Derivatives of higher orders
In this lesson, we will learn how to find higher-order derivatives, as well as write the general formula for the “nth” derivative. In addition, the Leibniz formula for such a derivative will be considered and, by popular demand, higher-order derivatives of implicit function. I suggest that you immediately take a mini-test:
Here is the function: 
and here is its first derivative:
In case you have any difficulties/misunderstandings about this example, please start with two basic articles of my course: How to find the derivative? and Derivative of a compound function. After mastering elementary derivatives, I recommend that you read the lesson The simplest problems with a derivative, on which we have dealt, in particular with second derivative.
It is not difficult even to guess that the second derivative is the derivative of the 1st derivative:
In principle, the second derivative is already considered a derivative of a higher order.
Similarly: the third derivative is the derivative of the 2nd derivative:
The fourth derivative is the derivative of the 3rd derivative: 
Fifth derivative: 
, and it is obvious that all derivatives of higher orders will also be equal to zero:
In addition to Roman numeration, the following designations are often used in practice:
, while the derivative of the “nth” order is denoted by . In this case, the superscript index must be enclosed in brackets.- to distinguish the derivative from the "y" in the degree.
Sometimes there is an entry like this: 
- third, fourth, fifth, ..., "nth" derivatives, respectively.
Forward without fear and doubt:
Example 1
Given a function. Find .
Solution: what can you say ... - forward for the fourth derivative :) 
It is no longer customary to put four strokes, so we move on to numerical indices:
Answer:
Okay, now let's think about this question: what to do if, according to the condition, it is required to find not the 4th, but, for example, the 20th derivative? If for the derivative of the 3-4-5th (maximum, 6-7th) order, the solution is drawn up quite quickly, then we will “get” to the derivatives of higher orders, oh, how not soon. Do not write down, in fact, 20 lines! In such a situation, you need to analyze several found derivatives, see the pattern and draw up a formula for the “nth” derivative. So, in Example No. 1, it is easy to understand that with each subsequent differentiation, an additional “triple” will “jump out” before the exponent, and at any step the degree of the “triple” is equal to the number of the derivative, therefore:
Where is an arbitrary natural number.
And indeed, if , then exactly the 1st derivative is obtained: 
, if - then 2nd: etc. Thus, the twentieth derivative is determined instantly: - and no "kilometer sheets"!
Warming up on our own:
Example 2
Find features. Write the order derivative
Solution and answer at the end of the lesson.
After an invigorating warm-up, we will consider more complex examples in which we will work out the above solution algorithm. For those who have read the lesson Sequence limit, it will be a little easier:
Example 3
Find for function .
Solution: to clarify the situation, we find several derivatives:
We are not in a hurry to multiply the resulting numbers! ;-) 
Perhaps enough. ... I even overdid it a little.
In the next step, it is best to write the formula for the "nth" derivative (as soon as the condition does not require this, then you can get by with a draft). To do this, we look at the results obtained and identify the patterns with which each next derivative is obtained.
First, they sign. Interleaving provides "flasher", and since the 1st derivative is positive, the following factor will enter the general formula: 
. An equivalent option will do, but personally, as an optimist, I love the plus sign =)
Secondly, in the numerator "winds" factorial, and it “lags behind” the number of the derivative by one unit:
And thirdly, the power of “two” grows in the numerator, which is equal to the number of the derivative. The same can be said about the degree of the denominator. Finally: 
For verification purposes, let's substitute a couple of values \u200b\u200b"en", for example, and: 
Great, now to make a mistake is just a sin:
Answer: 
A simpler function for a do-it-yourself solution:
Example 4
Find features.
And a trickier problem:
Example 5
Find features.
Let's repeat the procedure one more time:
1) First we find several derivatives. Three or four is usually enough to catch patterns.
2) Then I strongly recommend compiling (at least on draft)"nth" derivative - it is guaranteed to protect against errors. But you can do without, i.e. mentally estimate and immediately write down, for example, the twentieth or eighth derivative. Moreover, some people are generally able to solve the problems under consideration orally. However, it should be remembered that "quick" methods are fraught, and it is better to play it safe.
