 Hi, I'm Zor, welcome to Indizor Education. I have explained basically the concept of partial derivative for function of two arguments. Well, we primarily will consider the functions of two arguments, although the theory actually actually is very easily extended to functions of multiple variables. But let's talk about functions of two arguments. So we have explained the theory behind it, behind this concept. And I would like to basically exemplify it with a few very, very easy examples. Now, I do suggest you to actually do it yourself first before even listening to whatever I'm saying and then check the answers. It's on the website, it's on unizor.com. That's where the whole course actually is located. So every lecture has notes and all these problems are in the notes. So do it yourself. There are answers to each of the six problems in the notes and then check if you are right. Okay, now I will do exactly the same and if you for whatever reason failed to get the same answers, well, let's just listen to the lecture and see what happens. All right. Just one small comment, which I usually say in the very beginning. This course is presented on unizor.com. It's for teenagers and high school students who would like to enhance their knowledge in mathematics and develop creativity, analytical thinking, logic, et cetera. All right, so the problems. All these problems are, okay, let's take derivatives. Now all the functions are functions of two arguments. So I will take two derivatives, function by one argument and then function by another argument. Okay, so let's do it. It's really very quick thing. Number one, function is square root of x, y. Now I'm interested in partial derivative by x. Now what does it mean to take partial derivative by x? It means assume that y is actually a constant which makes this the function of one argument, which is x and y is a constant, right? And then we know how to take derivative of this function. This is basically the same as square root of x times constant, right? So first we use the chain rule. First is outer function is square root, which is something in the power of one half. And this is the power function, which means our derivative is the power, which is one half times whatever is under the square root will be in the power of one half minus one, which is minus one half, which means it will be one over one over x, y. And then we will, according to the chain rule, we will go into the function under the square root, which is x times constant. And derivative of this is equal to this particular constant, right, which is y. So the whole thing is y divided by two square root of x, y. Now similarly, let's take the partial derivative by y. That means we assume that x is a constant. And then basically all these considerations are exactly the same. So it's square root of some constant times y. So that would be one half, one over one over x, y, which is x, y to the power of minus one half, right? Times derivative of the inner function, which is since y is a variable and x is a constant, so derivative by y would be x. So the whole answer is x divided by two square root of x, y. That's it, very simple. What's most important when you're taking derivatives, partial derivatives, you have to really know exactly which is your variable and which is the constant. So in this case, whenever we are differentiating by x, we assume that the y is constant. And whenever we are differentiating by y, we assume that x is constant. And by the way, if it's a function of n arguments and we are trying to make a partial derivative by one of them, we assume that all others are constants. So we are always dealing with a function of one argument. In this case, argument is x. In this case, argument is y. This is the argument by which we are differentiating. And all other arguments are just constant. You have to pay attention to them in as much as you pay attention to different constants. Next, e to the power of x, y. Okay, so again, differentiating by x gives me what? y is a constant. So we have again a composition of two function. Inner function is some kind of a constant, y multiplied by x. Outer function is e to the power of something. So first, we are differentiating the outer function, which is e to the power of x, y, right? Because the derivative of e to some power is e to that sum power. Now we are multiplying by the derivative of the inner function. Now inner function is x, which is variable times y, which is a constant. And that's why whenever we are differentiating by x, it's just the constant y. So this is the answer. And whenever we are differentiating by y, it's kind of symmetrical. We start from the same thing. The derivative of the outer function times derivative of the inner function, y is a variable, x is a constant in this case, right? So the derivative is x. So these are two derivatives, partial derivatives. Next. OK. Number one, differentiating by x. So y is a constant. So it's like 1 over x squared plus 1, or 1 over x squared plus 25, or whatever else. Now the derivative is, well, first since is something to the power of minus 1, right? It's x squared plus y squared to the power of minus 1. So derivative of the power function is we have to put minus 1 times power minus 1, right? So power was minus 1, so it will be minus 2. So it will be 1 over x squared plus y squared, right? That's what power minus 2 means. And then since, again, we are applying the chain rule, derivative of the inner function. Inner function is x squared plus y squared. y is a constant, so we can differentiate without it, because the derivative is 0. So we have only derivative of x squared, which is 2x. So it's 2x. So the whole thing is minus 2x divided by x squared plus y squared squared. Now differentiating by y, well, I'm sure you understand this is a completely symmetrical kind of things. And the result would be very similar to this, except it would be y instead of x squared plus y squared, right? Because the beginning will be the same, and at the end, we have to multiply by the derivative of y squared, which is 2y. That's it. You see, these are relatively easy exercises, because if you know how to differentiate functions one argument, you automatically know how to differentiate function of multiple arguments partially, because whenever you're differentiating partially, you just have only one argument anyway, and all others are constants. So these are just exercises. sine x divided by y squared. OK, partial derivative by x. So x is variable. y is constant. So that basically means that y as a constant goes outside, and then we will multiply by derivative of the sine, which is cosine of x. Cosine x divided by y squared. Now, derivative by y. You see, now it's not symmetrical at all, right? So right now, we have to do separately one from another. OK, now, differentiating by y. That means that the sine x is actually a constant, right? So as a constant, it goes out times the derivative of y of 1 over y squared, or y to the power of minus 2, which means it would be minus 2 y to the power of minus 3. So that's what it is. So it's minus 2 sine x divided by 1 cube, y cube. Next, arc tangent of x square root of y. OK, again, this is non-symmetrical kind of things. All right, differentiating by x. So we have a composition of two functions. Outer function is arc tangent. Inner function is a constant. Square root of y is a constant, right? Multiply by variable x. OK, so the chain rule, first derivative of arc tangent is 1 divided by 1 plus this square, which is x square y. So if you don't remember, again, I'm just saying arc tangent of, let's say, t. Derivative is 1 over 1 plus t square. So I just happened to remember it. How to derive it, we did it when we were considering derivatives of regular functions. OK, times derivative of the inner function. So what's the inner function? It's a multiplication of a constant by x. So if we want to have a derivative y, x is just this constant. That's it. Now, by y, it's slightly different. Well, first of all, we have to start from the same thing. We start from the derivative of arc tangent, which is 1 over 1 plus argument square. Now we have to multiply by the derivative of the inner function. x is a constant, and y is the variable, right, in this case. So x goes to the top. It's just a multiplier. And derivative of square root of y, this is 2 square root of y in the denominator, right? Again, square root of y derivative is 1 over 2 square root of y. So that's the answer. And my last problem, which is, again, relatively easy as everything else, you see, whenever you're doing differentiation, you don't have any kind of creativity thing, as you do have with integrals. So these can be done relatively mechanically as long as you understand and remember regular derivatives of regular functions. So my last example is this. Now, differentiation of z by x. Now, if x is a variable and y is a constant, well, if you remember, we used to use a to the power of x. But that's exactly the same thing, a is a constant or y is a constant, just different letter. And the derivative of this exponential function is itself times logarithm of the base of my exponent. Now, if you want to differentiate by y, now x is a constant. Now it's just a power function. Since x is a constant, y is variable. It's a power function. Power function is you have the power x times power minus 1. That's it. So all these examples are just to emphasize how easy it is to deal with partial derivatives. It's exactly the same as with derivatives of regular functions of one argument. Again, I do suggest you to do all these exercises yourself. All these examples are on unizord.com. And check if you had the same answers as presented on the website. Other than that, we have finished. Thank you very much and good luck.