 Hi, I'm Zor. Welcome to Unizor education. I would like to talk today about basic properties of partial derivatives. This lecture is part of the course of advanced mathematics for high school students and teenagers in general. It's offered on Unizor.com website. I do suggest you to watch the lecture from there. You have to just go through the bunch of menus like calculus, partial derivatives, basic properties, because every lecture on this website has notes, very detailed notes, and for students who want to challenge themselves, there are exams which you can take, not for every topic, but for many of them. And it's absolutely free, so no advertisement to them. So basically go to this Unizor.com to watch this lecture. That's my better recommendation. All right, so properties of partial derivatives. Well, first of all, let me start from a trivial fact. Partial derivative is a derivative by one particular variable of multivariable function, which means other variables are fixed, which means that the properties which we have learned about regular derivatives should all be transferred without any modifications to the partial derivatives, because we are actually taking a derivative by one particular variable at a time. So in particular, let me just go one by one. For instance, we have two functions of two variables, and we are talking about their sum. Now, partial derivative of this by one particular variable, since y is basically a constant and x is the only variable we are basically differentiating by, it should be equal to the sum of derivatives. It's kind of obvious. It follows immediately from the corresponding property of the regular derivatives. Now, as an example, for instance, you have function logarithm of x squared plus y squared plus xy, and you would like to differentiate it by x. Well, that's basically sum of two functions, and it's equal to this plus this. Now, here we are differentiating by x. Well, this is actually a composition function. I do have a separate property for the composition, but I will just look ahead. Basically, it's logarithm of something, which means the derivative would be equal to one over that something, times derivative of the inner function, which is 2x. So it's 2x here. Plus derivative of xy by x, y is a constant, so it's just a multiplication. So it's derivative is equal to y. Next property. So I will do very fast these trivial properties because we have to spend some time on more specific for multivariable functions. Okay, next one is multiplication by constant. So if you have multiplication by constant, obviously it's constant multiplied by an example. For instance, you have function 2 logarithm of x square plus y square by dx. It's two times partial derivative of this, which we have already obtained in the previous example, which is this, that's from logarithm and inner function is 2x, so 2x. So basically it's 4x. That's the answer. Next is product. Again, the same thing. Remember the property for the product, first time derivative of the second or second time derivative of the first. It's exactly the same thing with partial derivative because we are differentiating by one particular variable. So this would look like this. So it's the first one times the second derivative plus the second one times derivative of the first. You want an example? Here's an example. x minus y times x square plus xy plus y square. I specifically chosen this one because we can actually multiply and simplify it and get the derivative differently, then we can compare the results, right? So if I will do d of this by dx, it's equal to the first one, which is x minus one, times derivative of this one by x partial derivative, which is what? 2x plus y, right? Plus the second one, which is x square plus xy plus y square, times derivative of the first one. Well, x minus y derivative by x is one, so it just remains. And what's this equal to? Minus 2x square plus x square. So it's minus, no, no, plus 2x square plus x square. So it's 3x square. Okay, so I have 3x square. Now, xy I have minus 2xy and plus xy, right? So this is minus 2xy and this is plus xy and another plus xy. So there is no xy plus 2xy and minus 2xy. Minus 2xy and plus xy, right? Okay. Now, y square. This is minus y square and this is plus y square. So there is no y square. So this is the own answer. Now, on the other hand, we can multiply x minus y times x square plus xy plus y square. And what do we have? We have x cube minus y cube. x cube minus y cube. Now, x square y with a minus, x square y with a plus. X, y square, this and this is a plus and this and this was a minus. So everything cancels except these two. And derivative of this by x, well, y is a constant, right? So it's 3x square, which is exactly what we have. Nice checking. Now, the last one is a chain rule. Again, completely equivalent to what we have with functions of one argument. So what if x is a function of t, y? And we want to differentiate by t. y is a constant. y doesn't really depend on anything. We are basically having a function of two arguments t and y, right? And x is just another function. So again, y is a constant. So we can completely disregard it. So it's differentiated, the derivative of f of x of t by dx times dx of t by dt. This is, again, completely in accordance with the chain rule for regular functions of one argument. f of x of t, and we would like to have the derivative of this. It's a derivative of f for x of x of t times derivative of x of t. Exactly the same thing. So that's the chain rule. Now, this is the chain rule for one argument. But we will make it a little bit more complex when we want to differentiate in case both arguments x and y depend on t. But that's in the future. Well, future in this lecture. All