3) At the final stage, we check the "nth" derivative - we take a pair of values "en" (better than neighboring ones) and perform a substitution. And even more reliable is to check all the derivatives found earlier. Then we substitute in the desired value, for example, or, and carefully comb the result.
Brief solution of the 4th and 5th examples at the end of the lesson.
In some tasks, in order to avoid problems, you need to do a little magic on the function:
Example 6
Solution: I don’t want to differentiate the proposed function at all, since it will turn out to be a “bad” fraction, which will make it very difficult to find subsequent derivatives.
In this regard, it is advisable to perform preliminary transformations: we use difference of squares formula and logarithm property 
:
Quite a different matter:
And old friends: 
I think everything is being looked at. Note that the 2nd fraction is signed, but the 1st is not. We construct the order derivative: 
Control: 
Well, for beauty, we take the factorial out of brackets:
Answer: 
An interesting task for an independent solution:
Example 7
Write the order derivative formula for the function
And now about the unshakable mutual responsibility, which even the Italian mafia will envy:
Example 8
Given a function. Find
The eighteenth derivative at the point . Just.
Solution: first, obviously, you need to find . Go: 
They started from the sine, and they came to the sine. It is clear that with further differentiation, this cycle will continue to infinity, and the following question arises: how best to “get” to the eighteenth derivative?
The “amateur” method: we quickly write down the numbers of subsequent derivatives on the right in the column: 
In this way:
But it works if the order of the derivative is not too large. If you need to find, say, the hundredth derivative, then you should use divisibility by 4. One hundred is divisible by 4 without a remainder, and it is easy to see that such numbers are located on the bottom line, therefore: .
By the way, the 18th derivative can also be determined from similar considerations:
The second line contains numbers that are divisible by 4 with a remainder of 2.
Another, more academic method is based on sine periodicity and reduction formulas. We use the ready-made formula "nth" derivative of the sine 
, into which the desired number is simply substituted. For instance:
(reduction formula 
)
;
(reduction formula 
)
In our case:
(1) Since the sine is a periodic function with a period, then the argument can be painlessly "unscrewed" 4 periods (i.e.).
The order derivative of the product of two functions can be found by the formula:
In particular: 
You don’t need to remember anything specially, because the more formulas you know, the less you understand. Much better to know Newton's binomial, since Leibniz's formula is very, very similar to him. Well, those lucky ones who get the derivative of the 7th or higher orders (which is really unlikely) will be forced to do so. However, when the time comes to combinatorics- you still have to =)
Let's find the third derivative of the function . We use the Leibniz formula:
In this case: 
. Derivatives are easy to click verbally: 
Now we carefully and CAREFULLY perform the substitution and simplify the result:
Answer:
A similar task for an independent solution:
Example 11
Find features
If in the previous example the solution "on the forehead" still competed with the Leibniz formula, then here it will already be really unpleasant. And even more unpleasant - in the case of a higher order of the derivative:
Example 12
Find the derivative of the specified order
Solution: the first and essential remark - to decide like this, probably, it is not necessary =) =)
Let's write down the functions and find their derivatives up to the 5th order inclusive. I assume that the derivatives of the right column have become oral for you: 
In the left column, the “live” derivatives quickly “ended” and this is very good - in the Leibniz formula, three terms will be zeroed:
I will dwell again on the dilemma that appeared in the article on complex derivatives: to simplify the result? In principle, you can leave it like that - it will be even easier for the teacher to check. But he may require to bring the decision to mind. On the other hand, simplification on one's own initiative is fraught with algebraic errors. However, we have an answer obtained in a "primal" way =) (see link at the beginning) and I hope it's correct:
Great, it all worked out.
Answer: 
Happy task for self-solving:
Example 13
For function :
a) find by direct differentiation;
b) find by the Leibniz formula;
c) calculate.
