A lot of theory has been written about geometric meaning. I will not go into the derivation of the function increment, I will remind you of the main thing for completing tasks:
The derivative at the point x is equal to the slope of the tangent to the graph of the function y = f (x) at this point, that is, it is the tangent of the angle of inclination to the X axis.
Let's immediately take the task from the exam and begin to understand it:
Task number 1. The figure shows function graph y = f(x) and the tangent to it at the point with abscissa x0. Find the value of the derivative of the function f(x) at the point x0.
Who is in a hurry and does not want to understand the explanations: build up to any such triangle (as shown below) and divide the standing side (vertical) by the lying (horizontal) and you will be happy if you don’t forget about the sign (if the straight line decreases (→ ↓), then the answer should be with a minus, if the straight line increases (→), then the answer must be positive!)
You need to find the angle between the tangent and the X axis, let's call it α: draw a straight line parallel to the X axis anywhere through the tangent to the graph, we get the same angle.
It is better not to take the point x0, because you will need a large magnifying glass to determine the exact coordinates.
Taking any right-angled triangle (3 options are suggested in the figure), we find tgα (the angles are equal, as corresponding), i.e. we obtain the derivative of the function f(x) at the point x0. Why so?
If we draw tangents at other points x2, x1, etc. tangents will be different.
Let's go back to 7th grade to build a straight line!
The equation of a straight line is given by the equation y = kx + b , where
k - tilt relative to the X axis.
b is the distance between the point of intersection with the Y-axis and the origin.
The derivative of a straight line is always the same: y" = k.
At whatever point on the line we take the derivative, it will be unchanged.
Therefore, it remains only to find tgα (as mentioned above: we divide the standing side by the lying side). We divide the opposite leg by the adjacent one, we get that k \u003d 0.5. However, if the graph is decreasing, the coefficient is negative: k = −0.5.
I advise you to check second way:
Two points can be used to define a straight line. Find the coordinates of any two points. For example, (-2;-2) and (2;-4):
Substitute in the equation y = kx + b instead of y and x the coordinates of the points:
-2 = -2k + b
Solving this system, we get b = −3, k = −0.5
Conclusion: The second method is longer, but in it you will not forget about the sign.
Answer: - 0.5
Task number 2. The figure shows derivative graph functions f(x). Eight points are marked on the x-axis: x1, x2, x3, ..., x8. How many of these points lie on the intervals of increasing function f(x) ?
If the graph of the function is decreasing - the derivative is negative (and vice versa).
If the graph of the function increases, the derivative is positive (and vice versa).
These two phrases will help you solve most of the problems.
Look carefully a drawing of a derivative or a function is given to you, and then choose one of two phrases.
We construct a schematic graph of the function. Because we are given a graph of the derivative, then where it is negative, the graph of the function decreases, where it is positive, it increases!
It turns out that 3 points lie on the areas of increase: x4; x5; x6.
Answer: 3
Task number 3. The function f(x) is defined on the interval (-6; 4). The picture shows graph of its derivative. Find the abscissa of the point where the function takes on the largest value.
I advise you to always build how the function graph goes, with such arrows or schematically with signs (as in No. 4 and No. 5):
Obviously, if the graph increases to -2, then the maximum point is -2.
Answer: -2
Task number 4. The figure shows a graph of the function f(x) and twelve points on the x-axis: x1, x2, ..., x12. At how many of these points is the derivative of the function negative?
The task is inverse, given the graph of the function, you need to schematically build what the graph of the derivative of the function will look like, and calculate how many points will lie in the negative range.
Positive: x1, x6, x7, x12.
Negative: x2, x3, x4, x5, x9, x10, x11.
Another type of task, when asked about some terrible "extremes"? It will not be difficult for you to find what it is, but I will explain for the graphs.
Task number 5. The figure shows a graph of the derivative of the function f(x), defined on the interval (-16; 6). Find the number of extremum points of the function f(x) on the segment [-11; 5].
Note the range from -11 to 5!
Let's turn our bright eyes to the plate: the graph of the derivative of the function is given => then the extrema are the points of intersection with the X axis.
Answer: 3
Task number 6. The figure shows a graph of the derivative of the function f (x), defined on the interval (-13; 9). Find the number of maximum points of the function f(x) on the interval [-12; 5].
Note the range from -12 to 5!
You can look at the plate with one eye, the maximum point is an extremum, such that before it the derivative is positive (the function increases), and after it the derivative is negative (the function decreases). These points are circled.
The arrows show how the graph of the function behaves.
Answer: 3
Task number 7. The figure shows a graph of the function f(x) defined on the interval (-7; 5). Find the number of points where the derivative of the function f(x) is equal to 0.
