Let the function u = f(x, y, z) continuous in some area D and has continuous partial derivatives in this region. Let us choose a point in the considered area M(x,y,z) and draw a vector from it S, whose direction cosines are cosα, cosβ, cosγ. On the vector S at a distance Δ s from its beginning we find a point M 1 (x+Δ x, y+Δ y, z+Δ z), where
Let's represent the full increment of the function f as:
Where 
After dividing by Δ s we get:
Insofar as 
The previous equality can be rewritten as:
Gradient.
Definition The limit of the relation at is called function derivative u = f(x, y, z) in the direction of the vector S and is denoted.
In this case, from (1) we obtain:
(2)
Remark 1. Partial derivatives are a special case of the directional derivative. For example, when 
we get:
Remark 2. Above, the geometric meaning of the partial derivatives of a function of two variables was defined as the slope coefficients of the tangents to the lines of intersection of the surface, which is the graph of the function, with the planes x = x 0 and y = y 0. In a similar way, we can consider the derivative of this function with respect to the direction l at the point M(x 0, y 0) as the slope of the line of intersection of the given surface and the plane passing through the point M parallel to the O axis z and direct l.
Definition A vector whose coordinates at each point of some area are the partial derivatives of the function u = f(x, y, z) at this point is called gradient functions u = f(x, y, z).
Designation: grad u = .
gradient properties.
1. Derivative with respect to the direction of some vector S equals the projection of the vector grad u per vector S . Proof. Unit direction vector S has the form eS =(cosα, cosβ, cosγ), so the right side of formula (4.7) is the scalar product of the vectors grad u and e s , that is, the specified projection.
2. Derivative at a given point in the direction of the vector S has the largest value equal to |grad u| if this direction is the same as the direction of the gradient. Proof. Denote the angle between the vectors S and grad u through φ. Then it follows from property 1 that |grad u|∙cosφ, (4.8) therefore, its maximum value is reached at φ=0 and is equal to |grad u|.
3. Derivative with respect to the direction of a vector perpendicular to the vector grad u, equals zero.
Proof. In this case, in formula (4.8) 
4. If z = f(x,y) is a function of two variables, then grad f= directed perpendicular to the level line f (x, y) = c, passing through this point.
Extrema of functions of several variables. A necessary condition for an extremum. Sufficient condition for an extremum. Conditional extreme. Method of Lagrange multipliers. Finding the largest and smallest values.
Definition 1. Dot M 0 (x 0, y 0) called maximum point functions z = f(x, y), if f (x o , y o) > f(x, y) for all points (x, y) M 0.
Definition 2. Dot M 0 (x 0, y 0) called minimum point functions z = f(x, y), if f (x o , y o) < f(x, y) for all points (x, y) from some neighborhood of the point M 0.
Remark 1. The maximum and minimum points are called extremum points functions of several variables.
Remark 2. The extremum point for a function of any number of variables is defined in a similar way.
Theorem 1(necessary extremum conditions). If a M 0 (x 0, y 0) is the extremum point of the function z = f(x, y), then at this point the first-order partial derivatives of this function are equal to zero or do not exist.
Proof.
Let's fix the value of the variable at counting y = y 0. Then the function f(x, y0) will be a function of one variable X, for which x = x 0 is the extremum point. Therefore, by Fermat's theorem or does not exist. The same assertion is proved for .
Definition 3. Points belonging to the domain of a function of several variables, at which the partial derivatives of the function are equal to zero or do not exist, are called stationary points this function.
Comment. Thus, the extremum can be reached only at stationary points, but it is not necessarily observed at each of them.
Theorem 2(sufficient conditions for an extremum). Let in some neighborhood of the point M 0 (x 0, y 0), which is a stationary point of the function z = f(x, y), this function has continuous partial derivatives up to the 3rd order inclusive. Denote Then:
1) f(x, y) has at the point M 0 maximum if AC-B² > 0, A < 0;
2) f(x, y) has at the point M 0 minimum if AC-B² > 0, A > 0;
3) there is no extremum at the critical point if AC-B² < 0;
4) if AC-B² = 0, additional research is needed.
Example. Let's find the extremum points of the function z=x² - 2 xy + 2y² + 2 x. To search for stationary points, we solve the system 
. So, the stationary point is (-2,-1). Wherein A = 2, AT = -2, With= 4. Then AC-B² = 4 > 0, therefore, an extremum is reached at the stationary point, namely the minimum (since A > 0).
Conditional extreme.
