education. I think it's about time to start doing certain exercises and problems with derivatives, because basically whatever I wanted to talk about derivatives from the theoretical perspective, basically I finished. Now, this lecture and all others are part of the advanced mathematics course for teenagers and high school students. It's presented on Unizor.com website. I suggest you to watch all these lectures of this course from the website because every lecture has very detailed notes. There are no advertising on the website and the site is free, so it's much better to watch it from there and read everything, all the notes. Besides, you have an option to sign in into the website, in which case you will be offered certain exams. You can take any exam as many times as you want, so everything would be much more resembling the real educational process. All right, so today we will discuss six different problems related to basically just taking derivative. It's just exercise of how to take derivatives. So let's do it one by one. Number one is just regular polynomial. So let's say you have a polynomial of n's degree. Now, you remember that if you have a function, its derivative is this. We have proven it before. Now, this is a sum of certain number of functions and derivative of sum is equal to sum of derivatives, right? So if I would like to make a derivative of this, and I'm using this particular notation. There are many notations for derivatives. I choose this one in this particular case. Then it's equal to sum of derivative of all of these. Now, this is a factor multiplier, constant multiplied by some function. And derivative of a constant multiplied by a function is equal to that constant multiplied by derivative of a function. So it's equal to an times derivative of x to the power of n. And this in turn equal to, there is one nuance here. So derivative of x to the power of n is n x to the power n minus one. Now, we cannot put n is equal to zero because if n is equal to zero, then this is a constant. And derivative of a constant is zero. So the first member of this sum, which used to be a zero, basically times x to the power of zero, which is one. So just a zero, it would be zero. And so you better start from one, this summation. That would be the result of it. That's it. So we have covered all the polynomials in one shot. Now all different exercises where polynomials are involved, they are all part of this general derivative of a polynomial. And these are the easiest kinds of derivative exercises, whichever possible, right? So the power function is the simplest one. Now next one, I have this. Now we all know this particular identity. So if two functions are equal, their derivative must be equal. So I'm just trying to prove that if I will take derivative of this and it will be the same as derivative of that, right? So that's exercise basically of how to take derivative of trigonometric functions. And also there is a product of two functions. So that's another exercise. All right, so we have to prove that derivative of this is equal to derivative of that. Excuse me. Well, what is this? This is a combined function, right? It's a function of function. First function is equal to 2x. Second function is equal to sin of x. And now we are talking about g of f of x. It's a compound function, right? Now you remember how we derive, how we find the derivative in case of compound function. So first we forget about inner function and consider this to be a function of some argument. Let's say y, if you wish. And just take a derivative as if it was a plain function, sin x. Now the derivative of sin of x is cosine. But now instead of x, we put 2x because that's the real argument. And inner function has to be also differentiated. And its derivative should be multiplied by this expression. So the derivative of 2x is 2, right? So basically that's what we have. We have 2 cosine of 2x. Now let's go from this side. Well, first of all obviously 2 is a multiplier. And multiplier always goes out of the derivative automatically. So we will have the derivative of this is equal to 2 times derivative of sin x times cosine x. Now derivative of a product. Derivative of a product of two functions is the first function times derivative of the second. Plus second function times derivative of the first. So what do we have here? First function sin times derivative of the second. Derivative of the cosine is minus sin. So we have minus sin square of x. Plus derivative of the first function times the second. Derivative of the sin is a cosine times cosine is a cosine square. Now let's compare what we have. We have here and this. Are they the same? Yes they are. Because cosine of 2x is equal to cosine square minus sin square. That's an identity of the trigonometry. So we have identical results. And obviously we expected that. But what's important is that here to get to this one we were using one set of rules for derivation, for differentiation. Primarily the compound functions. And in this case we were using a derivative of a product of two functions. So two completely different rules into basically different cases. They lead to the same result because we started from different representation of the same thing. That's my exercise number two. Okay. Let's move on. By the way with trigonometry it's interesting, right? The derivative of a sin is cosine. Derivative of the cosine is minus sin. So it's kind of close except the sin, right? Now there is another set of functions. They are called hyperbolic sin and hyperbolic cosine. Here is their definition. Sin hyperbolic of x. By definition is this. e to the power of x minus e to the power of minus x divided by 2. And the cosine hyperbolic is equal to, by definition, e to the power of x plus e to the power of minus x divided by 2. Now why are they called sin and cosine even with the word hyperbolic