 Hi, I'm Zor. Welcome to Unizor education. Well, let's continue talking about properties of derivatives. In the previous lecture, we have learned that the derivative of the linear combination of two functions is a linear combination of derivatives. Very simple thing, right? Well, in particular, you can obviously say that if you have some of two functions, it's also a linear combination, right? With coefficients one is equal to some of their derivatives. Very simple thing. Derivative of a sum is equal to sum of derivatives. Well, it would be great if we can say exactly the same about the product, right? That the product of derivatives derivative of a product is equal to product of derivatives, but this is not true. And I'm going to prove right now a completely different formula which shows that with the product it's not as simple as as it is with linear combination. Okay, so and by the way, I do suggest you to watch this lecture from Unizor.com and look at the notes for this lecture because notes are very detailed and whatever I'm writing right now everything is in the notes like textbook. So we have to find a derivative of the function which is a product of two other functions. Now, as usually, I will basically use the notation like delta of h of x which is equal to h of x plus delta x minus h of x and same thing delta f of x is f of x plus delta x minus f of x and and delta g of x is g of x plus delta x minus g of x. So my derivative of the f is actually limit of delta f of x divided by delta x as delta x goes to zero and same thing with g and h, right? So right now our purpose is to find the ratio between delta h of x and delta x and take the limit. So what is delta h of x? Well, it's the h of x plus delta x which is f of x plus delta x times g of x plus delta x minus minus h of x which is f of x times g of x. Well, that's not easy to subtract. You know with linear functions, with linear combination of functions, I can regroup them and delta of the combined function would be linear combination of delta of individual components. Here, I cannot regroup anything. It doesn't work this way because this is the multiplication and this is subtraction. So what should I do? Okay, here is a little trick. Let's subtract and add the same thing which is f of x plus delta x times g of x. I subtract it and then I will add Why is it better? Here is why. Because here I can factor out f of x plus x plus delta x and what do I have in parentheses? g of x plus delta x minus g of x, right? Which is delta g of x. Correct? Now, here I can factor out g of x and what will be multiplied by f of x plus delta x minus f of x, which is delta f of x. Now these two, as delta x goes to zero, are infinitesimals. These are not. These are just functions. These are values of the function around point x. This is exactly at point x and this is not exactly at point x, but it's in a very, very close proximity of the point x. And as delta x goes to zero, f of x plus delta x would actually be closer and closer to f of x, but not to zero, presumably. Alright? So, as I know this, I can say the following. That delta h of x divided by delta x limit as delta x goes to zero is. Now I have kind of linear combination because this is easy. This is a little bit more difficult. But now the limit of sum is sum of limits, right? So it's limit of f of x plus delta x times delta g of x divided by delta x plus limit of g of x times delta f of x divided by delta x, right? So if I divide this by delta x, my delta x is here, grouped with delta g of x and this is a separate thing of delta x. And plus limit of this thing, and again, I put delta x under delta f of x. So this purely artificial trick of plus and minus sum expression gives me the ability to basically express the derivative of function h of x limit, basically, as a sum of two limits. So what are these limits? Well, let's just think about it. Limit of f of x plus delta x as delta x goes to zero is equal to f of x because f of x is differentiable function. Therefore, it's continuous function. And obviously if we are approaching argument to value x, then the value of function is approaching f of x. That's the consequence of f being a continuous function. Now, this limit now is a product of two different things. So this goes to f of x, as I was just saying, and limit of delta g of x divided by delta x, as delta x goes to zero, is g derivative of x, right? That's the definition. Now, in this case, g of x completely independent of delta x, right? It's a constant, basically. From the limits perspective, g of x is just a constant. So this thing is equal to g of x can be factored out. And what will be left limit of delta f of x divided by delta x? That will be derivative of function f of x. So what do we have as a result? Well, as a result, we will have limit of the product of two variables, each one having the limits is equal to the product of their limits, which is limit of this, which is f of x, limit of this, which is derivative of g of x plus g of x goes as a factor out, and limit of this is f of x. And this is the final formula. So as you see, the derivative of a product is not the product of derivatives. However, it can be expressed in a nice and symmetrical form, which looks like this. So it's the first function times derivative of the second plus the second times derivative of the first. You can remember it probably even this way. So it's a little bit more complicated than with a linear combination of the functions. There everything is straightforward. Here it's not, and you do have to remember, basically. Formulas like this must be remembered. Okay, now let's use this function, I mean this formula to derive a few derivatives. x cubed times cosine x. Derivative of this. So this is the product of two functions, x cubed and cosine x. So it's the first function, which is x cubed times derivative of the second, which is sine of x with a minus sign, right? Plus the second, which is cosine x times derivative of the first one, three x square. So that's the result. Next, d of three x cubed times two to the power of x divided by dx. Okay, use different notation. d by dx. Okay, well obviously three is a factor which can always be factored out. Then I will have this product. So the derivative would be three. First function, which is x cubed times derivative of the second plus the second times derivative of the first, right? Three is outside and this is what's left. Well, equals to three times x cube x cube. Derivative of two to the power of x is natural logarithm of two times two to the power of x. Plus the same three, two to the power of x and the derivative of x cube is three x square. Well, obviously you can simplify it a little bit, but this is basically the result. Next. Okay, now this seems to be like one function, right? There is no product of two functions, but the problem is I have two x here. I don't know what's the derivative of sine of two x. I know with the derivative of sine or a cosine, but not sine of two x. I mean obviously I can probably start from the beginning and try to make some kind of regular calculations of the limit whenever delta x goes to zero, etc. Similarly to whatever I did with whenever I was deriving the derivative of the sine. At the same time, I can use the trigonometry to convert it into this. Sine of two x is two sine x cosine x, right? So two goes out. Now I have a product of two functions, sine and cosine. So it's the first function times the derivative of the second one. So it's sine x times derivative of the cosine, which is minus sine. So it's minus sine square now. Plus second function, which is a cosine x, times the derivative of the first. The derivative of the sine is cosine, so it's square. So it's two. And what is a cosine square minus sine square? Well, that's the cosine two x. Now whenever I will talk in one of the next lectures about composition of the functions, because this is the composition function two x, x multiplied by two, it's one function. And another function is a sine. So it's a composition of two functions. And if I will use the rules of the composition of these two functions, we will have exactly the same result. But that would be in one of the next lectures. Right now we're talking only about the product of two functions, and I was using this rule to get to this formula. And in the lecture about composition of two functions, I will derive exactly the same formula differently. Okay, and I also have another example, which is also on the product, I'll use dx, notation. So it's x cubed five to the power of x and cosine x. So now we have a composition. Sorry, not the position, product. Product of three functions. But I can use this rule by grouping basically functions. So first I will group it this way. So it's equal to the first function by derivative of the second one. So it's x cubed five to the power of x derivative of cosine is minus sine minus sine x plus cosine x times derivative of x cubed times five to the power of x. And now I have only two two functions, which I'm multiplying, right? So I can use first function times derivative of the second one, which is natural logarithm of five times five to the power of x plus second function times derivative of the x cube, which is 3x square. So that's basically the result. I mean, obviously I can simplify it a little bit, which I do by the way in my notes. But this is how the three functions multiplied by each other can be... You can use this the same rule for product of two functions, just grouping the functions in separate groups. And obviously you can expand it to any number of functions which are multiplied one by another. So that concludes my examples series. And this is basically how we can derive, how we can make, we can take a derivative of product of two functions. I do suggest you to read all the notes for this lecture, especially this little trick which I made to reduce the derivative of the product of two functions into something which basically is expressed only in terms of derivatives of the components of this product. That's it for today. Thank you very much and good luck.