Let the function u = f (x, y, z) continuous in some region D and has continuous partial derivatives in this region. Let us select a point in the area under consideration M(x,y,z) and draw a vector from it S, direction cosines of which are cosα, cosβ, cosγ. On the vector S at a distance Δ s from its beginning we will find a point M 1 (x+Δ x, y+Δ y, z+Δ z), Where
Let's imagine the full increment of the function f as:
Where 
After dividing by Δ s we get:
Because the 
the previous equality can be rewritten as:
Gradient.
Definition The limit of the ratio at is called derivative of a function u = f (x, y, z) in the direction of the vector S and is designated .
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 angular coefficients of tangents to the lines of intersection of the surface, which is the graph of the function, with planes x = x 0 And y = y 0. In a similar way, we can consider the derivative of this function in direction l at the point M(x 0, y 0) as the angular coefficient of the line of intersection of a given surface and a plane passing through a point M parallel to the O axis z and straight l.
Definition A vector whose coordinates at each point of a certain region 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 to vector S . Proof. Unit direction vector S looks like eS =(cosα, cosβ, cosγ), therefore the right-hand 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 greatest value equal to |grad u|, if this direction coincides with the direction of the gradient. Proof. Let us denote the angle between the vectors S and grad u through φ. Then from property 1 it follows that |grad u|∙cosφ, (4.8) therefore, its maximum value is achieved at φ=0 and is equal to |grad u|.
3. Derivative in the direction of a vector perpendicular to the vector grad u, is equal to 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 extremum. Lagrange multiplier method. 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 a point M 0.
Note 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 determined in a similar way.
Theorem 1(necessary conditions for an extremum). If M 0 (x 0, y 0)– 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, y 0) 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 statement is proved similarly 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 only be reached 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. Let us 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, further research is needed.
Example. Let's find the extremum points of the function z = x² - 2 xy + 2y² + 2 x. To find stationary points, we solve the system 
. So, the stationary point is (-2,-1). Wherein A = 2, IN = -2, WITH= 4. Then AC–B² = 4 > 0, therefore, at a stationary point an extremum is reached, namely a minimum (since A > 0).
Conditional extremum.
Definition 4. If the function arguments f (x 1 , x 2 ,…, x n) are 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. Extremum of the function f (x 1 , x 2 ,…, x n) when conditions (1) are met, it 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) related by the equation φ (x,y)= 0, defining some curve in the O plane xy. Reconstructing perpendiculars to plane O from each point of this curve xy until it intersects with the surface z = f (x,y), we obtain a spatial curve lying on the surface above the curve φ (x,y)= 0. The task 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 determine the necessary conditions for a conditional extremum for a function of two variables by first introducing the following definition:
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 λi – some are constant, called Lagrange function, and the numbers λ i– indefinite Lagrange multipliers.
Theorem(necessary conditions for a conditional extremum). Conditional extremum of a function z = f (x, y) in the presence of the coupling equation φ ( x, y)= 0 can only be achieved at stationary points of the Lagrange function L (x, y) = f (x, y) + λφ (x, y).
Scalar field a part of space (or all space) is called, each point to which the numerical value of some scalar quantity corresponds.
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 quantity U does not depend on time, but depends on the position (coordinates) of point M in space, that is, it is a function of three variables, it is called field function. And conversely, every function of three variables u=f(x, y, z) specifies some scalar field.
The flat scalar field function depends on two variables z=f(x, y).
Consider the 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 and 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 following limit is called if it exists:
where are the direction cosines of the vector; α, β, γ are the angles formed by the vector with the coordinate axes, if .
For a function of two variables, these formulas take the form:
or 
,
because .
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 this function in the direction of this vector:
.
Consequence. The directional derivative has the greatest value if this direction coincides with the direction of the gradient (justify yourself using the definition of the scalar product and assuming that ).
Conclusions:
1. The gradient is a vector showing the direction of the greatest increase in the function at a given point and having a module numerically equal to the rate of this increase:
.
2. Directional derivative is the rate of change of a 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 with respect to the direction of this vector coincides with the corresponding partial derivative.
For example, if , then .
Example
Function given 
, dot A(1, 2) and vector.
Find: 1) ;
Solution
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]
Self-test questions:
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 the scalar field at a given point?
5. What is the directional derivative called?
6. Formulate the rules for finding extrema of a function of two variables.
Option 1
Task No. 1
A) 
; b) 
;
V) ; G) 
.
Task No. 2 Examine a function for continuity: find the discontinuity points of the function and determine their type. Construct a schematic graph of the function.
Task No. Given a complex number Z. Required: write the number Z in algebraic and trigonometric forms. .
Task No. 4.
1) y = 3x 5 – sinx, 2) y = tgx, 3) y = , 4) .
Task No. 5. Investigate a function using differential calculus methods and, using the results of the study, construct a graph. .
Task No. 6. The function z=f(x,y) is given. Check whether the identity F≡0 holds? 
Task No. 7 Given a function Z=x 2 +xy+y 2, point and vector. Find:
1) grad z at the point A;
2) derivative at a point A in the direction of the vector .
Option 2
Task No. 1 Calculate the limits of functions without using L'Hopital's rule.
A) 
; b) 
;
V) 
; G) 
.
Task No. 2 Examine a function for continuity: find the discontinuity points of the function and determine their type. Construct a schematic graph of the function.
Task No. 3 Given a complex number Z. Required: write the number Z in algebraic and trigonometric forms.
Task No. 4. Find the first order derivatives of these functions.
Introducing the concept of partial derivative of a function of several 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 definition 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 definition 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 x-axis or parallel to the ordinate). 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), and the displacement will be Δ l(see Figure 5.6).
When moving in a given direction, the z function will increase Δ l z = f(x + Δx; y + Δy) – f(x,y), called the increment of the function z in a given direction l.
Derivative of z l` in the direction l
functions of two variables
z = f(x,y) is the limit of the ratio of the function increment in this direction to the displacement value Δ l as 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 a direction l
It can be proven 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 point (6; -3) (see Figure 5.7).
To do this, first find the partial derivatives of this function at point (3; 1): z x ` = (2x + y)/(x 2 + xy) = (2*3 + 1)/(3 2 + 3*1) = 7 /12;
z y ` = x/(x 2 + xy) = 3/(3 2 + 3*1) = 3/12 = 1/4.
Note that Δx = 6 – 3 = 3; Δy = -3 – 1 = -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.
Gradient function
From the school mathematics course we know 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 be extended to n-dimensional space (instead of two coordinates there will be n coordinates).
Gradient grad z of the function z = f(x 1, x 2, ...x n) is the vector of partial derivatives of the function at a point, i.e. vector with coordinates 
.
It can be proven that the gradient of a function characterizes the direction of the fastest growth of the level of a function at a point.
For example, for the function z = 2x 1 + x 2 (see Figure 5.8), the gradient at any point will have coordinates (2; 1). You can construct it 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 so on. .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).
From Figure 5.8 it is clearly seen that the level of the function increases in the direction of the gradient, since the constructed level lines correspond to the level values 4 > 3 > 2.
Figure 5.8 - Gradient of function z = 2x 1 + x 2
Let's 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) = 10 – solid line).
Figure 5.9 - Gradients of the function z = 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)) = (-4; - 2). Note that the point (0.5; 1) lies on the level line 1/(x 1 x 2) = 2, because z = f(0.5; 1) = 1/(0.5*1) = 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). Let's 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 another 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 equal to
(-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 function level (towards the level line 1/(x 1 x 2) = 10 > 2).
It can be proven that the gradient is always perpendicular to the level line (level surface) passing through a given point.