You can look at the above table (the derivative is zero, which means these are extremum points). And in this problem, the graph of the function is given, which means you need to find number of inflection points!
And you can, as usual: we build a schematic graph of the derivative.
The derivative is zero when the graph of functions changes its direction (from increasing to decreasing and vice versa)
Answer: 8
Task number 8. The picture shows derivative graph function f(x) defined on the interval (-2; 10). Find the intervals of increasing function f(x). In your answer, indicate the sum of integer points included in these intervals.
Let's build a schematic graph of the function:
Where it increases, we get 4 integer points: 4 + 5 + 6 + 7 = 22.
Answer: 22
Task number 9. The picture shows derivative graph function f(x) defined on the interval (-6; 6). Find the number of points f(x) where the tangent to the graph of the function is parallel to or coincides with the line y = 2x + 13.
We are given a graph of the derivative! This means that our tangent must also be “translated” into a derivative.
Tangent derivative: y" = 2.
Now let's build both derivatives:
The tangents intersect at three points, so our answer is 3.
Answer: 3
Task number 10. The figure shows the graph of the function f (x), and the points -2, 1, 2, 3 are marked. At which of these points is the value of the derivative the smallest? Please indicate this point in your answer.
The task is somewhat similar to the first one: to find the value of the derivative, you need to build a tangent to this graph at a point and find the coefficient k.
If the line is decreasing, k< 0.
If the line is increasing, k > 0.
Let's think about how the value of the coefficient will affect the slope of the straight line:
With k = 1 or k = − 1, the graph will be in the middle between the x and y axes.
The closer the straight line to the X-axis, the closer the coefficient k to zero.
The closer the line is to the Y-axis, the closer the coefficient k is to infinity.
At point -2 and 1 k<0, однако в точке 1 прямая убывает "быстрее" больше похоже на ось Y =>that is where the smallest value of the derivative will be
Answer: 1
Task number 11. The line is tangent y = 3x + 9 to the graph of the function y = x³ + x² + 2x + 8 . Find the abscissa of the point of contact.
The line will be tangent to the graph when the graphs have a common point, like their derivatives. Equate the equations of the graphs and their derivatives:
Solving the second equation, we get 2 points. To check which one is suitable, we substitute each of the x's into the first equation. Only one will do.
I don’t want to solve a cubic equation at all, but a square one for a sweet soul.
That's just what to write down in response, if you get two "normal" answers?
When substituting x (x) into the original graphs y \u003d 3x + 9 and y \u003d x³ + x² + 2x + 8, you should get the same Y
y= 1³+1²+2×1+8=12
Right! So x=1 will be the answer
Answer: 1Task number 12. The line y = − 5x − 6 is tangent to the graph of the function ax² + 5x − 5 . Find a .
Similarly, we equate the functions and their derivatives:
Let's solve this system with respect to the variables a and x :
Answer: 25
The task with derivatives is considered one of the most difficult in the first part of the exam, however, with a small amount of attentiveness and understanding of the issue, you will succeed, and you will raise the percentage of completion of this task!
In task No. 7 of the profile level of the USE in mathematics, it is necessary to demonstrate knowledge of the function of the derivative and the antiderivative. In most cases, simply defining the concepts and understanding the meanings of the derivative is sufficient.
The figure shows a graph of a differentiable function y = f(x). Nine points are marked on the x-axis: x 1 , x 2 , …, x 9 . Among these points, find all points where the derivative of the function y = f(x) is negative. In your answer, indicate the number of points found.
1. On the graph, the function periodically increases, periodically decreases.
2. In those intervals where the function decreases, the derivative has negative values.
3. These intervals contain points x 3 , x 4 , x 5 , x 9 . There are 4 such points.
The figure shows the graph of the function y \u003d f (x). Points -2, -1, 2, 4 are marked on the x-axis. At which of these points is the value of the derivative the largest? Please indicate this point in your answer.
1. The function has several intervals of decreasing and increasing.
2. Where the function decreases. The derivative has a minus sign. Such points are among those indicated. But there are points on the graph where the function increases. Their derivative is positive. These are the points with abscissas -2 and 2.
3. Consider a graph at points with x=-2 and x=2. At the point x = 2, the function goes up steeper, which means that the tangent at this point has a larger slope. Therefore, at the point with the abscissa 2. The derivative has the greatest value.
The line is tangent to the graph of the function 
. Find a.
