Write formulas for differentiating basic elementary functions. Formulas and rules for differentiation (finding the derivative)
Table of derivatives of elementary functions
Definition 1
The calculation of the derivative is called differentiation.
Denote the derivative $y"$ or $\frac(dy)(dx)$.
Remark 1
To find the derivative of a function, according to the basic rules, differentiation is converted into another function.
Consider the table of derivatives. Let us pay attention to the fact that functions after finding their derivatives are transformed into other functions.
The only exception is $y=e^x$, which turns into itself.
Derivative Differentiation Rules
Most often, when finding a derivative, it is required not only to look at the table of derivatives, but first to apply the rules of differentiation and the proof of the derivative of the product, and only then use the table of derivatives of elementary functions.
1. The constant is taken out of the sign of the derivative
$C$ is a constant (constant).
Example 1
Differentiate the function $y=7x^4$.
Solution.
Find $y"=(7x^4)"$. We take out the number $7$ for the sign of the derivative, we get:
$y"=(7x^4)"=7(x^4)"=$
using the table, you need to find the value of the derivative of the power function:
$=7 \cdot 4x^3=$
We transform the result to the form accepted in mathematics:
Answer:$28x^3$.
2. The derivative of the sum (difference) is equal to the sum (difference) of the derivatives:
$(u \pm v)"=u" \pm v"$.
Example 2
Differentiate the function $y=7+x-5x^3+4 \sin x-9\sqrt(x^2)+\frac(4)(x^4) -11\cot x$.
Solution.
$y"=(7+x-5x^5+4 \sin x-9\sqrt(x^2)+\frac(4)(x^4) -11\cot x)"=$
apply the rule of differentiation of the derivative sum and difference:
$=(7)"+(x)"-(5x^5)"+(4 \sin x)"-(9\sqrt(x^2))"+(\frac(4)(x^4) )"-(11\cot x)"=$
note that when differentiating, all powers and roots must be transformed to the form $x^(\frac(a)(b))$;
we take all the constants out of the sign of the derivative:
$=(7)"+(x)"-(5x^5)"+(4\sin x)"-(9x^(\frac(2)(5)))"+(4x^(-4) )"-(11\cot x)"=$
$=(7)"+(x)"-5(x^5)"+4(\sin x)"-9(x^(\frac(2)(5)))"+4(x^( -4))"-11(\cot x)"=$
having dealt with the rules of differentiation, some of them (for example, like the last two) are applied simultaneously in order to avoid rewriting a long expression;
we have obtained an expression from elementary functions under the sign of the derivative; Let's use the table of derivatives:
$=0+1-5 \cdot 5x^4+4\cos x-9 \cdot \frac(2)(5) x^(-\frac(3)(5))+12x^(-5)- 11 \cdot \frac(-1)(\sin^2 x)=$
transform to the form accepted in mathematics:
$=1-25x^4+4 \cos x-\frac(18)(5\sqrt(x^3))+\frac(12)(x^5) +\frac(11)(\sin^2 x)$
Note that when finding the result, it is customary to convert terms with fractional powers into roots, and with negative ones into fractions.
Answer: $1-25x^4+4 \cos x-\frac(18)(5\sqrt(x^3))+\frac(12)(x^5) +\frac(11)(\sin^2 x )$.
3. The formula for the derivative of the product of functions:
$(uv)"=u" v+uv"$.
Example 3
Differentiate the function $y=x^(11) \ln x$.
Solution.
First we apply the rule for calculating the derivative of the product of functions, and then we use the table of derivatives:
$y"=(x^(11) \ln x)"=(x^(11))" \ln x+x^(11) (\lnthx)"=11x^(10) \ln x+x^ (11) \cdot \frac(1)(x)=11x^(10) \ln x-\frac(x^(11))(x)=11x^(10) \ln x-x^(10)=x ^(10) (11 \ln x-1)$.
Answer: $x^(10) (11 \ln x-1)$.
4. The formula for the derivative of a private function:
$(\frac(u)(v))"=\frac(u" v-uv")(v^2)$.
Example 4
Differentiate the function $y=\frac(3x-8)(x^5-7)$.
Solution.
$y"=(\frac(3x-8)(x^5-7))"=$
according to the rules of priority of mathematical operations, we first perform division, and then addition and subtraction, so we first apply the rule for calculating the derivative of the quotient:
$=\frac((3x-8)" (x^5-7)-(3x-8) (x^5-7)")((x^5-7)^2) =$
apply the rules of derivatives of the sum and difference, open the brackets and simplify the expression:
$=\frac(3(x^5-7)-5x^4 (3x-8))((x^5-7)^2) =\frac(3x^5-21-15x^5+40x^ 4)((x^5-7)^2) =\frac(-12x^5+40x^4-21)((x^5-7)^2)$ .
