Hello. Conveyor saw derivation rules. I have seen this exact derivative. This is not a difficult thing at all. I'm sure you remember. Have you looked a little. But here, too, I reminded I'll move on. Here again, with problems resolved here we go. Our problem is the following. Simply to the problem. Yet such a paraboloid. Now you've memorized it. zero when x is zero y's four intermediate compound. As the top four in the top-down e started, paraboloid. z x squared plus y squared at zero when you have made as x and y equals four two radii of a circle in the plane. We say that y is equal to x between y'yl get the following equation. minus x plus one. I choose a very simple function, but for real simple functions of this kind of life will not. Finding full derivative thereof with respect to x we want. Where is it? at x is equal to one. Derivative function we can find the full course The. But at some point we want to calculate. And this is what it means to work in also, what it means in terms of geometry We also want to know. Necessarily in this space x and y, x y z in space may be a surface. May be a potential energy. May be a thermodynamic functions. E, an electric field on may be potential. But we know that the geometric meaning other kinds of thinking that physical We understand what it means in environments. Now was our rule. Dedicating full-derived partial derivative. A full derivative of f with respect to x of all the changes This partial with respect to x changes with a change from a partial to y something that happens. But y is a function of the x-over as When the change in the y with respect to x enters. Now that we diol chain derivative rule. To apply it in place of f with respect to x partial derivatives needed. Given here for. This partial derivative with respect to x minus two x's. According to the partial derivative of y minus two years. Then we need d to d *. Se but here, the x and y independently in f. But as one condition of y with respect to x function has been provided for herein. So d to d * a minus. These are preliminary. EUR replace them by taking away minus two x minus d * d to two years, which is also a negative for I would get it. We find that x is equal to one if we want to y when x is equal to zero in a happens. I.e., x is zero and y is equal in a We find exact derivative. d been looking for for functions. Or here is not complete this function We can not talk about derivatives. The geometrical meaning, but that it also saw We'll see you again on the sample well. We have a very simple function that x instead of y for placing the we can do. Z that is given to us. Y squared minus x squared minus four. But in terms of y given x. We could take it into place. So this is a function of x and y at x L x We keep it in place when you put in y means y squared minus x squared minus four instead of minus one minus x squared plus one. Edit your one of them in the plane z x is a function of variables. Derivatives here are already full. Not the other derivatives. Because of one variable. Therefore, the curve of this little f over it but it's true that this curve is selected as the exact derivative also on line, there is this After making placement is the only is equal to the variable functions derivative. It is the same thing. Here we make account see two negative four x. It seems to be a little different, but I have turned negative in here at two x two x minus four x plus two. We also ensure that the same be it. We also looked at x is equal to a derivative in Although these two minus four x is x minus two, so instead of a sheep chain We found the value to be derived mitigated We find the same thing as needed. Now it up once the geometric meaning Let's dig into. There are two things given to us. A surface equation. BI right of this equation. Where is this correct? x b y plane is correct. As b is correct. It's above the surface like this kind of coming down a paraboloid. But it's true equation in three dimensions If we consider all our this, this equation that provides all our geometrically in it at any point We draw a vertical line through When this bi plane as obtained happens. BI is built right onto a cylinder plane. We have seen this in the first chapter. Therefore, they both have When the paraboloid directly with the surface of the said vertical plane section occurs. Such paraboloid, ie bi watermelon Think of it as your half-bent. When you cut the vertical blades that bi to obtain such a parabolic edge. If this common interest that such a parabola parabola slightly I tried to boot according to the scale. That what is happening in the x-y plane If we consider x and y equals z in the plane is zero. As you can see that z is equal to zero in the x-y plane cross-section of a paraboloid circle x squared plus y squared equals four ie a circle with a radius of two. On the other hand y equals x plus one minus this kind of true. And more than one minus the slope of a This is true. Here on this parabola, so that this on the estimation of these two paraboloid here as a three- size in perspective, we see becomes parabolic curve. If it takes off, so as z f x This large y After loading the obtained If we take it as a function of this. I.e., on the surface, on which the We take the slope of the curve is. And the same thing when it accounts for two types of involved. So here's something that will be remembered always full derivative only a function of two variables is not enough. Because then you only have partial derivatives. As we say, but also a y get function. Then the intersection of these two surfaces We find the slope of the curve on is fully differential. Let's look at a second example. But simply a function of selected genes where x Cosine t, y as well as two plus sine t Let's take. This is a line in the plane. After a bit we'll see what it is. Built on the cylinder means that The intersection of these surfaces. Formula is easy to apply. Now find the total derivative with respect to t we want. We want to find the derivative function fully. This derivative is equal to pi divided by three