Definition 4. If the function arguments f (x 1 , x 2 ,…, x n) bound by additional conditions in the form m equations ( m< n) :
φ 1 ( x 1, x 2,…, x n) = 0, φ 2 ( x 1, x 2,…, x n) = 0, …, φ m ( x 1, x 2,…, x n) = 0, (1)
where the functions φ i have continuous partial derivatives, then equations (1) are called connection equations.
Definition 5. Function extremum f (x 1 , x 2 ,…, x n) under conditions (1) is called conditional extremum.
Comment. We can offer the following geometric interpretation of the conditional extremum of a function of two variables: let the arguments of the function f(x,y) are related by the equation φ (x, y)= 0, defining some curve in the plane O hu. Having restored from each point of this curve perpendiculars to the plane O hu before crossing the surface z = f (x, y), we obtain a spatial curve lying on the surface above the curve φ (x, y)= 0. The problem is to find the extremum points of the resulting curve, which, of course, in the general case do not coincide with the unconditional extremum points of the function f(x,y).
Let us define the necessary conditional extremum conditions for a function of two variables by introducing the following definition beforehand:
Definition 6. Function L (x 1 , x 2 ,…, x n) = f (x 1 , x 2 ,…, x n) + λ 1 φ 1 (x 1 , x 2 ,…, x n) +
+ λ 2 φ 2 (x 1 , x 2 ,…, x n) +…+λ m φ m (x 1 , x 2 ,…, x n), (2)
where λ i - some constants, called Lagrange function, and the numbers λ i– indefinite Lagrange multipliers.
Theorem(necessary conditional extremum conditions). Conditional extremum of the function z = f(x, y) in the presence of the constraint equation φ ( x, y)= 0 can only be reached at stationary points of the Lagrange function L (x, y) = f (x, y) + λφ (x, y).
Consider the function u(x, y, z) at the point М(x, y, z) and the point М 1 (x + Dx, y + Dy, z + Dz).
Let's draw a vector through the points M and M 1 . The angles of inclination of this vector to the direction of the coordinate axes x, y, z will be denoted by a, b, g, respectively. The cosines of these angles are called direction cosines vector .
The distance between the points M and M 1 on the vector will be denoted by DS.
where the quantities e 1 , e 2 , e 3 are infinitely small at .
From geometrical considerations it is obvious:
Thus, the above equalities can be represented as follows:
Note that s is a scalar value. It only determines the direction of the vector.
From this equation follows the following definition:
The limit is called derivative of the function u(x, y, z) in the direction of the vector at the point with coordinates (x, y, z).
Let us explain the meaning of the above equalities with an example.
Example 9.1. Calculate the derivative of the function z \u003d x 2 + y 2 x at the point A (1, 2) in the direction of the vector. In (3, 0).
Decision. First of all, it is necessary to determine the coordinates of the vector .
We find the partial derivatives of the function z in general form:
The values of these quantities at point A: 
To find the direction cosines of the vector, we perform the following transformations:
=
An arbitrary vector directed along a given vector is taken as a value, i.e. determining the direction of differentiation.
From here we obtain the values of the direction cosines of the vector:
cosa = ; cosb=-
Finally we get: 
- the value of the derivative of the given function in the direction of the vector .
If a function u = u(x, y, z) is given in some domain D and some vector whose projections on the coordinate axes are equal to the values of the function u at the corresponding point
,
then this vector is called gradient functions u.
In this case, we say that a field of gradients is given in the region D.
Theorem: Let the function u = u(x, y, z) be given and the gradient field
.
Then the derivative with respect to the direction of some vector is equal to the projection of the vector gradu onto the vector .
Proof: Consider a unit vector and some function u = u(x, y, z) and find the scalar product of the vectors and degrees.
The expression on the right side of this equality is the derivative of the function u in the direction s.
Those. 
. If the angle between the vectors degrees and denoted by j, then the scalar product can be written as the product of the modules of these vectors and the cosine of the angle between them. Taking into account the fact that the vector is unit, i.e. its modulus is equal to one, we can write:
The expression on the right side of this equality is the projection of the vector grad u to vector .
The theorem has been proven.
To illustrate the geometric and physical meaning of the gradient, let's say that the gradient is a vector showing the direction of the fastest change of some scalar field u at some point. In physics, there are such concepts as temperature gradient, pressure gradient, etc. Those. the direction of the gradient is the direction of the fastest growth of the function.
In terms of geometric representation, the gradient is perpendicular to the level surface of the function.
1) The case of a function of two variables. The direction is given by a vector. We choose a unit vector that specifies the direction on the plane: 
. This vector forms an angle with the positive direction of the OX axis. The directional derivative of a function of two variables is called the expression 
.