Answer:$\frac(-12x^5+40x^4-21)((x^5-7)^2)$.
Example 5
Let us differentiate the function $y=\frac(x^7-2x+3)(x)$.
Solution.
The function y is a quotient of two functions, so we can apply the rule for calculating the derivative of a quotient, but in this case we get a cumbersome function. To simplify this function, you can divide the numerator by the denominator term by term:
$y=\frac(x^7-13x+9)(x)=x^6-13+\frac(9)(x)$.
Let us apply to the simplified function the rule of differentiation of the sum and difference of functions:
$y"=(x^6-13+\frac(9)(x))"=(x^6)"+(-13)"+9(x^(-1))"=6x^5+ 0+9 \cdot (-x^(-2))=$
$=6x^5-\frac(9)(x^2)$.
Answer: $6x^5-\frac(9)(x^2)$.
Let the function y = f(x) be defined in the interval X. derivative function y \u003d f (x) at the point x o is called the limit
=
.
If this limit finite, then the function f(x) is called differentiable at the point x o; moreover, it turns out to be necessarily and continuous at this point.
If the considered limit is equal to (or - ), then provided that the function at the point X o is continuous, we will say that the function f(x) has at a point X o infinite derivative.
The derivative is denoted by the symbols
y , f (x o), , .
Finding the derivative is called differentiation functions. The geometric meaning of the derivative is that the derivative is the slope of the tangent to the curve y=f(x) at a given point X o ; physical sense - in that the derivative of the path with respect to time is the instantaneous speed of the moving point during rectilinear motion s = s(t) at the moment t o .
If a With is a constant number, and u = u(x), v = v(x) are some differentiable functions, then the following differentiation rules hold:
1) (c) " = 0, (cu) " = cu";
2) (u+v)" = u"+v";
3) (uv)" \u003d u "v + v" u;
4) (u / v) "= (u" v-v "u) / v 2;
5) if y = f(u), u = (x), i.e. y = f((x)) - complex function, or superposition, composed of differentiable functions and f, then , or
6) if for the function y = f(x) there exists an inverse differentiable function x = g(y), and 0, then .
Based on the definition of the derivative and the rules of differentiation, one can compile a list of tabular derivatives of the basic elementary functions.
1. (u )" = u 1 u" ( R).
2. (a u)" = a u lna u".
3. (e u)" = e u u".
4. (log a u)" = u"/(u ln a).
5. (ln u)" = u"/u.
6. (sin u)" = cos u u".
7. (cos u)" = - sin u u".
8. (tg u)" = 1/ cos 2 u u".
9. (ctg u)" = - u" / sin 2 u.
10. (arcsin u)" = u" / .
11. (arccos u)" = - u" / .
12. (arctg u)" = u"/(1 + u 2).
13. (arcctg u)" = - u"/(1 + u 2).
Let us calculate the derivative of the exponential expression y=u v , (u>0), where u and v essence of the function X having derivatives at a given point u",v".
Taking the logarithm of the equality y=u v , we obtain ln y = v ln u.
Equating derivatives with respect to X from both parts of the obtained equality using rules 3, 5 and the formula for the derivative of the logarithmic function, we will have:
y"/y = vu"/u + v" ln u, whence y" = y (vu"/u + v" ln u).
(u v)"=u v (vu"/u+v" log u), u > 0.
For example, if y \u003d x sin x, then y" \u003d x sin x (sin x / x + cos x ln x).
If the function y = f(x) is differentiable at a point x, i.e. has a finite derivative at this point y", then = y "+, where 0 at х 0; hence y = y" х + x.
The main part of the function increment, linear with respect to x, is called differential functions and is denoted by dy: dy \u003d y "x. If we put y \u003d x in this formula, then we get dx \u003d x" x \u003d 1x \u003d x, therefore dy \u003d y "dx, i.e. a symbol for the notation for the derivative can be thought of as a fraction.
Function increment y is the increment of the ordinate of the curve, and the differential d y is the increment of the ordinate of the tangent.
Let us find for the function y=f(x) its derivative y = f (x). The derivative of this derivative is called second order derivative functions f(x), or second derivative, and denoted 
.
The following are defined and denoted in the same way:
third order derivative
-
,
fourth order derivative -
and generally speaking nth order derivative
-
.