1. The coordinates of the tangent point satisfy both equations: the tangent and the function. So we can equate the equations. We get:
2. We simplify the equality by moving all the terms in one direction:
3. There must be one solution at the point of contact, so the discriminant of the resulting equation must be equal to zero. This is the condition for the uniqueness of the root of the quadratic equation.
4. We get:
If the task is solved correctly, then you get 1 point.
Approximately 5 minutes.
02.01.2020
Rare daughters-in-law can boast that they have even and friendly relations with their mother-in-law. Usually the opposite happens
DERIVATIVE-derivative of the function y = f(x) defined on some interval ( a, b) at the point x this interval is called the limit to which the ratio of the increment of the function tends f at that point to the corresponding increment of the argument as the increment of the argument approaches zero.
The derivative is usually denoted as follows:
Other notations are also widely used:
Let the point M moves in a straight line. Distance s moving point, counted from some initial position M 0 , depends on time t, i.e. s is a function of time t: s= f(t). Let at some point in time t moving point M was at a distance s from the starting position M 0, and at some next moment t+ D t was in a position M 1 - on distance s+ D s from the initial position ( see pic.).
Thus, for a period of time D t distance s changed by the value D s. In this case, we say that during the time interval D t magnitude s received increment D s.
The average speed cannot in all cases accurately characterize the speed of moving a point. M at the time t. If, for example, the body at the beginning of the interval D t moved very quickly, and at the end very slowly, then the average speed will not be able to reflect the indicated features of the movement of the point and give an idea of the true speed of its movement at the moment t. To more accurately express the true speed using the average speed, you need to take a smaller period of time D t. It most fully characterizes the speed of movement of a point at the moment t the limit to which the average speed tends at D t® 0. This limit is called the speed of movement at a given moment:
Thus, the speed of movement at a given moment is the limit of the ratio of the increment of the path D s to the time increment D t when the time increment tends to zero. Because
The construction of tangents is one of those problems that led to the birth of differential calculus. The first published work on differential calculus, written by Leibniz, was titled A new method of maxima and minima, as well as tangents, for which neither fractional nor irrational quantities are an obstacle, and a special kind of calculus for this.
Let the curve be the graph of the function y =f(x) in a rectangular coordinate system ( cm. rice.).
For some value x function matters y =f(x). These values x and y point on the curve M 0(x, y). If the argument x give increment D x, then the new value of the argument x+ D x corresponds to the new value of the function y+ D y = f(x + D x). The corresponding point of the curve will be the point M 1(x+ D x,y+ D y). If we draw a secant M 0M 1 and denote by j angle formed by a secant with a positive axis direction Ox, it is directly seen from the figure that
If now D x tends to zero, then the point M 1 moves along the curve, approaching the point M 0 and angle j changes with change D x. At Dx® 0 the angle j tends to some limit a and the line passing through the point M 0 and the component with the positive direction of the abscissa axis, angle a, will be the desired tangent. Its slope:
Hence, f´( x) = tga
those. derivative value f´( x) for a given value of the argument x equals the tangent of the angle formed by the tangent to the graph of the function f(x) at the corresponding point M 0(x,y) with positive axis direction Ox.
Definition. If the function y = f(x) has a derivative at the point x = x 0, then the function is differentiable at this point.
If the function y = f(x) is differentiable at some point x = x 0, then it is continuous at this point.
Thus, at discontinuity points, the function cannot have a derivative. The converse conclusion is false, i.e. from the fact that at some point x = x 0 function y = f(x) is continuous, it does not follow that it is differentiable at this point. For example, the function y = |x| continuous for all x(–Ґ x x = 0 has no derivative. There is no tangent to the graph at this point. There is a right tangent and a left tangent, but they do not coincide.
Some theorems on differentiable functions. Theorem on the roots of the derivative (Roll's theorem). If the function f(x) is continuous on the segment [a,b], is differentiable at all interior points of this segment and at the ends x = a and x = b vanishes ( f(a) = f(b) = 0), then inside the segment [ a,b] there is at least one point x= With, a c b, in which the derivative fў( x) vanishes, i.e. fў( c) = 0.
Finite increment theorem (Lagrange's theorem). If the function f(x) is continuous on the segment [ a, b] and is differentiable at all interior points of this segment, then inside the segment [ a, b] there is at least one point With, a c b that
f(b) – f(a) = fў( c)(b– a).