No, I'm not a sadist at all - point "a" here is quite simple =)
But seriously, the “direct” solution by successive differentiation also has the “right to life” - in some cases its complexity is comparable to the complexity of applying the Leibniz formula. Use as you see fit - this is unlikely to be grounds for not counting the assignment.
Short solution and answer at the end of the lesson.
To raise the final paragraph you need to be able to differentiate implicit functions:
Higher order derivatives of implicit functions
Many of us have spent long hours, days and weeks of our lives studying circles, parabola, hyperbole– and sometimes it even seemed like a real punishment. So let's take revenge and differentiate them properly!
Let's start with the "school" parabola in its canonical position:
Example 14
An equation is given. Find .
Solution: the first step is familiar:
The fact that the function and its derivative are expressed implicitly does not change the essence of the matter, the second derivative is the derivative of the 1st derivative: 
However, there are rules of the game: derivatives of the 2nd and higher orders are usually expressed only through "x" and "y". Therefore, we substitute into the resulting 2nd derivative: 
The third derivative is the derivative of the 2nd derivative:
Similarly, let's substitute: 
Answer:
"School" hyperbole in canonical position- for independent work:
Example 15
An equation is given. Find .
I repeat that the 2nd derivative and the result should be expressed only through "x" / "y"!
Short solution and answer at the end of the lesson.
After children's pranks, let's look at German pornography @ fia, let's look at more adult examples, from which we learn another important solution:
Example 16
Ellipse himself.
Solution: find the 1st derivative: 
And now let's stop and analyze the next moment: now we have to differentiate the fraction, which is not at all encouraging. In this case, of course, it is simple, but in real-life problems there are only a couple of such gifts. Is there a way to avoid finding the cumbersome derivative? Exists! We take the equation and use the same technique as when finding the 1st derivative - we “hang” strokes on both parts: 
The second derivative must be expressed only through and , so now (right now) it is convenient to get rid of the 1st derivative. To do this, we substitute into the resulting equation: 
To avoid unnecessary technical difficulties, we multiply both parts by: 
And only at the final stage we draw up a fraction: 
Now we look at the original equation and notice that the result obtained can be simplified:
Answer:
How to find the value of the 2nd derivative at some point (which, of course, belongs to the ellipse), for example, at the point 
? Very easy! This motif has already been encountered in the lesson about normal equation: in the expression of the 2nd derivative you need to substitute 
:
Of course, in all three cases, you can get explicitly given functions and differentiate them, but then mentally prepare to work with two functions that contain roots. In my opinion, the solution is more convenient to carry out "implicitly".
Final example for self solution:
Example 17
Find implicit function
The Leibniz formula for calculating the nth derivative of the product of two functions is given. Its proof is given in two ways. An example of calculating the derivative of the nth order is considered.
ContentSee also: Derivative of the product of two functions
Leibniz formula
Using the Leibniz formula, you can calculate the nth derivative of the product of two functions. It looks like this:
(1)
,
where
are binomial coefficients.
The binomial coefficients are the coefficients of the expansion of the binomial in powers of and :
.
Also the number is the number of combinations from n to k .
Proof of the Leibniz formula
We apply the formula for the derivative of the product of two functions:
(2)
.
Let us rewrite formula (2) in the following form:
.
That is, we consider that one function depends on the x variable, and the other depends on the y variable. At the end of the calculation, we assume . Then the previous formula can be written as:
(3)
.
Since the derivative is equal to the sum of the terms, and each term is the product of two functions, then to calculate the derivatives of higher orders, you can consistently apply the rule (3).
Then for the nth order derivative we have:
.
Given that and , we get the Leibniz formula:
(1)
.
Proof by induction
We present the proof of the Leibniz formula by the method of mathematical induction.
Let's rewrite the Leibniz formula:
(4)
.
For n = 1 we have:
.
This is the formula for the derivative of the product of two functions. She is fair.
Let us assume that formula (4) is valid for the nth order derivative. Let us prove that it is valid for the derivative n + 1 -th order.
Differentiate (4):
;
.
So we found:
(5)
.
Substitute in (5) and take into account that:
.
This shows that formula (4) has the same form for the derivative n + 1
-th order.