in t We want to find the value. I can understand it as a geometric we want. So that's just like a machine account let's not If we understand the issues we are more creative said. Now we've found it. Other than that of differential infinitesimal where t divided by partial derivatives of the two parts comprising of the total derivatives were getting full. Now you've done it before. E derivative of f with respect to x here As you can see x minus one-half. x the derivative of x with respect to t by t t minus sine derivative. Plus a negative partial derivative of f to y divided by two years. Where y is the cosine of the derivative with respect to t t. As you can see the term bi bi obtained mixed We are. There are both t and x. But it is known as a function of x t. It is known also as a function of y t. Therefore, take them We can place. Minus one-half times this dilemma the x's minus plus one-half the value of that times x cos t was. You're installing and sinus t minus here Because bi minus minus one remains. instead of y t are writing two plus sinus. T multiplied by the cosine. The cosine t. As you can see here right here sinus t exist. Cosine has t. Here are t cosine, sine t exist. One plus one minus one divided by two divided by two to one where these terms is more simple. This two and two minus one divided by the product of here is a coming and the cosine t remains. So as a result of the partial derivatives of t, as a result of a total derivative minus the cosine t. In that the value pi account thirds Let If we say that this is obviously very easy. PI and sixty degrees divided by three. Sixty-one-half degree from the cosine. So we find a negative divided by two. Now, in the following also we can do. This function given z. t as a function of x, y, t given function. So it took them over the We could place. Four minus one divided by four times the cosine square t t plus two plus sine squared. Because here we have y squared. If we arrange them in a very simple algebraic this process. See here comes the cosine square. Sine-squared is coming. They are a doing. Comes to the conclusion that such transactions. If we take the derivative of it really is b z t as t, we find the negative cosine. Chain derivatives in the previous page with rules t directly as minus cosine We found. Here again, I repeat. In the examples, not the mathematical complexity We choose as simple functions but stretch for him to place it at It is possible. But in real life problems, technology problems such simple tasks in may not be. Rather than chaining derivation rules will be effective. Here we already find the same result we see. The following geometry. This is a paraboloid again. I think I love you but paraboloids I love these because. Easy to draw. Students interested audience in terms of you'd be so easy to follow. This is a paraboloid facing down. Given by the following parametric representation If we consider the circle, x sine cosine T'Su T'Su minus two years. Where y instead of t we make sine cosine t t plus one square sine-squared We're going. We know that one. Place them de x squared plus y two minus one equals the square. So the equation of the circle b. This circle is equal to two years in bi feature the center of it. A radius. So if we think that in the x-y plane circle two in radius, the radius of the center in two a. See if we consider our equals zero z 16 is equal to zero at the thought of going here. x squared plus y squared equals 16. So a circle with a radius of four. This is the radius of the circle four. such a situation occurs in the x-y plane. Looking at this in perspective, wherein The four radius the x and y plane of the circle to cut the paraboloid. This is in the center with a diameter of two cylinder. When you take up this cylinder is this paraboloid for such a thing to be cut. Here too the theta equals pi divided by three values were also. When rice theta divided by three and one-half y's here also two divided by two plus one point three roots will be found. He it geometrically wherein we can see. So, on this surface that we cylinder intersection When that emerged on the ellipse We are progressing vBulletin accounts slope. Our function was ever simple. For example, we take a little complicated functions such we take a function wherein y equals x square pattern see end of the world so much but not e, x and y are in such which is something will be a function of mixed sine times these things will be a lot of stuff. Characterized in that said function. This is a composite function. cube to the cube of x plus y plus x and y to say to pi'yl This function becomes the sine together. So we found the function z is equal to Fi. But both the function of x and y. Therefore, to calculate the exact derivative the full derivative with respect to x can be calculated and xa apology by I wish both to y the partial derivatives can be calculated. Chained to them, with derivatives rules According to the partial derivative of f with respect to x full derivative According to u because for a univariate function. But with respect to x and y by while others are partial derivatives of derivatives. Here try this for these accounts Or of the function z because, well if you want x to y, and partial derivatives are account. Y is equal to zero at x equals a thereof values we'd like to find work rather than take x zero will put. will put in place a year. These values are obtained. But it also f y x is equal to the square full account derivative We can not think of. It also means that you take the full derivative Instead obtained by placing As you can d f d u d after full derivative of income with respect to x. Of processing other than the derivatives with respect to x at This is obtained. This derivative is obtained by placing The results in this happening and here We find negative pie. Bi is a closed function other functions. See here a relationship between X, y is there. So the only independent variable but as a function implicit function given. It may be an ellipse that. If we say that the pros get y minus the top of this ellipse