and the letter h here? Well, because their properties, quite frankly, are really close to regular sin and regular cosine. And let me just illustrate it by taking derivative. So what's the derivative of sin? Hyperbolic sin. Well, obviously you divide by 2 that retains. The derivative of e to the power of x is e to the power of x. Derivative of e to the power of minus x, you can consider it to be the compound function. So you have minus. You have e to the power of minus x times derivative of an inner function, which is minus x, which gives me minus 1. So that would be plus, which obviously happens to be hyperbolic cosine of x. You see, very much like regular trigonometric functions. Now let's do the hyperbolic cosine. Okay, the derivative is e to the power of x plus e to the power of minus x times the derivative of the inner function, which is minus 1, which gives me minus here, which happens to be hyperbolic of sin. So almost the same like in trigonometry. Because in trigonometry, there is a minus sign. The derivative of the cosine is minus sign. In this particular case, they are completely symmetrical, so to speak. And that's not only the resemblance. You remember that in a regular trigonometry, sin square of x plus cosine square of x is equal to 1, right? How about here? Just out of curiosity. It's not related to differentiations or any other property which we are discussing right now. But just for fun, let's do it. The hyperbolic sin square of x is equal to square of that is 4, right? Square of e to the power of x is e to the power of 2x minus 2 times this times that. And they are canceling each other. So it's minus 2 and plus e to the power of minus x. Now what's the cosine hyperbolic of square? Also 4. And here we have e to the power of 2x plus 2 plus e to the power of minus x. And what can I say? I can say that cosine hyperbolic square minus sin hyperbolic square is equal to, this is all canceling out, 2 minus minus 2, that's 4 divided by 4, 1. You see, again, very close to whatever we have with trigonometry except the sine. So that's basically two main things. The differentiation is symmetrical and this formula instead of plus we have a minus, but again it's resembling. So this resemblance actually was sufficient to call these hyperbolic or sine and hyperbolic or cosine. So that's my exercise with hyperbolic functions. Now let's go back to trigonometry. Let's consider second and co-second and the co-second. Now, again, we will consider them to be, in both cases, compound functions. So the first function, let's talk about this one, would be cosine of x and g of x would be 1 over x. Now g of f of x is equal to 1 over cosine of x. So we can actually use the rules for differentiation of these compound functions. And the derivative of this is minus 1 over x square, right? Because this is x to the power of minus 1 and you can consider this to be like a regular power function, which means its derivative would be minus x to the power of minus 1, which is this. Okay, so the derivative of 1 over cosine x is equal to, first we differentiate 1 over x, but instead of x we substitute cosine square, which is minus 1 over cosine square of x. Times derivative of the inner function. Inner function is cosine and its derivative is minus sine of x. So the result is minus and minus sine divided by cosine square. Or if you wish you can transform it a little bit, sine over cosine would be tangent and another 1 over cosine is second, so it's tangent times second if you want to. Tangent times 1 over cosine is second. Okay, so basically differentiation of second is multiplication by tangent, if you wish. I mean it's just kind of an accident I would say. Alright, now let's talk about cosine. Same thing, same approach. First we differentiate 1 over x function, but instead of x we substitute sine of x. So it would be 1 over minus 1 over and instead of x it would be sine square x. Times derivative of the inner function, which is sine. So it's time derivative of sine, which is cosine. And again you can transform it into what? Minus cosine over sine that's cotangent. And there is one more 1 over sine which is cosecant. So multiplication by cotangent and minus sine would be an operation of differentiation of cosine, cosecant. Alright, that's for second and cosecant. Okay, a little bit more work with tangent and cotangent. Tangent x, by definition tangent is sine of x divided by cosine of x. Now some people remember the rule for differentiation of the ratio of two functions. Quite honestly I don't. I do remember multiplication but not division. I remember if you have multiplication of two functions it's one times derivative of another plus another derivative of the first one. But for some reason I never remember this division of two functions. I can always consider this to be a multiplication, right? It's sine times 1 over cosine x. If you wish it's sine of x times second of x, right? Now second derivative of second was what? It's multiplication by tangent, right? It was tangent of x. We just did it second ago. Second of x. Okay. So this is multiplication. So it's the first times derivative of the second which is this. Times tangent of x and again second plus. So the first times derivative of the second. Now derivative of the first which is cosine of x and then second one without change. So that's the result. How can we transform it? Well, second is 1 over cosine, right? So this is actually 1. So let's just go to original sine and cosine. So sine, tangent is sine over cosine. Second is 1 over cosine. And this is 1. So what do we see here? Sine over cosine and another. So it's 1 plus tangent square of x. That's the derivative of the tangent. Okay, let's go to cotangent. By the way, I didn't mention it before, but I think it would be much better if you would do all these exercises yourself and then check against the answers which