So, formula (4) is valid for n = 1
. From the assumption that it is true for some number n = m, it follows that it is true for n = m + 1
.
The Leibniz formula has been proven.
Example
Calculate the nth derivative of a function
.
We apply the Leibniz formula
(2)
.
In our case
;
.
According to the table of derivatives, we have:
.
We apply the properties of trigonometric functions:
.
Then
.
This shows that differentiation of the sine function leads to its shift by . Then
.
We find derivatives of the function .
;
;
;
,
.
Since for , only the first three terms in the Leibniz formula are nonzero. Finding binomial coefficients.
;
.
According to the Leibniz formula, we have:
.
The text of the work is placed without images and formulas.
The full version of the work is available in the "Job Files" tab in PDF format
"Me too, Newton's binomial!»
from The Master and Margarita
“Pascal's triangle is so simple that even a ten-year-old child can write it out. At the same time, it conceals inexhaustible treasures and links together various aspects of mathematics that at first glance have nothing in common with each other. Such unusual properties allow us to consider Pascal's triangle one of the most elegant schemes in all of mathematics.
Martin Gardner.
Objective: generalize the formulas of abbreviated multiplication, show their application to solving problems.
Tasks:
1) study and systematize information on this issue;
2) analyze examples of problems for the use of Newton's binomial and formulas for the sum and difference of degrees.
Research objects: Newton's binomial, formulas for the sum and difference of degrees.
Research methods:
Working with educational and popular science literature, Internet resources.
Calculations, comparison, analysis, analogy.
Relevance. A person often has to deal with problems in which it is necessary to count the number of all possible ways to arrange some objects or the number of all possible ways to carry out some action. Different paths or options that a person has to choose add up to a wide variety of combinations. And a whole branch of mathematics, called combinatorics, is busy looking for answers to the questions: how many combinations are there in one or another case.
Representatives of many specialties have to deal with combinatorial quantities: scientist-chemist, biologist, designer, dispatcher, etc. The growing interest in combinatorics in recent years is due to the rapid development of cybernetics and computer technology.
Introduction
When they want to emphasize that the interlocutor exaggerates the complexity of the tasks that he faced, they say: “I also need Newton’s binomial!” Say, here's Newton's binomial, it's difficult, but what problems do you have! Even those people whose interests have nothing to do with mathematics have heard about Newton's binomial.
The word "binomial" means a binomial, i.e. the sum of two terms. From the school course, the so-called abbreviated multiplication formulas are known:
( a+ b) 2 = a 2 + 2ab + b 2 , (a+b) 3 = a 3 +3a 2 b+3ab 2 +b 3 .
A generalization of these formulas is a formula called Newton's binomial formula. The formulas for factoring the difference of squares, the sum and difference of cubes are also used at school. Do they have a generalization for other degrees? Yes, there are such formulas, they are often used in solving various problems: proving divisibility, reducing fractions, approximate calculations.
The study of generalizing formulas develops deductive-mathematical thinking and general mental abilities.
SECTION 1. NEWTON'S BINOMIAL FORMULA
Combinations and their properties
Let X be a set consisting of n elements. Any subset Y of the set X containing k elements is called a combination of k elements from n , and k ≤ n .
The number of different combinations of k elements out of n is denoted C n k . One of the most important formulas of combinatorics is the following formula for the number C n k:
It can be written after obvious abbreviations as follows:
In particular,
This is quite consistent with the fact that in the set X there is only one subset of 0 elements - the empty subset.
The numbers C n k have a number of remarkable properties.
The formula С n k = С n - k n is valid, (3)
The meaning of formula (3) is that there is a one-to-one correspondence between the set of all k-member subsets from X and the set of all (n - k)-member subsets from X: to establish this correspondence, it is enough for each k-member subset of Y match its complement in the set X.
The formula С 0 n + С 1 n + С 2 n + ... + С n n = 2 n is valid (4)
The sum on the left side expresses the number of all subsets of the set X (C 0 n is the number of 0-member subsets, C 1 n is the number of single-member subsets, etc.).