Example 3.15. Calculate the derivative of the function y=(3x 3 -2x+1)sin x.
Solution. By rule 3, y"=(3x 3 -2x+1)"sin x + (3x 3 -2x+1)(sin x)" = = (9x 2 -2)sin x + (3x 3 -2x +1) cos x.
Example 3.16 . Find y", y = tg x + .
Solution. Using the rules for differentiating the sum and the quotient, we get: y"=(tgx + )" = (tgx)" + ()" = + 
=
.
Example 3.17. Find the derivative of a complex function y= , u=x 4 +1.
Solution. According to the rule of differentiation of a complex function, we get: y "x \u003d y " u u" x \u003d () " u (x 4 +1)" x \u003d (2u +. Since u \u003d x 4 +1, then (2 x 4 + 2+ 
.
In all the formulas below, letters u and v differentiable functions of the independent variable are denoted x: , 
, but in letters a, c, n- permanent:
1. 
3. 
4. 
5. 
6. 
The remaining formulas are written both for functions of an independent variable and for complex functions:
8. 
9. 
11. 
12. 
13. 
14. 
15. 
16. 
17. 
7a. 
8a. 
9a. 
11a. 
12a. 
13a. 
16a. 
17a. 
When solving the examples below, detailed notes are made. However, one should learn to differentiate without intermediate entries.
Example 1 Find the derivative of a function 
.
Solution. This function is the algebraic sum of functions. We differentiate it using formulas 3, 5, 7 and 8:
Example 2 Find the derivative of a function 
Solution. Applying formulas 6, 3, 7 and 1, we get
Example 3 Find the derivative of a function 
and calculate its value at 
Solution. This is a complex function with an intermediate argument . Using formulas 7a and 10, we have
.
Example 4 Find the derivative of a function 
.
Solution. This is a complex function with an intermediate argument . Applying formulas 3, 5, 7a, 11, 16a, we get
Example 5 Find the derivative of a function 
.
Solution. We differentiate this function by formulas 6, 12, 3 and 1:
Example 6 Find the derivative of a function 
and calculate its value at .
Solution. First, we transform the function using the properties of logarithms:
Now we differentiate by formulas 3, 16a, 7 and 1:
.
Let us calculate the value of the derivative at .
Example 7 Find the derivative of the function and calculate its value at .
Solution. We use formulas 6, 3, 14a, 9a, 5 and 1:
.
Calculate the value of the derivative at :
.
The geometric meaning of the derivative.
The derivative of a function has a simple and important geometric interpretation.
If the function 
differentiable at a point X, then the graph of this function has a tangent at the corresponding point, and the slope of the tangent is equal to the value of the derivative at the point under consideration.
The slope of the tangent drawn to the graph of the function 
at point ( X 0 , at 0), is equal to the value of the derivative of the function at x = x 0 , i.e. 
.
The equation for this tangent has the form
Example 8. Write an equation for a tangent to a function graph 
at point A (3.6).
Solution. To find the slope of the tangent, we find the derivative of this function:
X= 3:
The tangent equation has the form
, or 
, i.e. 
Example 9 Compose the equation of the tangent drawn to the graph of the function at the point with the abscissa x=2.
Solution. First, find the ordinate of the touch point. Since point A lies on the curve, its coordinates satisfy the equation of the curve, i.e.
; 
.
The equation of the tangent drawn to the curve at the point has the form 
. To find the slope of the tangent, we find the derivative:
.
The slope of the tangent is equal to the value of the derivative of the function at X= 2:
The tangent equation is:
, 
, i.e. 
The physical meaning of the derivative. If the body moves in a straight line according to the law s=s(t), then for a period of time (from the moment t until the moment 
) it will go some way. Then there is the average speed of movement for a period of time .
speed body movements at a given time t is called the limit of the ratio of the path to the increment of time, when the increment of time tends to zero:
.
Therefore, the time derivative of the path s t equal to the speed of the rectilinear motion of the body at a given time:
.
The rate of physical, chemical and other processes is also expressed using the derivative.
Function derivative is equal to the rate of change of this function for a given value of the argument X:
Example 10 The law of movement of a point along a straight line is given by the formula 
(s - in meters, t - in seconds). Find the speed of the point at the end of the first second.
Solution. The speed of a point at a given time is equal to the derivative of the path s by time t:
,
So, the speed of the point at the end of the first second is 9 m/s.