right, so these are elementary properties of partial derivatives, which are completely following, 100% following from the corresponding properties of the regular derivatives. Now let's talk about something which is more specific to function of two arguments. Now, specificity starts when we have both arguments changing, right? Let me just recall the definition of the regular derivative. So if you have regular function, the derivative is equal to increment of the function divided by increment of the argument when increment of the argument goes to zero, right? Now, from this, we actually derived something similar, this type of symbolics, which this means basically one infinitesimal variable. And this is another infinitesimal variable. And their ratio, we understand in the sense that this ratio is actually converges to this number. And the difference between this and this is some kind of infinitesimal variable of a higher order, right? So now, what if you have a function of two arguments? Now, both arguments can actually be incremented. So let me just try to play exactly the same game with two variables. Let me start with this, which is immediately following from this, right? So if this limit is this, then this thing is approximately equal to derivative, the increment of the function approximately equals to derivative at the point times increment of the argument. And this precision is greater whenever delta x is smaller. And whenever delta x goes to zero, this actually goes to equality, right? Okay, now, now let's consider we have a function of two arguments. And we do something like this. We increment both of them and see how it changes the increment of the function. I would like the same kind of approach, how increment of the function can be expressed in terms of increment of the arguments. So this is how it is implemented in the function of one argument. And now I'm trying to do something similarly. Again, increment of the function in terms of increment of the argument. Again, as an approximation, which is as, which goes more and more precise as increments of the arguments go to zero. All right, so let me just transform it slightly. I will do it this way. I will subtract this and add this. I didn't really change anything, just subtract it and add exactly the same expression. Now, in this particular case, what happens? I have only one argument changing. Only x, while this one, y plus delta y, remains the same. So I can use actually this formula, because now I have basically a function of one argument, x, y is fixed. So y is not incremented, only x is incremented. So I can say that this is equal to derivative by this argument. Well, in case, since I'm dealing with a function of two arguments, I have to use the partial derivative of the function f of x, y plus delta y by dx times delta x. But now this is approximation, because I'm approximating this with this. So I'm taking derivative by x and multiplying it by increment of the argument. For a smooth function, we're obviously assuming that functions are differentiable, etc. That's true. Now, how about this thing? Sorry, the second one, this thing. Here, I have exactly the same situation. My x remains the same and function is incremented only because y argument is incremented. So I can also approximate it with derivative by y in this particular case, right? This is derivative by y times delta y. So this is an approximation and this is approximation for this and this is approximation for this using this property, which is true for functions of one variable. So if my delta x and my delta y are infinitesimals and are basically going to zero, then this becomes more and more precise equality. Now, here I'm talking about two infinitesimals, right? This is infinitesimal and this is infinitesimal. So what does it mean that they're getting closer and closer? Well, let me just repeat this for regular infinitesimal variables. So let's assume you have two infinitesimal variables, a, n, and b, n. What does it mean that they're getting almost the same and they are more and more the same as n goes to infinity considering both of them are infinitesimal? It means the following. Their ratio is approaching one. So the ratio between them, if it's approaching to one, it means that they are of the same degree of infinitesimalness. So that's what basically this means. The ratio between this and this is going to one. That's what we are actually saying when we are using the following. Instead of this, I will write differential because this is a differential actually, right? It's increment, but whenever we're changing delta x and delta y to zero, we're talking about differential. So this differential is equal to this. So I'm replacing delta with g assuming that d is now an infinitesimal variable. And now I have an equality here in this sense because this is an infinitesimal and this is infinitesimal. Now, and the ratio between them is actually converges to one. And that's what we are actually having. And this is called a total differential of the function of two arguments. So you are partially deriving by x times dx. They're partial derivative by y and dy. That's what a total differential means. Now, you can actually say that this is a definition of total differential. I did not really derive. I basically defined this total differential, but I was trying to explain that it does make sense. It makes sense in terms of this is infinitesimal, this is infinitesimal increment of the function of two arguments expressed in terms of infinitesimal increments of the argument and partial derivatives. By the way, I forgot to tell, I had this f of x comma y plus f of x comma y plus delta y, right, in this particular case. And I changed it to y because when delta y