For any k, 1≤ k≤ n , the equality
C k n \u003d C n -1 k + C n -1 k -1 (5)
This equality is easy to obtain using formula (1). Indeed,
1.2. Derivation of Newton's binomial formula
Consider the powers of the binomial a +b .
n = 0, (a +b ) 0 = 1
n = 1, (a +b ) 1 = 1a+1b
n = 2(a +b ) 2 = 1a 2 + 2ab +1 b 2
n = 3(a +b ) 3 = 1 a 3 + 3a 2 b + 3ab 2 +1 b 3
n = 4(a +b ) 4 = 1a 4 + 4a 3 b + 6a 2 b 2 +4ab 3 +1 b 4
n=5(a +b ) 5 = 1a 5 + 5a 4 b + 10a 3 b 2 + 10a 2 b 3 + 5ab 4 + 1 b 5
Note the following regularities:
The number of terms of the resulting polynomial is one greater than the exponent of the binomial;
The exponent of the first term decreases from n to 0, the exponent of the second term increases from 0 to n;
The degrees of all monomials are equal to the degrees of the binomial in the condition;
Each monomial is the product of the first and second expressions in various powers and a certain number - the binomial coefficient;
Binomial coefficients equidistant from the beginning and end of the expansion are equal.
A generalization of these formulas is the following formula, called Newton's binomial formula:
(a + b ) n = C 0 n a n b 0 + C 1 n a n -1 b + C 2 n a n -2 b 2 + ... + C n -1 n ab n -1 + C n n a 0 b n . (6)
In this formula n can be any natural number.
We derive formula (6). First of all, let's write:
(a + b ) n = (a + b )(a + b ) ... (a + b ), (7)
where the number of brackets to be multiplied is n. From the usual rule for multiplying a sum by a sum, it follows that expression (7) is equal to the sum of all possible products, which can be composed as follows: any term of the first of the sums a + b multiplied by any term of the second sum a+b, on any term of the third sum, etc.
From what has been said, it is clear that the term in the expression for (a + b ) n match (one-to-one) strings of length n, composed of letters a and b. Among the terms there will be similar terms; it is obvious that such members correspond to strings containing the same number of letters a. But the number of lines containing exactly k times the letter a, is equal to C n k . Hence, the sum of all terms containing the letter a with a factor exactly k times is equal to С n k a n - k b k . Since k can take the values 0, 1, 2, ..., n-1, n, formula (6) follows from our reasoning. Note that (6) can be written shorter: (8)
Although formula (6) is called Newton's name, in reality it was discovered even before Newton (for example, Pascal knew it). Newton's merit lies in the fact that he found a generalization of this formula for the case of non-integer exponents. It was I. Newton in 1664-1665. derived a formula expressing the degree of the binomial for arbitrary fractional and negative exponents.
The numbers C 0 n , C 1 n , ..., C n n , included in formula (6), are usually called binomial coefficients, which are defined as follows:
From formula (6) one can obtain a number of properties of these coefficients. For example, assuming a=1, b = 1, we get:
2 n = C 0 n + C 1 n + C 2 n + C 3 n + ... + C n n ,
those. formula (4). If we put a= 1, b = -1, then we will have:
0 \u003d C 0 n - C 1 n + C 2 n - C 3 n + ... + (-1) n C n n
or С 0 n + C 2 n + C 4 n + ... = C 1 n + C 3 n + + C 5 n + ... .
This means that the sum of the coefficients of the even terms of the expansion is equal to the sum of the coefficients of the odd terms of the expansion; each of them is equal to 2 n -1 .
The coefficients of the terms equidistant from the ends of the expansion are equal. This property follows from the relation: С n k = С n n - k
An interesting special case
(x + 1) n = C 0 n x n + C 1 n x n-1 + ... + C k n x n - k + ... + C n n x 0
or shorter (x +1) n = ∑C n k x n - k .
1.3. Polynomial theorem
Theorem.
Proof.