Example 11. A body thrown vertically upward moves according to the law , where v 0 - initial speed, g is the free fall acceleration. Find the speed of this movement for any moment of time t. How long will the body rise and to what height will it rise if v0= 40 m/s?
Solution. The speed at which a point is moving at a given time t equal to the derivative of the path s by time t:
.
At the highest point of the ascent, the velocity of the body is zero:
, 
, , 
, 
With.
Over 40/ g seconds the body rises to a height
, 
m.
Second derivative.
Function derivative in general is a function of X. If we calculate the derivative of this function, then we get the second-order derivative or the second derivative of the function .
Second derivative functions is called the derivative of its first derivative 
.
The second derivative of a function is denoted by one of the symbols - , , . In this way, 
.
Derivatives of any order are defined and denoted in a similar way. For example, a third order derivative:
or 
,
Example 12. 
.
Solution. First we find the first derivative
Example 13 Find the second derivative of a function 
and calculate its value at x=2.
Solution. First we find the first derivative:
Differentiating again, we find the second derivative:
Let us calculate the value of the second derivative at x=2; we have
The physical meaning of the second derivative.
If the body moves in a straight line according to the law s = s(t), then the second derivative of the path s by time t equal to the acceleration of the body at a given time t:
Thus, the first derivative characterizes the speed of some process, and the second derivative characterizes the acceleration of the same process.
Example 14 The point moves in a straight line according to the law 
. Find the speed and acceleration of the movement 
.
Solution. The speed of the body at a given time is equal to the derivative of the path s by time t, and acceleration is the second derivative of the path s by time t. We find:
; then ;
; then 
Example 15 The speed of rectilinear motion is proportional to the square root of the path traveled (as, for example, in free fall). Prove that this motion occurs under the action of a constant force.
Solution. According to Newton's law, the force F causing the movement is proportional to the acceleration, i.e.
or 
According to the condition 
. Differentiating this equality, we find
Therefore, the acting force 
.
Applications of the derivative to the study of a function.
1) The condition for the function to increase: A differentiable function y = f(x) monotonically increases on the interval X if and only if its derivative is greater than zero, i.e. y = f(x) f'(x) > 0. This condition geometrically means that the tangent to the graph of this function forms an acute angle with a positive direction to the x-axis.
2) The condition for the function to decrease: A differentiable function y = f(x) monotonically decreases on the interval X if and only if its derivative is less than zero, i.e.
y = f(x)↓ f'(x) This condition geometrically means that the tangent to the graph of this function forms an obtuse angle with the positive direction of the x-axis)
3) The condition of the constancy of the function: A differentiable function y = f(x) is constant on the interval X if and only if its derivative is equal to zero, i.e. y = f(x) - constant f'(x) = 0 . This condition geometrically means that the tangent to the graph of this function is parallel to the oX axis, i.e. α \u003d 0)
Function extremes.
Definition 1: The point x \u003d x 0 is called minimum point function y = f(x), if this point has a neighborhood, for all points of which (except for the point itself) the inequality f(x)> f(x 0)
Definition 2: The point x \u003d x 0 is called maximum point function y = f(x) if this point has a neighborhood for all points of which (except for the point itself) the inequality f(x)< f(x 0).
Definition 3: The minimum or maximum point of a function is called a point extremum. The value of the function at this point is called extreme.
Remarks: 1. The maximum (minimum) is not necessarily the maximum (smallest) value of the function;
2. A function can have several maximums or minimums;
3. A function defined on a segment can reach an extremum only at the interior points of this segment.
5) Necessary condition for an extremum: If the function y \u003d f (x) has an extremum at the point x \u003d x 0, then at this point the derivative is equal to zero or does not exist. These points are called critical points of the 1st kind.
6) Sufficient conditions for the existence of the extremum of the function: Let the function y \u003d f (x) be continuous on the interval X and have inside this interval as a critical point of the 1st kind x \u003d x 0, then:
a) if this point has a neighborhood in which for x< х 0 f’(x) < 0, а при x>x 0 f’(x) > 0, then x = x 0 is a point minimum functions y = f(x);
b) if this point has a neighborhood in which for x< х 0 f’(x) >0, and for x> x 0
f'(x)< 0, то х = х 0 является точкой maximum functions y = f(x);
c) if this point has such a neighborhood that in it both to the right and to the left of the point x 0 the signs of the derivative are the same, then there is no extremum at the point x 0.
The intervals of decreasing or increasing functions are called intervals. monotony.
Definition1: The curve y = f(x) is called convex down on the interval a< х <в, если она лежит выше касательной в любой точке этого промежутка и кривая у = f(x) называется convex up on the interval a< х <в, если она лежит ниже касательной в любой точке этого промежутка.