2) The case of a function of three variables. Let a unit vector be given that forms angles with the axes OX, OY and OZ, respectively. If we designate the coordinates of the vector as , then by the formula for the cosine of the angle between two vectors and we get . Likewise, . Τᴀᴋᴎᴍ ᴏϬᴩᴀᴈᴏᴍ, the unit vector making angles with the axes OX, OY and OZ, has coordinates . The directional derivative of a function of three variables is called the expression
.
Definition.Gradient functions are usually called a vector 
. For this reason, the derivative of a function in the direction given by the unit vector can be calculated by the formula 
, where on the right in the formula is the scalar product of the gradient of the function and the unit direction vector.
The main property of the gradient: among all possible directions, the derivative in direction takes the largest, and positive, value in the direction of the gradient. This property follows from the definition of the scalar product. Since the positivity of the derivative means the growth of the function, the direction of the gradient at the point is ϶ᴛᴏ the direction of the greatest growth of the function.
Partial derivatives of higher orders.
Any partial derivative of a function of variables 
itself is also a function of variables. The partial derivative of the partial derivative of a function of many variables is called second order partial derivative functions . In this case, if the variables with respect to which the derivatives are taken first from the function and then from the function do not coincide, such a partial derivative is usually called mixed. Second order partial derivative notation: 
. In the case when and are continuous functions in a neighborhood of some point, at this point.
Similarly, partial derivatives of any order are introduced.
EXAMPLE
Hosted on ref.rf
Find from function . We have 
.
In order to calculate the same derivative using MAXIMs, we use the command diff(log(x+3*y),x,2,y,1).
Higher order differentials.
By analogy with derivatives, differentials of higher orders are introduced, that is, differentials from differentials. Consider a function of three variables . The differential of this function is the expression . Note that the derivatives included in the last expression are functions of , and the differentials of the variables do not depend on . For this reason, under the condition of continuity of mixed derivatives, the second-order differential has the form
In the last formula, we have used the property of equality of mixed derivatives. It is easy to see that the formula for the second-order differential is similar to the formula for the second degree of the sum of three terms. It is not difficult to calculate the differentials of the second and third orders of the function of two variables: ,
An exercise. To find for the function at the point (1,1).
Taylor formula for a function of many variables.
As in the case of functions of one variable, for functions of many variables, Taylor's formula gives the relationship between the increment of a function at a point and its differentials at the same point:
where 
.
In particular, for a function of two variables we have:
Here 
.
Directional derivative. - concept and types. Classification and features of the category "Directional derivative." 2017, 2018.
Consider the function u(x, y, z) at the point М(x, y, z) and the point М1(x + Dx, y + Dy, z + Dz). Let's draw a vector through the points M and M1. The angles of inclination of this vector to the direction of the coordinate axes x, y, z will be denoted by a, b, g, respectively. The cosines of these angles are called direction cosines of the vector. ... .
Consider the function u(x, y, z) at the point М(x, y, z) and the point М1(x + Dx, y + Dy, z + Dz). Let's draw a vector through the points M and M1. The angles of inclination of this vector to the direction of the coordinate axes x, y, z will be denoted by a, b, g, respectively. The cosines of these angles are called direction cosines of the vector. ... .
An important characteristic of the scalar field U(M) is the rate of change of the field function in the indicated direction. If this direction coincides with the direction of one of the coordinate axes, then we will get the value of the corresponding partial derivative. From vector algebra... .
Let the function U = F (X, Y, Z) be continuous in some domain D and have continuous partial derivatives in this domain. We choose a point M(X,Y,Z) in the area under consideration and draw a vector S from it, the direction cosines of which are cosA, cosB, cosG. On the vector S at a distance DS from its origin... .
The derivative of a function at a point along the direction is called the limit where if the limit exists. If the function is differentiable, then the directional derivative is calculated by the formula (1) where are the direction cosines of the vector In particular, if is a function of two variables,... .
scalar field. Level surfaces. ELEMENTS OF MATHEMATICAL FIELD THEORY Main Stages in the Development of Mathematical Physics Mathematical physics emerged as an independent science in the late 18th and early 19th centuries. It is in this...
Introducing the concept of a partial derivative of a function of many variables, we incremented the variables individually, leaving all other arguments unchanged. In particular, if we consider a function of two variables z = f(x, y), then either the variable x was given an increment Δx, and then in the domain of the function there was a transition from a point with coordinates (x, y) to a point with coordinates (x + Δx ;y); or the variable y was given an increment Δy, and then in the domain of the function there was a transition from a point with coordinates (x, y) to a point with coordinates (x; y + Δy) (see Figure 5.6). Thus, the point at which we took the partial derivative of the function moved in directions parallel to the coordinate axes on the plane (either parallel to the abscissa axis or parallel to the ordinate axis). Let us now consider the case when the direction can be taken arbitrarily, i.e. increments are given to several variables at once. For the case of a function of two variables, we will move to the point (x + Δx; y + Δy), while the displacement will be Δ l(see figure 5.6).
