Математика. Материалы курса

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What interesting things are we going to discuss today? It’s the differentiation of a complex function and a function given implicitly. We have also spoken about such concepts for a function of a single variable.

So, what are complex functions? First, we are given a finite set of functions, we have written them into the system, each of them is a function of m variables (t₁…t_m), and we will assume that they are differentiable at the point with upper indices 0, we denoted it t⁰. They are differentiable, i.e. they have differentials at this point. And the function f is considered as the function of variables (x₁…x_n), we will assume that it is differentiable at the point (x₁⁰…x_n⁰), where these numbers are defined as the values of the system function at the point t⁰. So, the function f is also differentiable, that is, it also has a differential.

What are complex functions? This is z=f(x_1,…x_n), where each of these variables is the function of the variables (t₁…t_m), that is, z depends on the variables t. This is a complex function obtained using the f function and using the system functions. What are we trying to find out now? This complex function will also be differentiable at t⁰. And our task is to find the formula for the differential of this function.

So, by the condition z, the function f(x_1,…x_n) is differentiable at a point, so there is a differential at this point, and its formula is well known to us. Partial derivatives for each variable are multiplied by the differential of this variable, and we find the sum of n on the terms. All functions of the system are also differentiable at the point t₀, which means that the differentials of these functions also exist and are calculated using the corresponding formulas. Here, each differential is the sum of m terms. So, z is the function of the variable (t_1,…t_m), by substituting the values of the differentials of the variables (x₁…x_m) instead of the values of partial differentials in the brackets (we have just written it down). We obtain the following expression: we open the brackets and see that there are n terms containing dt₁ multiplier. We group these terms, as well as the terms containing dt_2, multiplier, and group the terms containing the differential of the variable t_m.

What do we get? We get the expression containing m terms. But we know the differential formula. The multiplier before the differential t₁ is the value of the partial derivative of z with respect to t₁. The next group of terms in parentheses forms the differential z with respect to the variable t_2, and so on. As a result, we can write down the values of the partial derivatives for each variable.

Let’s take a closer look at a less complex example. Let z be the function of the variables x and y. In turn, x and y are functions of the variables u and v. We get a very complex function, where z is the dependent variable, and the independent variables are u and v. How can we find partial derivatives of z(u, v). Let’s draw a diagram like this. Look, it will be easier for us to use the formula: z depends on x and y, in turn, each of these variables depends on u and v.

Look, if we want to differentiate z by u, we have two trails to the variable u from z (via x and y). Note how we label these trails (see the video). In this case, the partial derivatives of ∂z over ∂u are just formed from these notations: the upper trail gives us 2 co-factors + we multiply 2 co-factors by 2 trails. Similarly, if we differentiate by v, the transition from x to v will change only at the last stage: from y to v will be ∂x by ∂v, ∂y by ∂v. And the formula is written in a similar way.

Let’s turn to the next question. What is an implicit function?

Some equation of the form (*) that connects three variables (x, y, z) is given. We assume that the set of solutions to this equation is not empty, that is, there are triples of numbers (x, y, z) that satisfy this equation. In which case do we say that this equation defines an implicit function? What is the function z=f(x, y) that is given by this equation? What is its domain of definition?

The domain of definition of such a function is the set of all possible pairs (x, y) for which there exists a z satisfying the equation (*), by the condition, this set is not empty. So, the set of such (x, y) exists, the function is defined.

And what are the values of this function? Each value of z from the point in the domain of definition is chosen, generally speaking, arbitrarily, it exists and can be not one, so the equation (*) very often sets infinitely many functions that are set implicitly. Most often, a function is defined in this way when z cannot be expressed in terms of x and y, so the differentiation problem is denoted. Let’s move on to this task.

Let’s try to find the partial derivative of the function given implicitly by the variable x, that is, we are looking for z’ by x assuming that z is the function of the variables x and y. From the equation (*), we pass to the equation (**) differentiating the first one by the alternating x. What do we see? In this equation (**), we differentiate by x assuming that x and y are independent variables, and z is the function of the variables x and y. Look, on the left, we differentiate a complex function of three variables, in which z is also the function of variables x and y, that is, in reality, it is already the function of variables (x, y), a complex function.

Let’s draw a diagram like this. Look, at this stage, z turns out to be the function of the variables x and y, x also depends on x, y. We can assume that it is x+0*y, just as y is considered 0*x+y. So, these are functions of variables x, y. How do we write the value on these transitions? The transition from the variable u to x, y, z is clear. Let’s make the next transition from x to x. It is not written here, but the derivative will be equal to 1, and y with respect to x will be equal to 0: y is an independent variable that when differentiating with respect to x gives us 0. What do we get as a result?

We calculate: ∂u by ∂x is 0, ∂x by ∂x is 1, and ∂z by ∂x∂ is z’ by x. Moving on to the strokes, we get the following equation (see the video). Let’s look again at the equation (**), we see that the condition (**) is equivalent to the fact that u’ by x is zero. Substituting the expression obtained, we get the following. We just need to express z’ by x.

So, the partial derivatives of the function z with respect to x, given implicitly, are determined by the following formula (see the video).

Similarly, by differentiating with respect to y, we obtain the value of the partial derivative z’ with respect to y.

Let’s see where this approach can be used. It often turns out that the surface is given by an equation that connects the variables x, y, z, and we cannot express z. There is a surface, but while making an equation of the tangent plane, there is a problem at the stage of calculating partial derivatives because z from this equation cannot be expressed. But we remember that in this case this function z(x, y) is set implicitly, and its partial derivatives are calculated using the formulas.

As a result, substituting the tangent plane into the equation, we get the following equation. Multiplying both parts of this equation by F’ with respect to z at p₀, and moving all the terms to the left side, we get the equation of the tangent plane for the given surface at p₀. It is clear that, in this case, the value of partial derivatives at the point p₀ is the coordinates of the normal vector, so the normal equation can be written in this form. If at least one of the partial derivatives of the denominator is equal to 0, we turn to the parametric setting of this line and write the normal equation in the following form (see the video).