Definition 2: The intervals in which the graph of the function is convex up or down are called swell at intervals function graph.
A sufficient condition for the curve to be convex. The graph of the differentiable function Y = f(x) is convex up on the interval a< х <в, если f”(x) < 0 и convex down, if f”(x) > 0.
Definition 1: The points at which the second derivative is zero or does not exist are called critical points of the second kind.
Definition 2: The point of the graph of the function Y = f(x), separating the intervals of the convexity of the opposite directions of this graph, is called the point inflection.
inflection point
Example: Given a function y \u003d x 3 - 2x 2 + 6x - 4. Investigate the function for intervals of monotonicity and extremum points. Determine the direction of the convexity and inflection points.
Solution: 1. Find the domain of the function: D(y) = ;
2. Find the first derivative: y’ = 3x 2 - 4x+ 6;
3. Let's solve the equation: y' = 0, 3x 2 - 4x+ 6 = 0, D 0, then this equation has no solution, therefore there are no extremum points. y' , then the function increases over the entire domain of definition.
4. Find the second derivative: y” = 6x - 4;
5. Solve the equation: y” = 0, 6x - 4 = 0, x =
Answer: ( ; - ) - inflection point, the function is convex upwards at x and convex upwards at x
Asymptotes.
1. Definition: The asymptote of a curve is a straight line to which the graph of a given function approaches indefinitely.
2. Types of asymptotes:
1) Vertical asymptotes. The graph of the function y = f(x) has a vertical asymptote if . The vertical asymptote equation has the form x = a
2) Horizontal asymptotes. The graph of the function y = f(x) has a horizontal asymptote if 
. The horizontal asymptote equation is y = b.
Example 1: For the function y = find the asymptotes.
3) Oblique asymptotes. The straight line y = kx + b is called the oblique asymptote of the graph of the function y = f(x) if . The values of k and b are calculated by the formulas: k = ; b = .
Solution: 
, then y = 0 is the horizontal asymptote;
(since x - 3 ≠ 0, x ≠ 3), then x = 3 is the vertical asymptote. 
,t. i.e. k = 0, then the curve has no oblique asymptote.
Example 2: For the function y = find the asymptotes.
Solution: x 2 - 25 ≠ 0 with x ≠ ± 5, then x \u003d 5 and x \u003d - 5 are horizontal asymptotes;
y = , then the curve has no vertical asymptote;
k = ; b = , i.e. y = 5x - oblique asymptote.
Examples of constructing function graphs.
Example 1 .
Investigate the function and build a graph of the function y \u003d x 3 - 6x 2 + 9x - 3
1. Find the domain of the function: D(y) = R
y (- x) \u003d (- x) 3 - 6 (- x) 2 + 9 (-x) - 3 \u003d - x 3 - 6x 2 - 9x - 3 \u003d - (x 3 + 6x 2 + 9x + 3), i.e.
(y \u003d x 5 - x 3 - odd, y \u003d x 4 + x 2 - even)
3. Is not periodic.
4. Find the points of intersection with the coordinate axes: if x \u003d 0, then y \u003d - 3 (0; - 3)
if Y = 0, x is hard to find.
5. Find the asymptotes of the graph of the function: There are no vertical asymptotes, because there are no x values for which the function is indefinite; y = , i.e., there are no horizontal asymptotes;
k = , i.e., there are no oblique asymptotes.
6. We examine the function for intervals of monotonicity and its extrema: y’ = 3x 2 - 12x + 9,
y'= 0, 3x 2 - 12x + 9 = 0 x 1 = 1; x 2 = 3 - critical points of the 1st kind.
Let's determine the signs of the derivative: y'(0) = 9 > 0; y'(2) = - 3< 0; y’(4) = 9 > 0
y max = y(1) = 1, (1;1) - maximum point; y min \u003d y (3) \u003d - 3, (3; - 3) - minimum point, function y for x and y 
.
7. We examine the function for intervals of convexity and inflection points:
y” = (y’)’ = (3x 2 - 12x + 9)’ = 6x - 12, y” = 0, 6x - 12 = 0 x = 2 - critical point of the 1st kind.
Let's determine the signs of the second derivative: y”(0) = - 12< 0; y”(3) = 6 > 0
Y(2) = - 1 (2; - 1) - inflection point, the function is convex up at x and convex down at x.