When moving in this direction, the function z will receive an increment Δ l z = f(x + Δx; y + Δy) – f(x,y), called the increment of the function z in the given direction l.
Derivative z l` in direction l
functions of two variables
z = f(x,y) is the limit of the ratio of the increment of the function in this direction to the amount of displacement Δ l when the latter tends to zero, i.e. .
Derivative z l` characterizes the rate of change of the function in the direction l.
The concept of directional derivative can be generalized to functions with any number of variables.
Figure 5.6 - Moving a point in the direction l
It can be proved that z l` = z x `cos α + z y `cos β, where α and β are the angles formed by the direction of movement of the point with the coordinate axes (see Figure 5.6).
For example, let's find the derivative of the function z = ln (x 2 + xy) at the point
(3; 1) in the direction going from this point to the point (6; -3) (see figure 5.7).
To do this, first find the partial derivatives of this function at the point (3; 1): z x ` = (2x + y)/(x 2 + xy) = (2*3 + 1)/(3 2 + 3*1) = 7 /12;
z y ` \u003d x / (x 2 + xy) \u003d 3 / (3 2 + 3 * 1) \u003d 3/12 \u003d 1/4.
Note that Δx = 6 – 3 = 3; Δy \u003d -3 - 1 \u003d -4; (Δ l) 2 = 9 + 16 = 25;
|Δ l| = 5. Then cos α = 3/5; cos β = -4/5; z l` =
z x `cos α +
z y `cos β = (7/12)*(3/5) - (1/4)*(4/5) = (7/4)*(1/5) - (1/4)*(4 /5) = (7*1 - 1*4)/(4*5) = 3/20.
function gradient
It is known from a school mathematics course that a vector on a plane is a directed segment. Its beginning and end have two coordinates. The vector coordinates are calculated by subtracting the start coordinates from the end coordinates.
The concept of a vector can also be extended to an n-dimensional space (instead of two coordinates there will be n coordinates).
Gradient grad z of the function z = f(х 1 , х 2 , …х n) is the vector of partial derivatives of the function at the point, i.e. vector with coordinates 
.
It can be proved that the gradient of a function characterizes the direction of the fastest growth of the level of the function at a point.
For example, for the function z \u003d 2x 1 + x 2 (see Figure 5.8), the gradient at any point will have coordinates (2; 1). It can be built on a plane in various ways, taking any point as the beginning of the vector. For example, you can connect point (0; 0) to point (2; 1), or point (1; 0) to point (3; 1), or point (0; 3) to point (2; 4), or t .P. (see figure 5.8). All vectors constructed in this way will have coordinates (2 - 0; 1 - 0) =
= (3 – 1; 1 – 0) = (2 – 0; 4 – 3) = (2; 1).
Figure 5.8 clearly shows that the level of the function grows in the direction of the gradient, since the constructed level lines correspond to the level values 4 > 3 > 2.
Figure 5.8 - Gradient function z \u003d 2x 1 + x 2
Consider another example - the function z = 1/(x 1 x 2). The gradient of this function will no longer always be the same at different points, since its coordinates are determined by the formulas (-1 / (x 1 2 x 2); -1 / (x 1 x 2 2)).
Figure 5.9 shows the level lines of the function z = 1 / (x 1 x 2) for levels 2 and 10 (the straight line 1 / (x 1 x 2) = 2 is indicated by a dotted line, and the straight line
1 / (x 1 x 2) \u003d 10 - solid line).
Figure 5.9 - Gradients of the function z \u003d 1 / (x 1 x 2) at various points
Take, for example, the point (0.5; 1) and calculate the gradient at this point: (-1 / (0.5 2 * 1); -1 / (0.5 * 1 2)) \u003d (-4; - 2). Note that the point (0.5; 1) lies on the level line 1 / (x 1 x 2) \u003d 2, because z \u003d f (0.5; 1) \u003d 1 / (0.5 * 1) \u003d 2. To depict the vector (-4; -2) in Figure 5.9, we connect the point (0.5; 1) with the point (-3.5; -1), because
(-3,5 – 0,5; -1 - 1) = (-4; -2).
