 Hi, I'm Zor. Welcome to Unisor Education. Continue talking about partial derivatives. This lecture is part of the advanced course of mathematics for high school students and teenagers. It's presented on unisor.com. I suggest you to watch this lecture from the website because it has notes for each lecture, very detailed notes, so you can just use it as a textbook, plus it has exams. If you are willing to subject yourself to tests, and the site is completely free and there are no advertisements, so it's outside the financial realm. Now, the previous lecture was about examples of partial derivatives. So for a certain number of functions, I was derived the first derivative by the first argument and by the second argument. All functions were the functions of two arguments. Now, today I would like to continue taking basically derivative from the derivative. In the area of regular derivatives for the function of one argument, it's called the second derivative. It's derivative from the first derivative. Now, in the realm of partial derivative, since we have two different arguments, I can have the first derivative by one argument, let's say by x, and then I can take the second derivative either by x or by y. Similarly, I can take the first derivative by y and then the second derivative by x or by y. Now, the interesting fact, which I did mention in one of the previous lectures, was that if you are taking the second mixed derivative, which means one argument is x and another is y, or one argument is y and then x, the results in a very broad category of functions should be the same. In other words, the second derivative, which is first by y and then by x, or we're getting the first derivative by x and the second by y, result should be the same. So now this lecture is to demonstrate that this is the fact. And what I'm going to do is I'm going to use the results of the first lecture, the previous lecture, where I just presented all these examples. And for each function, I had both derivative by x and derivative by y. What I will do now, I will take the derivative by y from the derivative by x, basically getting this. And then I will take derivative by x from whatever I got for a derivative by y. And that will compare the results. They must be the same, right? So let me just illustrate that this is the fact. So my first function was, and I'm going to use the results of the previous lecture. So dz by dx is equal to y divided by 2 square root xy. dz by dy is equal to x 2 square root of xy. Now, this is what has been the subject of the previous lecture. I derived both of these. Now what I will do, I will have the second derivative from this one, but now by y. So it would be d2 dy dx from y divided by 2 square root of xy. Let's calculate what this is. Well, to make my life easier, I would put it as 1 half x to the power minus 1, 2, 1 half, right? This is square root of x in the denominator. So it's minus 1 half times y to the power of 1 half. Now, I am getting the derivative by y right now. So this is a constant. So it's only by y. So the result would be by y. And the result would be, well, this is a constant. So it's 2 square root of x stays. Now this would be 1 half. So it's 2 would be here. And y to the power of minus 1 half. So it's square root of y, right? Am I right? Oh, sorry. Why do I have 2 in the numerator? It's 1 half. So which is equal to 1 over 4 square root of xy, right? Now this is my first derivative by y. Now I will take the partial derivative of this by x. So I need 2 d2 dx dy from x divided by 2 square root of xy equals 2. So again, let me just rewrite it. It would be x to the power of 1 half times y to the power of minus 1 half, right? And I am differentiating by x now. So 2 remains y to the power of minus 1 half is square root of y. And all I have to do is differentiate square root of x basically, which is 1 over 2 square root of x. So it's 4 square root of xy, the same thing as here. So as you see, the mixed derivatives first by x then by y or first by y then by x are the same. Okay, next. So this is basically both for illustration that the mixed derivatives are the same and also just another practice in differentiating. Really kind of a straightforward thing, but it requires certain accuracy. So my function is e to the power xy. Now my first derivative by x is equal to y e to the power xy. My first derivative by y is equal to x e to the power xy. Now I would like to differentiate this by y and this by x and compare the results. So this is differentiating by y. So it's dy dy dx of y e xy. So we are differentiating by y. So we are basically considering this as a product of two functions. So it's derivative of this times this plus derivative of this times that, right? So derivative of y times this. So x is a constant. So that's derivative of y is 1, so it's e to the power xy plus y and derivative of this. Derivative of exponent is this exponent times x1 plus xy times x to the power of xy. Now this we have to differentiate by x. Same thing, differentiating by x by parts. No, not by parts. Sorry, that's integral. It's the derivative of a product. So it's this one times this, which is e to the power xy. Now y is constant. This is variable plus x times derivative of e to the power of xy by x. So that's y times e to the power xy. And I have exactly the same result as before. So again my mixed derivatives are the same. Okay. Two down, four to go. Okay, now my first derivative by x is equal to minus 2x divided by x squared plus y squared squared. And my derivative by y equals minus 2y minus 2y divided by x squared plus y squared squared. Now this I will differentiate by y. So x is a constant actually. So d2 by dy dx equals 2. So x is a constant, right? So minus 2x stays as a multiplier. Now here I have some expression to the power of minus 2. Now if I have something, power minus 2, I will have minus 3. I will have minus 2 times the power of minus 3 times derivative of the inner function, which is, well, I'm differentiating by y right now. So x can be omitted. So it's 2y. 8xy divided by x squared plus y squared cubed. Now this I have to differentiate by x. Again, minus 2y is just a multiplier. That's a