8. Additional points:
| X | - 1 | |
| at | - 19 |
9. Let's build a graph of the function:
Investigate the function and plot the function y =
1. Find the domain of the function: 1 - x ≠ 0, x ≠ 1, D(y) = .
2. Find out if the given function is even or odd: 
,
y(- x) ≠ y(x) is not even and y(- x) ≠ - y(x) is not odd
3. Is not periodic.
4. Find the points of intersection with the coordinate axes: x \u003d 0, then y \u003d - 2; y = 0, then 
, i.e. (0; - 2); ().
5. Find the asymptotes of the graph of the function: since x ≠ 1, then the line x = 1 is the vertical asymptote;
Let the function y = f(x) be defined in the interval X. derivative function y \u003d f (x) at the point x o is called the limit
If this limit finite, then the function f(x) is called differentiable at the point x o; moreover, it turns out to be necessarily and continuous at this point.
If the considered limit is equal to ¥ (or - ¥), then provided that the function at the point x o is continuous, we will say that the function f(x) has at a point x o infinite derivative.
The derivative is denoted by the symbols
y ¢, f ¢(x o), , .
Finding the derivative is called differentiation functions. The geometric meaning of the derivative is that the derivative is the slope of the tangent to the curve y=f(x) at a given point x o; physical sense - in that the derivative of the path with respect to time is the instantaneous speed of the moving point during rectilinear motion s = s(t) at the moment t o .
If a With is a constant number, and u = u(x), v = v(x) are some differentiable functions, then the following differentiation rules hold:
1) (c) " = 0, (cu) " = cu";
2) (u+v)" = u"+v";
3) (uv)" \u003d u "v + v" u;
4) (u / v) "= (u" v-v "u) / v 2;
5) if y = f(u), u = j(x), i.e. y = f(j(x)) - complex function, or superposition, composed of differentiable functions j and f, then , or
6) if for a function y = f(x) there exists an inverse differentiable function x = g(y), and ¹ 0, then .
Based on the definition of the derivative and the rules of differentiation, one can compile a list of tabular derivatives of the basic elementary functions.
1. (u m)" = m u m - 1 u" (m О R).
2. (a u)" = a u lna × u".
3. (e u)" = e u u".
4. (log a u)" = u"/(u ln a).
5. (ln u)" = u"/u.
6. (sin u)" = cos u × u".
7. (cos u)" = - sin u × u".
8. (tg u)" = 1/ cos 2 u × u".
9. (ctg u)" = - u" / sin 2 u.
10. (arcsin u)" = u" / .
11. (arccos u)" = - u" / .
12. (arctg u)" = u"/(1 + u 2).
13. (arcctg u)" = - u"/(1 + u 2).
Calculate the derivative of the exponential expression
y=u v , (u>0), where u and v essence of the function X having derivatives at a given point u",v".
Taking the logarithm of the equality y=u v , we obtain ln y = v ln u.
Equating derivatives with respect to X from both parts of the obtained equality using rules 3, 5 and the formula for the derivative of the logarithmic function, we will have:
y"/y = vu"/u + v" ln u, whence y" = y (vu"/u + v" ln u).
(u v)"=u v (vu"/u+v" log u), u > 0.
For example, if y \u003d x sin x, then y" \u003d x sin x (sin x / x + cos x × ln x).
If the function y = f(x) is differentiable at a point x, i.e. has a finite derivative at this point y", then \u003d y "+a, where a®0 at Dx® 0; hence D y \u003d y" Dx + a x.
The main part of the function increment, linear with respect to Dx, is called function differential and is denoted dy: dy \u003d y "Dx. If we put y \u003d x in this formula, then we get dx \u003d x" Dx \u003d 1 × Dx \u003d Dx, therefore dy \u003d y "dx, i.e. the symbol for denoting the derivative can be considered like a fraction.
D function increment y is the increment of the ordinate of the curve, and the differential d y is the increment of the ordinate of the tangent.
Let us find for the function y=f(x) its derivative y ¢= f ¢(x). The derivative of this derivative is called second order derivative functions f(x), or second derivative, and is denoted.
The following are defined and denoted in the same way:
third order derivative - ,
fourth order derivative -
and generally speaking nth order derivative - .
Example 15 Calculate the derivative of the function y=(3x 3 -2x+1)×sin x.
Solution. By rule 3, y"=(3x 3 -2x+1)"×sin x + (3x 3 -2x+1)×(sin x)" =
= (9x 2 -2) sinx + (3x 3 -2x+1) cos x.
Example 16. Find y", y = tg x + .