Let's take another point on the same level line, for example, point (1; 0.5) (z = f(1; 0.5) = 1/(0.5*1) = 2). Calculate the gradient at this point
(-1/(1 2 *0.5); -1/(1*0.5 2)) = (-2; -4). To depict it in Figure 5.9, we connect the point (1; 0.5) with the point (-1; -3.5), because (-1 - 1; -3.5 - 0.5) = (-2; - 4).
Let's take one more point on the same level line, but only now in a non-positive coordinate quarter. For example, point (-0.5; -1) (z = f(-0.5; -1) = 1/((-1)*(-0.5)) = 2). The gradient at this point will be
(-1/((-0.5) 2 *(-1)); -1/((-0.5)*(-1) 2)) = (4; 2). Let's depict it in Figure 5.9 by connecting the point (-0.5; -1) with the point (3.5; 1), because (3.5 - (-0.5); 1 - (-1)) = (4 ; 2).
It should be noted that in all three cases considered, the gradient shows the direction of growth of the level of the function (toward the level line 1/(x 1 x 2) = 10 > 2).
It can be proved that the gradient is always perpendicular to the level line (level surface) passing through the given point.
scalar field called a part of space (or the whole space), each point, which corresponds to the numerical value of some scalar quantity.
Examples
A body that has a certain temperature value at each point is a scalar field.
An inhomogeneous body, each point of which corresponds to a certain density - a scalar density field.
In all these cases, the scalar value U does not depend on time, but depends on the position (coordinates) of the point M in space, that is, it is a function of three variables, it is called field function. And vice versa, any function of three variables u=f(x, y, z) defines some scalar field.
Planar scalar field function depends on two variables z=f(x, y).
Consider a scalar field u=f(x, y, z).
A vector whose coordinates are the partial derivatives of a function calculated at a given point is called gradient function at this point or the gradient of the scalar field.
Consider a vector with two points on it M 0 (x 0 , y 0 , z 0) and . Let's find the increment of the function in the direction:
Directional derivative the next limit is called if it exists:
where are the direction cosines of the vector ; α, β, γ are the angles that the vector forms with the coordinate axes, if .
For a function of two variables, these formulas take the form:
or 
,
as .
There is a relationship between the gradient and the directional derivative at the same point.
Theorem. The scalar product of the gradient of a function and a vector of some direction is equal to the derivative of the given function in the direction of this vector:
.
Consequence. The derivative with respect to the direction has the greatest value if this direction coincides with the direction of the gradient (justify yourself using the definition of the dot product and assuming that ).
Findings:
1. A gradient is a vector showing the direction of the greatest increase in the function at a given point and having a modulus numerically equal to the rate of this increase:
.
2. The derivative in direction is the rate of change of the function in the direction: if , then the function in this direction increases, if , then the function decreases.
3. If the vector coincides with one of the vectors, then the derivative in the direction of this vector coincides with the corresponding partial derivative.
For example, if , then .
Example
Given a function 
, dot A(1, 2) and vector .
Find: 1) ;
Decision
1) Find the partial derivatives of the function and calculate them at point A.
, .
Then 
.
2) Find the direction cosines of the vector:
Answer: ; .
Literature [ 1,2]
Questions for self-examination:
1. What is called a function of two variables, its domain of definition?
2. How are partial derivatives determined?
3. What is the geometric meaning of partial derivatives?
4. What is called the gradient of a scalar field at a given point?
5. What is called a directional derivative?
6. Formulate the rules for finding the extrema of a function of two variables.
Option 1
Task number 1
a) 
; b) 
;
in) ; G) 
.
Task number 2 Investigate a function for continuity: find breakpoints of the function and determine their type. Construct a schematic graph of the function.
Task number Given a complex number Z. Required: write the number Z in algebraic and trigonometric forms. .
Task number 4.
1) y \u003d 3x 5 - sinx, 2) y \u003d tgx, 3) y \u003d, 4) .
Task number 5. Investigate the function using the methods of differential calculus and, using the results of the study, build a graph. .
Task number 6. The function z=f(x,y) is given. Check if the identity F≡0 is fulfilled? 
Task number 7 Given a function Z=x2+xy+y2, point and vector . To find:
1) gradz at the point BUT;
2) derivative at a point BUT in the direction of the vector .
Option 2
Task number 1 Calculate the limits of functions without using L'Hopital's rule.
a) 
; b) 
;
in) 
; G) 
.
Task number 2 Investigate a function for continuity: find breakpoints of the function and determine their type. Build a schematic graph of the function.
Task number 3 Given a complex number Z. Required: write the number Z in algebraic and trigonometric forms.
Task number 4. Find the first order derivatives of these functions.