are provided on the website for each problem. Okay, cotangent. Now this is cosine of x divided by sine of x by definition of the cotangent, right? Okay, now I will do differently. I will also use the multiplication, but I will use it in this particular way. So if I would like to differentiate my cotangent of x, now this is the product of two functions. So it's a derivative of the first one times the second, which is minus sine of x times the second, which is sine of x. That's the easy part. The first one remains as is, and we have to differentiate the second one. Differentiate one over something. It's the combination of two functions, right? Compounded function. So first we do the derivative of one over something, which is one over minus one over sine square of x, right? Times product, times the derivative of the inner function, which is cosine of x derivative of the sine. And so what do we have here? Minus one minus cotangent square of x, right? Well, you probably can simplify it a little bit if you want, because this is cotangent is cosine over sine. Now this is sine over sine. So you will have minus sine square, minus cosine square divided by cosine, sorry, sine square, right? Minus one is minus sine square divided by sine square, and this is minus cotangent. Now this is minus, this is one, right? Minus one, actually, that's why minus. And what remains? One over sine square, one over sine is cos second, right? So you can write it cos second square of x. Just another representation. Similar transformation could have been done with a tangent. Personally I prefer something like this, but it doesn't really matter. Any representation is equivalent to each other. Okay, so we finished tangent and cotangent, and what's remained? Uh-huh, arc sine and arc cosine. And this is something else, which we have also addressed before. How can I find a derivative of this? Well, let's just recall the definition of what is arc sine. Well arc sine of x is an angle in radians, by the way. Sine of which is equal to x, right? So if I will take sine of arc sine x, I will get x, right? That's the definition basically of arc sine. Now, if these two functions are equivalent, identical, then I can take derivative of both of them, and I should have exactly the same result. Well, derivative of this one is one. Derivative of this is a compound function. First we have to differentiate the outer function, which is cosine of arc sine of x, times derivative of the inner function, dx of arc sine of x. This is what we have to define. We should really find out what this is, a derivative of arc sine. Now, I know this, and basically I know this as well. What is this? Cosine of an angle sine of which is equal to x. Okay, so there is an angle phi. I know that sine of phi is equal to x. And the question is what is cosine of phi? Well, obviously sine square plus cosine square are equal to 1, which means that my cosine of phi is equal to square root of 1 minus sine square of phi, which is square root of 1 minus x square, right? So, this is this part. Now, I did not actually go deeper into the sine of this, whether angle, you see sometimes sine can be positive, sometimes negative, cosine can be positive or negative. We have to really examine where exactly the angle is to make plus or minus with this square root. But considering that arc sine of x is an angle between minus pi over 2 and pi over 2, by definition, if you remember, this is exactly where cosine is positive. So, this is the correct formula. In which case the derivative of arc sine of x is equal to 1 over square root of 1 minus x square. Okay? And let's do the same with arc cosine. Okay, arc cosine of x. Well, again, this is an angle. Where is this angle? It's from 0 to pi. Again, by definition of arc cosine. You remember it's needed to have only one single value for an angle because there is a period, periodicity of the functions, cosine and sine. So, arc cosine produced the values from 0 to pi. Okay, now, what can I say about this? Well, again, let's start from the obvious. Cosine of arc cosine of x is equal to x. Let's take derivative of both. Now, this is by definition. Let's take derivative of both. Derivative of the cosine is minus sine times derivative of the inner function, which is exactly what we need to. And it's equal to derivative of the x, which is 1. Now, what is this? Well, exactly the same thing. Sine of arc cosine of x is equal to square root of 1 minus x square. But let's think about the sine. Do we need some kind of a sine? If this angle is from 0 to pi, sine is positive from 0 to pi. So, this is correct formula. So, which means, now, this minus actually stays, which means the dx of arc cosine of x is equal to minus 1 over 1 minus x square. So, what's interesting, by the way, is that this part is the same, but with arc sine is plus and with arc cosine is minus. Well, it happens. Okay, these are very, very basic manipulations with different functions, which normally people deal with in mathematics. And examples of some techniques which were used. Well, what kind of techniques? Well, first of all, we were using that the sum of functions, derivative of the sum of functions is equal to sum of derivatives. A constant multiplied by a function goes outside of the derivative. Derivative of a product is equal to first times derivative of a second plus second times derivative of the first. What else? We were using the compound functions, derivative and we were using this indirect method, which in turn is using compound functions to calculate the derivative. Well, that's basically the most important techniques which we were using. I do suggest you to go to website unizor.com and again, solve just by yourself if you didn't do it before all these six exercises and there are answers on the website so you can check yourself. Well, that's it. Thank you very much and good luck.