Theorem on the ratio of increments of two functions (Cauchy's theorem). If f(x) and g(x) are two functions continuous on the segment [a, b] and differentiable at all interior points of this segment, and gў( x) does not vanish anywhere inside this segment, then inside the segment [ a, b] there is such a point x = With, a c b that
Let the function y =f(x) is differentiable on some interval [ a, b]. Derivative values f ў( x), generally speaking, depend on x, i.e. derivative f ў( x) is also a function of x. When this function is differentiated, the so-called second derivative of the function is obtained f(x), which is denoted f ўў ( x).
derivative n- order of the function f(x) is called the derivative (of the first order) of the derivative n- 1- th and is denoted by the symbol y(n) = (y(n– 1))ў.
Function differential y = f(x), where x is an independent variable, is dy = f ў( x)dx, some function from x, but from x only the first factor can depend f ў( x), while the second factor ( dx) is the increment of the independent variable x and does not depend on the value of this variable. Because dy there is a function from x, then we can determine the differential of this function. The differential of the differential of a function is called the second or second-order differential of this function and is denoted d 2y:
d(dx) = d 2y = f ўў( x)(dx) 2 .
Differential n- order is called the first differential of the differential n- 1- order:
d n y = d(d n–1y) = f(n)(x)dx(n).
If the function depends not on one, but on several arguments x i(i changes from 1 to n,i= 1, 2,… n),f(x 1,x 2,… x n), then in differential calculus the concept of a partial derivative is introduced, which characterizes the rate of change of a function of several variables when only one argument changes, for example, x i. Partial derivative of the 1st order with respect to x i is defined as the ordinary derivative, it is assumed that all arguments except x i, keep constant values. For partial derivatives, we introduce the notation
Partial derivatives of the 1st order defined in this way (as functions of the same arguments) can, in turn, also have partial derivatives, these are partial derivatives of the second order, etc. Taken with respect to different arguments, such derivatives are called mixed. Continuous mixed derivatives of the same order do not depend on the order of differentiation and are equal to each other.
Anna Chugainova
derivative functions at a point is called the limit of the ratio of the increment of the function to the increment of the argument, provided that tends to zero.
Basic rules for finding the derivative
If - and - are differentiable functions at a point, (i.e. functions that have derivatives at a point), then:
Table of derivatives of basic functions
1. 8. 
2. 9. 
3. 
10.
5. 
12. 
6. 
13.
7. 
The rule of differentiation of a complex function. If and, i.e. , where and have derivatives, then
Differentiation of a function defined parametrically. Let the dependence of a variable on a variable be given parametrically by means of a parameter:
Task 3. Find derivatives of given functions.
1) 
Solution. Applying rule 2 for finding derivatives and formulas 1 and 2 of the table of derivatives, we obtain:
Solution. Applying rule 4 for finding derivatives and formulas 1 and 13 of the table of derivatives, we obtain:
.
Solution. Applying rule 3 for finding derivatives and formulas 5 and 11 of the table of derivatives, we obtain:
Solution. Assuming where, according to the formula for finding the derivative of a complex function, we get:
Solution. We have: Then, according to the formula for finding the derivative of a function given parametrically, we get:
4. Derivatives of higher orders. L'Hopital's rule.
The second order derivative of a function is called the derivative of its derivative, i.e. . The following notation is used for the second derivative: or, or.
1st order derivative of a function is called the derivative of its th-order derivative. For the derivative of the -th order, the following notation is used: or, or.
L'Hopital's rule. Let the functions and be differentiable in a neighborhood of a point, and the derivative does not vanish. If the functions and are either infinitesimal or infinitely large at the same time, and there is a limit of the ratio of at, then there is also a limit of the ratio of at. And
.
The rule also applies when
Note that in some cases, the disclosure of uncertainties of the form or may require repeated application of L'Hospital's rule.
View uncertainties, etc. elementary transformations are easily reduced to uncertainties of the form or.
Task 4. Find the limit using L'Hopital's rule.
Solution Here we have an indeterminacy of the form, since at. Let's apply L'Hospital's rule:
.
After applying L'Hopital's rule, we again got the uncertainty of the form, because at. Applying L'Hopital's rule again, we get:
.
5. Function research
a) Increasing and decreasing functions
The function is called increasing on the segment , if for any points and from the segment where, the inequality takes place. If the function is continuous on the interval and at, then it increases on the interval.
The function is called waning on the segment , if for any points and from the segment, where, the inequality takes place. If the function is continuous on the interval and at, then it decreases on the interval.
If a function is only increasing or only decreasing on a given interval, then it is called monotonous on the interval.
b) Function extremes
minimum point functions .
If exists -neighborhood of the point such that the inequality holds for all points in this neighborhood, then the point is called maximum point functions .
The maximum and minimum points of a function are called its extreme points.
The point is called stationary point if or does not exist.
If there is a -neighbourhood of the stationary point such that for and for, then - is the maximum point of the function.