Solution. Using the rules for differentiating the sum and the quotient, we get: y"=(tgx + )" = (tgx)" + ()" = + = .
Example 17. Find the derivative of a complex function y= ,
u=x 4 +1.
Solution. According to the rule of differentiation of a complex function, we get: y "x \u003d y " u u" x \u003d () " u (x 4 +1)" x \u003d (2u +. Since u \u003d x 4 +1, then
(2 x 4 +2+ .
Example 18.
Solution. Let us represent the function y= as a superposition of two functions: y = e u and u = x 2 . We have: y" x \u003d y " u u" x \u003d (e u)" u (x 2)" x \u003d e u ×2x. Substituting x2 instead of u, we get y=2x .
Example 19. Find the derivative of the function y=ln sin x.
Solution. Denote u=sin x, then the derivative of the complex function y=ln u is calculated by the formula y" = (ln u)" u (sin x)" x = .
Example 20. Find the derivative of the function y= .
Solution. The case of a complex function obtained as a result of several superpositions is exhausted by the successive application of Rule 5:
Example 21. Calculate the derivative y=ln .
Solution. Taking logarithms and using the properties of logarithms, we get:
y=5/3ln(x 2 +4) +7/3ln(3x-1)-2/3ln(6x 3 +1)-1/3tg 5x.
Differentiating both parts of the last equality, we get:
2.2. Limit analysis in economics. Function elasticity
In economic research, specific terminology is often used to refer to derivatives. For example, if f(x) is a production function that expresses the dependence of the output of any product on the costs of the factor x, then f"(x) called marginal product; if g(x) is a cost function, i.e., a function g(x) expresses the dependence of total costs on the volume of production x, then g"(x) called marginal cost.
Marginal Analysis in Economics- a set of methods for studying the changing values of costs or results when the volume of production, consumption, etc. changes. based on the analysis of their limiting values. For the most part, planning calculations based on ordinary statistical data are carried out in the form of summary indicators. In this case, the analysis consists mainly in the calculation of average values. However, in some cases, a more detailed study is necessary, taking into account the limiting values. For example, when determining the costs of grain production in a region for the future, it is taken into account that the costs may be different depending, all other things being equal, on the expected volumes of grain harvest, since on the worst lands again involved in cultivation, production costs will be higher than on area on average.
If the relationship between two indicators v and x is given analytically: v = f(x) - then average value represents the relation v/x, a ultimate- derivative.
Finding labor productivity. Let the function
u = u(t), expressing the amount of production u while working t. Let's calculate the amount of goods produced during the time
Dt \u003d t 1 - t 0: Du \u003d u (t 1) - u (t 0) \u003d u (t 0 + Dt) - u (t 0). Average labor productivity is the ratio of the amount of output produced to the time spent, i.e. z cf.= Du/Dt.
Worker productivity z(t 0) at the moment t 0 is called the limit to which z tends to cf. for Dt®0: . The calculation of labor productivity, therefore, is reduced to the calculation of the derivative: z (t 0) \u003d u "(t 0).
Production costs K of homogeneous products is a function of the quantity of production x. Therefore, we can write K = K(x). Assume that the quantity of production increases by D X. The production costs x + Dх correspond to the production costs K(x + Dх). Consequently, the increment in the amount of production D X corresponds to the increment of production costs DK = K(x + Dх) - K(x).
The average increment of production costs is DK/Dх. This is the increment in production costs per unit increment in the quantity of output.
The limit is called marginal cost of production.
If denoted by u(x) sales proceeds x units of goods, it is called marginal revenue.
With the help of the derivative, you can calculate the increment of the function corresponding to the increment of the argument. In many problems, it is more convenient to calculate the percentage increase (relative increase) of the dependent variable corresponding to the percentage increase of the independent variable. This brings us to the concept of the elasticity of a function (sometimes called relative derivative). So, let a function y = f(x) be given, for which there exists a derivative y ¢ = f ¢(x). Function elasticity y = f(x) with respect to variable x call the limit
It is denoted by E x (y) = x/y f ¢ (x) = .
Elasticity relatively x is the approximate percentage increase in the function (up or down) corresponding to a 1% increase in the independent variable. Economists measure the sensitivity, or sensitivity, of consumers to changes in the price of a product using the concept of price elasticity. Demand for some products is characterized by the relative sensitivity of consumers to price changes, small changes in price lead to large changes in the quantity purchased. The demand for such products is called relatively elastic or just flexible. For other products, consumers are relatively insensitive to price changes, that is, a significant change in price leads to only a small change in the number of purchases. In such cases, the demand relatively inelastic or just inelastic. Term perfectly inelastic demand means the extreme case where a change in price does not result in any change in the quantity demanded. An example is the demand of patients with acute diabetes for insulin or the demand of drug addicts for heroin. And vice versa, when, at the smallest price reduction, buyers increase their purchases to the limit of their capabilities, then we say that demand is perfectly elastic.