In order to obtain a monomial after opening the brackets, you need to choose those brackets from which it is taken, those brackets from which it is taken, etc. and those brackets from which it is taken. The coefficient of this monomial after reduction of similar terms is equal to the number of ways in which such a choice can be made. The first step of the sequence of choices can be done in ways, the second step - , the third - etc., the -th step - in ways. The desired coefficient is equal to the product
SECTION 2. Derivatives of higher orders.
The concept of derivatives of higher orders.
Let the function be differentiable in some interval. Then its derivative, generally speaking, depends on X, that is, is a function of X. Therefore, with respect to it, we can again raise the question of the existence of a derivative.
Definition . The derivative of the first derivative is called derivative of the second order or second derivative and is denoted by the symbol or, i.e.
Definition . The derivative of the second derivative is called the third order derivative or the third derivative and is denoted by the or symbol.
Definition . derivativen th order functions is called the first derivative of the derivative (n -1)-th order of this function and is denoted by the symbol or:
Definition . Derivatives of order higher than the first are called higher derivatives.
Comment. Similarly, one can obtain the formula n-th derivative of the function:
The second derivative of a parametrically defined function
If a function is given parametrically by equations, then to find the second order derivative, it is necessary to differentiate the expression for its first derivative as a complex function of an independent variable.
Since then
and considering that,
We get it, that is.
Similarly, we can find the third derivative.
Differential of sum, product and quotient.
Since the differential is obtained from the derivative by multiplying it by the differential of an independent variable, then, knowing the derivatives of the basic elementary functions, as well as the rules for finding derivatives, one can come to similar rules for finding differentials.
1 0 . The differential of a constant is zero.
2 0 . The differential of the algebraic sum of a finite number of differentiable functions is equal to the algebraic sum of the differentials of these functions .
3 0 . The differential of the product of two differentiable functions is equal to the sum of the products of the first function and the differential of the second and the second function and the differential of the first .
Consequence. The constant factor can be taken out of the sign of the differential.
2.3. Functions given parametrically, their differentiation.
Definition . A function is said to be parametrically defined if both variables X and y are defined each separately as single-valued functions of the same auxiliary variable - the parametert :
wheret changes within.
Comment . We present the parametric equations of a circle and an ellipse.
a) Circle centered at the origin and radius r has parametric equations:
b) Let's write the parametric equations for the ellipse:
By excluding the parameter t From the parametric equations of the lines under consideration, one can arrive at their canonical equations.
Theorem . If the function y from argument x is given parametrically by the equations, where and are differentiable with respect tot functions and then.
2.4. Leibniz formula
To find the derivative n th order of the product of two functions, the Leibniz formula is of great practical importance.
Let u and v- some functions from a variable X having derivatives of any order and y = UV. Express n-th derivative through derivatives of functions u and v .
We have consistently
It is easy to notice the analogy between the expressions for the second and third derivatives and the expansion of Newton's binomial in the second and third powers, respectively, but instead of the exponents there are numbers that determine the order of the derivative, and the functions themselves can be considered as "zero-order derivatives". Given this, we obtain the Leibniz formula:
This formula can be proved by mathematical induction.
SECTION 3. APPLICATION OF THE LEIBNIZ FORMULA.
To calculate the derivative of any order from the product of two functions, bypassing the sequential application of the formula for calculating the derivative of the product of two functions, we use Leibniz formula.
Using this formula, consider examples of calculating the nth derivative of the product of two functions.
Example 1
Find the second derivative of a function
By definition, the second derivative is the first derivative of the first derivative, i.e.
Therefore, we first find the first order derivative of the given function according to differentiation rules and using derivative table:
Now we find the derivative of the first order derivative. This will be the desired second-order derivative:
Answer:
Example 2
Find the th-order derivative of a function
Solution.
We will sequentially find derivatives of the first, second, third, and so on orders of the given function in order to establish a pattern that can be generalized to the -th derivative.