If there is a -neighborhood of the stationary point such that for and for, then -point of minimum of the function.
a) Curve direction. Inflection points
convex up on the interval , if it is located below the tangent drawn to the graph of the function at any point in this interval.
A sufficient condition for upward convexity of the graph of a function on an interval is the fulfillment of the inequality for any of the intervals under consideration.
The graph of a differentiable function is called convex down on the interval , if it is located above the tangent drawn to the graph of the function at any point in this interval.
A sufficient condition for downward convexity of the graph of a function on an interval is the fulfillment of the inequality for any of the intervals under consideration.
The point at which the direction of the convexity of the function graph changes is called inflection point.
A point where or does not exist is the abscissa of the inflection point if it has different signs to the left and right of it.
d) Asymptotes
If the distance from the point of the graph of a function to a certain straight line tends to zero with an infinite distance from the origin of the point, then the straight line is called asymptote of the graph of the function.
If there is a number such that, then the line is vertical asymptote.
If there are limits 
, then the line is oblique (horizontal at k=0) asymptote.
e) General study of function
1. Function scope
2. Points of intersection of the graph with the coordinate axes
3. Investigation of a function for continuity, even / odd and periodicity
4. Intervals of monotonicity of a function
5. Extremum points of the function
6. Convexity intervals and inflection points of the graph of a function
7. Asymptotes of the graph of a function
8. Graph of the function.
Task 5. Investigate the function and plot its graph.
Solution. 1) The function is defined on the entire number axis, except for the point where the denominator of the fraction vanishes. . We have: does not belong to the scope of this function. Therefore, the stationary points of this function are the points, the minimum value (as shown in the figure).
Showing the relationship of the sign of the derivative with the nature of the monotonicity of the function.
Please be extremely careful in the following. Look, the schedule of WHAT is given to you! Function or its derivative
Given a graph of the derivative, then we are only interested in function signs and zeros. No "knolls" and "hollows" are of interest to us in principle!
The figure shows a graph of a function defined on an interval. Determine the number of integer points where the derivative of the function is negative.
Solution:
In the figure, the areas of decreasing function are highlighted in color:
4 integer values fall into these areas of decreasing function.
The figure shows a graph of a function defined on an interval. Find the number of points where the tangent to the graph of the function is parallel or coincident with the line.
Solution:
Since the tangent to the function graph is parallel (or coincides) with a straight line (or, which is the same, ) having slope, equal to zero, then the tangent has a slope .
This in turn means that the tangent is parallel to the axis, since the slope is the tangent of the angle of inclination of the tangent to the axis.
Therefore, we find extremum points on the graph (maximum and minimum points), - it is in them that the functions tangent to the graph will be parallel to the axis.
There are 4 such points.
The figure shows a graph of the derivative of a function defined on the interval . Find the number of points where the tangent to the graph of the function is parallel or coincident with the line.
Solution:
Since the tangent to the graph of the function is parallel (or coincides) with a straight line, which has a slope, then the tangent has a slope.
This in turn means that at the points of contact.
Therefore, we look at how many points on the graph have an ordinate equal to .
As you can see, there are four such points.
The figure shows a graph of a function defined on an interval. Find the number of points where the derivative of the function is 0.
Solution:
The derivative is zero at the extremum points. We have 4 of them:
The figure shows a function graph and eleven points on the x-axis:. At how many of these points is the derivative of the function negative?
Solution:
On intervals of decreasing function, its derivative takes negative values. And the function decreases at points. There are 4 such points.
The figure shows a graph of a function defined on an interval. Find the sum of the extremum points of the function .
Solution:
extremum points are the maximum points (-3, -1, 1) and the minimum points (-2, 0, 3).
The sum of extreme points: -3-1+1-2+0+3=-2.
The figure shows a graph of the derivative of a function defined on the interval . Find the intervals of increasing function . In your answer, indicate the sum of integer points included in these intervals.
Solution:
The figure highlights the intervals on which the derivative of the function is non-negative.
There are no integer points on the small interval of increase, on the interval of increase there are four integer values : , , and .
Their sum:
The figure shows a graph of the derivative of a function defined on the interval . Find the intervals of increasing function . In your answer, write the length of the largest of them.
Solution:
In the figure, all the intervals on which the derivative is positive are highlighted, which means that the function itself increases on these intervals.
The length of the largest of them is 6.
The figure shows a graph of the derivative of a function defined on the interval . At what point on the segment does it take the greatest value.
Solution:
We look at how the graph behaves on the segment, namely, we are interested in derivative sign only .
The sign of the derivative on is minus, since the graph on this segment is below the axis.