constant. Now here I have basically similar thing. Minus 2 as a coefficient, x squared plus y squared the power of minus 3 times. I'm differentiating by x. So I have 2x here. And what do I have? Minus and minus it's plus. So 2 and 2 is 8xy, yx and x squared plus y squared to the power of minus 3, which is 3 in the denominator. Same thing. Good. Let's continue our struggle. Number 4. Sine x divided by y squared. Now my derivative by x is equal to cosine x divided by y squared. And derivative by y is equal to minus 2 sine x divided by y cubed. Now, by the way, the previous examples were symmetrical, x and y. Now this one is not. However, we still have to have the same result if we're differentiating this by y and this by x. So again, we differentiate this by y, which means that cosine of x is just a multiplier because x is a constant. And we have to basically differentiate y to the power of minus 2. So it would be minus 2y to the power of minus 3, right? Which is equal to minus 2 cosine x, y cubed. Now here, I have to differentiate by x. So 2 and y cubed are constants, so I only have sine. So it would be minus 2. Derivative from sine is a cosine and y cubed is remaining. The same result. That's fast and simple. Okay, now this is much more difficult. So we have to be very careful. Z is equal to arc tangent of x square root of y. My first derivative by x is equal to square root of y divided by 1 plus x squared y. And my first derivative by y is equal to x divided by 2 square root of y 1 plus x squared y. Okay, now we have to differentiate this by y and this by x and compare the results. Okay, so d2 by y, y to the power of 1 half times 1 plus x square y to the power of minus 1, right? That's easier for me. Equals. Now I'm differentiating by y. x is a constant. So it's a product. So it's first times second derivative or the second times first derivative. Okay, so y to the power of 1 half times derivative of this one by y, which is minus 1, 1 plus x square y to the power of minus 2 times derivative of inner function by y. That's x square plus, so what I did is the first times derivative of the second. Now the second times derivative of the first. The derivative of the first is 1 half y to the power minus 1 half times 1 plus x square y to the power of minus 1 equals 2. Okay, let me now convert it into regular fractions. So this is minus square root of y x square divided by 1 plus x square y square plus 1 over 2 square root of y 1 plus x square y. Am I right? I think I'm right. Now let me get to the common denominator. The common denominator would be 2 square root of y y plus x square y square. So I need to add here 2 square root of y and I will have minus 2 x square square root of y and square root of y is y. Now here I have to add 1 plus x square y because this is a square and this is a single. So plus 1 plus x square y. So what will be? Well, this is minus 2 x square y, this is plus, so that would be 1 minus x square y divided by 2 square root of y 1 plus x square y square. That's my answer. I'll put it here. 1 minus x square y divided by 2 square root of y 1 plus x square y square. Okay, now let's do it through a different procedure. So differentiate this by x. So it's d2 by dx dy. So, and here I will have x times 2 times y to the power of minus 1 half, right? Since it's here. 1 plus x square y power minus 1. Okay, so differentiating this by x is... Okay, y is the constant. So 2 y to the power of minus 1 half stays. y2, it's 1 half. Sorry, it's 1 half. Now here I have the derivative of this which is 1 times this. So I will have this plus... Now that's derivative of this times this plus x times derivative of this which is minus 2. No, sorry, it's minus 1 times square. And we are differentiating by x. So it's times 2 x y is a constant. Something like this, right? So what do I have? Okay, this is 2 square root of y. It remains. Now this is the common denominator. 1 plus x square y square. Now here this is 1 over 1 plus x square y. So to have it as a power of 2 I have to multiply it by 1, right? Like this. Minus 2 x square y. And what do I have? Half 1 minus x square 1, x square y. Exactly like this and exactly like this. So as we see it's exactly the same. Alright, that's good. Now the last one. My function was y to the power of x. So my first derivative is equal to y to the power of x logarithm y. And my first derivative by y is equal to xy to the power of x minus 1. Now this I have to differentiate by y. So d2 dy dx. So that's what? Differentiate by y. So that's a product. So first I differentiate this one and x is a constant. So that's xy to the power of x minus 1 times logarithm a. Plus y to the power of x times derivative of logarithm is 1 over i. Now y to the power of x divided by y is y to the power of x minus 1. So y to the power of x minus 1 goes outside. And inside I have x logarithm y plus 1. Now this one I have to differentiate by x. So again, that's a product. First derivative of this times this, which is this, plus x times derivative of this, which is y to the power of x minus 1 times logarithm y. And y to the power of x minus 1 outside, 1 plus x logarithm y, which is exactly the same as this one. So we did six different examples. In the previous lecture I have calculated partial derivatives by x and by y. In this lecture from every derivative by x I calculated derivative by y and from every derivative by y I calculated derivative by x. And in these two cases I got exactly the same result. Now this is not the proof that d2 dx dy is equal to d2 dy dx. However, it actually brings our feelings warmer and warmer that this probably is correct. We'll talk about this later. Anyway, that's it for today. I do suggest you to go to the website and notes contain all these examples. Try to do it yourself. So differentiate everything yourself. Make sure you do it accurately. You know, sometimes even if it's not very creative and basically taking derivative is not very creative. If you know more or less what's the derivatives of the main functions. But you need some accuracy and make sure that you have exactly the same results when you are differentiating in different orders. That's it. Thanks very much and good luck.