We find the first order derivative as derivative of the quotient:
Here the expression is called the factorial of a number. The factorial of a number is equal to the product of numbers from one to, that is,
The second derivative is the first derivative of the first derivative, that is
Third order derivative:
Fourth derivative:
Note the regularity: the numerator contains the factorial of a number that is equal to the order of the derivative, and the denominator contains an expression in the power by one greater than the order of the derivative, that is
Answer.
Example 3
Find the value of the third derivative of a function at a point.
Solution.
According to table of higher order derivatives, we have:
In this example, that is, we get
Note that a similar result could also be obtained by successively finding derivatives.
At a given point, the third derivative is:
Answer:
Example 4
Find the second derivative of a function
Solution. First, let's find the first derivative:
To find the second derivative, we differentiate the expression for the first derivative again:
Answer:
Example 5
Find if
Since the given function is a product of two functions, it would be advisable to apply the Leibniz formula to find the fourth-order derivative:
We find all derivatives and calculate the coefficients of the terms.
1) Calculate the coefficients for the terms:
2) Find the derivatives of the function:
3) Find the derivatives of the function:
Answer:
Example 6
The function y=x 2 cos3x is given. Find the derivative of the third order.
Let u=cos3x , v=x 2 . Then, according to the Leibniz formula, we find:
The derivatives in this expression are:
(cos3x)′=−3sin3x,
(cos3x)′′=(−3sin3x)′=−9cos3x,
(cos3x)′′′=(−9cos3x)′=27sin3x,
(x2)′=2x,
(x2)′′=2,
(x2)′′′=0.
Therefore, the third derivative of the given function is
1 ⋅ 27sin3x ⋅ x2+3 ⋅ (−9cos3x) ⋅ 2x+3 ⋅ (−3sin3x) ⋅ 2+1 ⋅ cos3x ⋅ 0
27x2sin3x−54xcos3x−18sin3x=(27x2−18)sin3x−54xcos3x.
Example 7
Find derivative n -th order function y=x 2 cosx.
We use the Leibniz formula, settingu=cosx, v=x 2 . Then
The remaining terms of the series are equal to zero, since(x2)(i)=0 for i>2.
Derivative n -th order cosine function:
Therefore, the derivative of our function is
CONCLUSION
The school studies and uses the so-called abbreviated multiplication formulas: squares and cubes of the sum and difference of two expressions and formulas for factoring the difference of squares, the sum and difference of cubes of two expressions. A generalization of these formulas is a formula called the Newton binomial formula and the formulas for factoring the sum and difference of powers. These formulas are often used in solving various problems: proving divisibility, reducing fractions, approximate calculations. Interesting properties of Pascal's triangle, which are closely related to Newton's binomial, are considered.
The paper systematizes information on the topic, gives examples of tasks for the use of Newton's binomial and formulas for the sum and difference of degrees. The work can be used in the work of a mathematical circle, as well as for independent study by those who are fond of mathematics.
LIST OF USED SOURCES
1. Vilenkin N. Ya. Combinatorics. - ed. "The science". - M., 1969
2. Nikolsky S.M., Potapov M.K., Reshetnikov N.N., Shevkin A.V. Algebra and beginning of mathematical analysis. Grade 10: textbook. for general education organizations basic and advanced levels - M.: Education, 2014. - 431 p.
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The solution of applied problems is reduced to the calculation of the integral, but it is not always possible to do this accurately. Sometimes it is necessary to know the value of a definite integral with some degree of accuracy, for example, to a thousandth.
There are tasks when it would be necessary to find the approximate value of a certain integral with the required accuracy, then numerical integration is used such as the Simposn method, trapezoids, rectangles. Not all cases allow us to calculate it with a certain accuracy.
This article considers the application of the Newton-Leibniz formula. This is necessary for the exact calculation of the definite integral. Detailed examples will be given, the change of variable in the definite integral will be considered, and we will find the values of the definite integral when integrating by parts.
Newton-Leibniz formula
Definition 1When the function y = y (x) is continuous from the segment [ a ; b ], and F (x) is one of the antiderivatives of the function of this segment, then Newton-Leibniz formula considered fair. Let's write it like this ∫ a b f (x) d x = F (b) - F (a) .