goes to zero, this goes to this. We are considering that the function is smooth enough. Now, what I need in this particular case in terms of smoothness is continuity of the partial derivative. If it's continuous, then this thing, as delta y goes to zero, you can replace it with this. So that's why I forgot to explain it, but it's kind of out of this thing. So this is the definition of the total differential of the function of two arguments. Now, let me switch to a particular case which I did kind of hinted before. What if you have this? So you have a function of two arguments, but each one of them is in turn depends on some parameter t. Well, I can actually say exactly the same thing as here. I can put this formula. This is a total differential. But now, what are the dx and dy? Total increment of the function of two arguments is equal to df of xy by dx. Now, what is dx? Now, dx is an increment now, not just an argument. Now, it's a function of the argument. And again, using the properties of the function of one argument, if you have, let's say, y is equal to y of x, dy is equal to y times dx, right? Increment of the function is equal to its derivative times increment of the argument. In our case, function is x and argument is t. So, we will do exactly the same thing here. Instead of dx, I will have x by t times dt plus, similarly, d of f of xy by dy times y derivative by t by t times dt. So, this is the total differential. And in this particular case, I can have a total derivative. If I will do instead of this, I will do this, df of x of t, y of t by dt. Since it's a function of one argument, right? I can have derivative by this particular argument, which is equal to, in this case, dt goes to the left, so we will have a partial derivative of this function by its first argument and derivative. I can use this one. It doesn't matter what kind of symbolic I'm using, this one or this one. It's all the same, right? Plus, similarly, df of xy by dy times dy by dt. And this is called a total derivative. So, we had total differential and total derivative of function of two arguments, when both arguments depend on the same parameter. Okay. Now, let me just give you one simple example, how it can be used. It's actually an example from physics. There is a formula, I think it was Clyperon, Clyperon's formula for ideal gases. So, if you have some kind of a reservoir with some kind of a piston, so you can, and here is your gas here. So, you can use this piston to squeeze it. Now, the gas has the following characteristics, volume, temperature and pressure. And the Clyperon's formula is, that's the formula. If you increase only the volume with constant temperature, the pressure should decrease, right? If you increase the volume, pressure should be decreased. If you increase the volume with the same temperature, pressure should increase. So, decrease volume, increase pressure. So, that's the relationship between P and V, between pressure and the volume. Now, if volume is the same and you start hitting the gas, then the pressure is again increasing, right? Because the molecules are moving faster. So, that's why with T increase, P should increase to make this constant. So, basically, I will use this to find out the dependency of the pressure on volume and temperature. So, let's say this constant is equal to C, whatever that C is, it's basically amount of gas here and whatever it is. It's a constant. Now, so, P in this case is equal to C, T divided by V. Now, let's assume that, number one, this piston is under our command. And also, we can have some kind of a temperature here to heat it up. And it's all under our control. So, volume is under our control. And it's some kind of a function of time. Let's say we are gradually squeezing it. And the temperature is under our control. So, we know these two functions. Now, the question is, how fast the pressure would change if I would start changing my volume and temperature according to these formulas, to these expressions, whatever that is. Maybe it's T times constant. Maybe it's a T square. Maybe it's some e to the power of T. Whatever the formula for this function is, doesn't matter. We know this, okay? This is the law by which we are conducting this experiment. Well, I shouldn't say law. It's not the law. It's rules according to which we conduct our experiment, okay? So, what is our function of two arguments? Well, that's P. And two arguments are T and V, both depending on time T. So, x is basically T, y is V, and f function is P. And we are interested in how the pressure would change with the time. So, we are interested in this particular expression, derivative of pressure by time. Well, let's just apply this formula. So, what do we have? We have partial derivative by first argument. First is, let's say, T. And the partial derivative by T would be what? C, V, right? Times derivative of T plus partial derivative by V. Well, that's actually minus C over V square, right? Times T. So, C and T are constant. And the derivative of 1 over V is minus 1 over V square times V. So, this is basically the formula. If we know the rules by which we are changing volume and the rules by which we are changing the temperature, then this is the formula by which we can come up with the changing of the pressure. At any given time, well, obviously, V is V of T. This is function of T. This is function of T, etc., right? So, this is how we are using this formula of total derivative to find out, in this particular case, how our pressure is changing with the time. Well, that's it. I do recommend you to read the notes for this lecture on Unisor.com. Well, other than that, basically, well, these, I would say, basic properties we have covered. So, thanks very much and good luck.