is at the point. This is going to hurt by geometry he not so important, but now we want to do here with respect to x of y derivative. It can be calculated in three ways. I once had found the formula b. Chained account directly derived We can. That's all we solve y, albeit mixed here y able to solve, then open function derivative directly from pick up. Now it accounts for three different pathways: We will. See this year minus three squared minus two x and y square equals zero If x is equal to a cove see that this year is going on. BI becomes quadratic functions. Here we can easily solve y. Single figures are already coming out correctly he selected values. Now let's do them. The first derivative of implicit function We had a formula. derivative of y with respect to x. This is the ratio of the partial derivatives tert is marked. To find this derivative of f with respect to x 're getting. We take the derivative of f by y. After that creating this rate feed, y, we find the home base. It is equal to x and y is equal to a three When establishing the value of the inclination by three We find that. As a second operation was done: take derivative with respect to x, let it complete. Zero right-hand side. So his full derivative, partial derivative, everything derivatives will be zero. On the left side, let it fully derivative. But the function of the x-y. We also remember. Now if we do it with respect to x y karen derivative of y d y d x happens twice. So one year base here. Now you have a product here. Minus two times. This time we take the derivative with respect to x, here, have a derivative with respect to x. But here a year multiplied by y in 's. Coming from x to y plus partial derivatives but the x's de to y, y is the derivative with respect to x need. I would get it. Wherein x square of the derivative with respect to x as we know, two x's coming. I would get it. When we arrange it so y is üssül There are two terms. Gather them together, y, which üssüsüz There are two terms together in them collect the previous year when I found home base We found the same formula is obtained. x is equal to one, the value in y is equal to three We found the same value again involved. Of course, this stretch of such a result possible. If there is a value in this way, this way, a we find the value you means that we're doing it wrong you a the problem is not defined correctly. We could do the following:Since this too mixed a non-function, the function y x As can be resolved here. Because a function of the second degree. If we use half the formula, minus b, b where x plus or minus the square root, b squared x squared, minus a, c. This is a known quadratic formula. He three x squared. Edit your x it y equals three. Now here's the square root of x squared four under involved. Take the square root of two, so y is equal to x x plus two x. This gives three x or y minus two x, then y giving equals minus x. a is equal to y and y de plus If we take the value of X, where y equals three We see that we are working. Where y equals three x inasmuch as the base year say that three years is equal to the base as found. Now here we can ask the following question:Which method is better? Now you can see where the quadratic enters. A function a little confused by it though You can not solve immediately. This process can not solve for You can not execute. Similarly, the implicit function You can always work with derivatives. Because we say that x and y for fi x it Thank the partial derivatives to y We place easily find. It is also very easily walkable because they take things to the partial derivative receive. As a result of the partial derivatives and full derivatives giving. He had an assignment:Sinus x y x equals and y is equal to one part in pie We want to find derivatives and also as given year is equal to the cosine of pi x divided by two on A derivative of x is equal to the full We want to find the value. As in the second portion if y were not given the full derivative just because we can not talk about x's and y's We have a function. Guidance as follows:y is equal to that of sheep Put something you see here by the cosine yes as one function is a function of one variable can achieve but also a sine cosine in there. Such product there. So going on a mixed terms. Be made, but the possibility of making mistakes is high. Instead of chaining when passing derived Remove account. Here we encourage you to do this problem for Search in the accounts and the results given. By accurately interim results I expect you to. Gene homework. Now as you can see here an X, one thing There where there is a sine the y are square. Here now you can not solve anything. You can not solve for y in terms of x. Therefore, as the open function of y Base You can not calculate. Here can be considered in two ways. By differentiating it directly make or implicit function i.e. that x and y are derivative function, say, a partial with respect to x derivatives, partial derivatives to y, rates Base gives y with a minus sign. Although chained directly derived you can get. That two of Y, Y is a base and from there For X, y is multiplied derivatives of these compounds as a function of work removable by As you can see, but here again this off function as a derivative can be calculated immediately. I also encourage you to get the results promise. I gave a little search information. Now I have seen quite a few things. Now I want to pause here again. Please skip over this example, Please understand, you do your homework. After this chain of derivation rules the more we will see an additional application. I've told you from the beginning. Important on its own tangent plane but the rules will be removed from the tangent plane also important. We have seen chained derivative rules, the full We saw derivative, We will see here again a new concept. Directional derivative. Gene variants will come from this chain. Is based on its tangent plane again. And also with a new concept called gradient will encounter. Now goodbye until next session Your stay. Please review them a little.