Function extremum
The function y=f(x) is called increasing (waning) in some interval if for x 1< x 2 выполняется неравенство f(x 1) < f (x 2) (f(x 1) >f(x2)).
If a differentiable function y = f(x) on a segment increases (decreases), then its derivative on this segment f ¢(x) > 0 (f ¢(x)< 0).
Dot x o called local maximum point (minimum) of the function f(x) if there is a neighborhood of the point x o, for all points of which the inequality f(x) £ f(x o) (f(x) ³ f(x o)) is true.
The maximum and minimum points are called extremum points, and the values of the function at these points are its extrema.
Necessary conditions for an extremum. If point x o is an extremum point of the function f(x), then either f ¢(x o) = 0, or f ¢(x o) does not exist. Such points are called critical, where the function itself is defined at the critical point. The extrema of a function should be sought among its critical points.
The first sufficient condition. Let x o- critical point. If f ¢ (x) when passing through the point x o changes the plus sign to minus, then at the point x o the function has a maximum, otherwise it has a minimum. If the derivative does not change sign when passing through a critical point, then at the point x o there is no extremum.
The second sufficient condition. Let the function f(x) have a derivative
f ¢ (x) in a neighborhood of a point x o and the second derivative at the very point x o. If f ¢(x o) = 0, >0 (<0), то точка x o is a local minimum (maximum) point of the function f(x). If =0, then one must either use the first sufficient condition or involve higher derivatives.
On a segment, the function y = f(x) can reach its minimum or maximum value either at critical points or at the ends of the segment.
Example 22. Find the extrema of the function f(x) = 2x 3 - 15x 2 + 36x - 14.
Solution. Since f ¢ (x) \u003d 6x 2 - 30x +36 \u003d 6 (x -2) (x - 3), then the critical points of the function x 1 \u003d 2 and x 2 \u003d 3. Extreme points can only be at these points. Since when passing through the point x 1 \u003d 2, the derivative changes sign from plus to minus, then at this point the function has a maximum. When passing through the point x 2 \u003d 3, the derivative changes sign from minus to plus, therefore, at the point x 2 \u003d 3, the function has a minimum. Calculating the values of the function in points
x 1 = 2 and x 2 = 3, we find the extrema of the function: maximum f(2) = 14 and minimum f(3) = 13.
Example 23. It is necessary to build a rectangular area near the stone wall so that it is fenced off with wire mesh on three sides, and adjoins the wall on the fourth side. For this there is a linear meters of the grid. At what aspect ratio will the site have the largest area?
Solution. Denote the sides of the site through x and y. The area of the site is S = xy. Let y is the length of the side adjacent to the wall. Then, by condition, the equality 2x + y = a must hold. Therefore, y = a - 2x and S = x(a - 2x), where 0 £ x £ a/2 (the length and width of the area cannot be negative). S ¢ = a - 4x, a - 4x = 0 for x = a/4, whence
y \u003d a - 2 × a / 4 \u003d a / 2. Since x = a/4 is the only critical point, let's check whether the sign of the derivative changes when passing through this point. For x< a/4 S ¢ >0, and for x >a/4 S ¢<0, значит, в точке x=a/4 функция S имеет максимум. Значение функции S(a/4) = a/4(a - a/2) = a 2 /8 (кв. ед).
Since S is continuous on and its values at the ends of S(0) and S(a/2) are equal to zero, then the found value will be the largest value of the function. Thus, the most favorable aspect ratio of the site under the given conditions of the problem is y = 2x.
Example 24. It is required to make a closed cylindrical tank with a capacity of V=16p » 50 m 3 . What should be the dimensions of the tank (radius R and height H) in order to use the least amount of material for its manufacture?
Solution. The total surface area of the cylinder is S = 2pR(R+H). We know the volume of the cylinder V = pR 2 H Þ H = V/pR 2 =16p/ pR 2 = 16/ R 2 . Hence, S(R) = 2p(R 2 +16/R). We find the derivative of this function:
S¢(R) = 2p(2R- 16/R 2) = 4p (R- 8/R 2). S ¢(R) = 0 for R 3 = 8, therefore,
R = 2, H = 16/4 = 4.
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