This formula is considered the basic formula of integral calculus.
To prove this formula, it is necessary to use the concept of an integral with the available variable upper limit.
When the function y = f (x) is continuous from the segment [ a ; b ] , then the value of the argument x ∈ a ; b , and the integral has the form ∫ a x f (t) d t and is considered a function of the upper limit. It is necessary to accept the notation of the function will take the form ∫ a x f (t) d t = Φ (x) , it is continuous, and the inequality of the form ∫ a x f (t) d t " = Φ " (x) = f (x) is valid for it.
We fix that the increment of the function Φ (x) corresponds to the increment of the argument ∆ x , it is necessary to use the fifth main property of a definite integral and obtain
Φ (x + ∆ x) - Φ x = ∫ ax + ∆ xf (t) dt - ∫ axf (t) dt = = ∫ ax + ∆ xf (t) dt = f (c) x + ∆ x - x = f(c) ∆x
where value c ∈ x ; x + ∆x .
We fix the equality in the form Φ (x + ∆ x) - Φ (x) ∆ x = f (c) . By definition of the derivative of a function, it is necessary to pass to the limit as ∆ x → 0, then we get a formula of the form located on [ a ; b ] Otherwise, the expression can be written
F (x) = Φ (x) + C = ∫ a x f (t) d t + C , where the value of C is constant.
Let's calculate F (a) using the first property of the definite integral. Then we get that
F (a) = Φ (a) + C = ∫ a a f (t) d t + C = 0 + C = C , hence C = F (a) . The result is applicable when calculating F (b) and we get:
F (b) = Φ (b) + C = ∫ abf (t) dt + C = ∫ abf (t) dt + F (a) , in other words, F (b) = ∫ abf (t) dt + F ( a) . Equality proves the Newton-Leibniz formula ∫ a b f (x) d x + F (b) - F (a) .
The increment of the function is taken as F x a b = F (b) - F (a) . With the help of notation, the Newton-Leibniz formula becomes ∫ a b f (x) d x = F x a b = F (b) - F (a) .
To apply the formula, it is necessary to know one of the antiderivatives y = F (x) of the integrand y = f (x) from the segment [ a ; b ] , calculate the increment of the antiderivative from this segment. Consider a few examples of calculations using the Newton-Leibniz formula.
Example 1
Calculate the definite integral ∫ 1 3 x 2 d x using the Newton-Leibniz formula.
Solution
Consider that the integrand of the form y = x 2 is continuous from the interval [ 1 ; 3 ] , then and is integrable on this segment. According to the table of indefinite integrals, we see that the function y \u003d x 2 has a set of antiderivatives for all real values \u200b\u200bof x, which means that x ∈ 1; 3 will be written as F (x) = ∫ x 2 d x = x 3 3 + C . It is necessary to take the antiderivative with C \u003d 0, then we get that F (x) \u003d x 3 3.
Let's use the Newton-Leibniz formula and get that the calculation of the definite integral will take the form ∫ 1 3 x 2 d x = x 3 3 1 3 = 3 3 3 - 1 3 3 = 26 3 .
Answer:∫ 1 3 x 2 d x = 26 3
Example 2
Calculate the definite integral ∫ - 1 2 x · e x 2 + 1 d x using the Newton-Leibniz formula.
Solution
The given function is continuous from the segment [ - 1 ; 2 ], which means that it is integrable on it. It is necessary to find the value of the indefinite integral ∫ x ex 2 + 1 dx using the method of summing under the differential sign, then we get ∫ x ex 2 + 1 dx = 1 2 ∫ ex 2 + 1 d (x 2 + 1) = 1 2 ex 2+1+C.
Hence we have a set of antiderivatives of the function y = x · e x 2 + 1 , which are valid for all x , x ∈ - 1 ; 2.
It is necessary to take the antiderivative at C = 0 and apply the Newton-Leibniz formula. Then we get an expression of the form
∫ - 1 2 x ex 2 + 1 dx = 1 2 ex 2 + 1 - 1 2 = = 1 2 e 2 2 + 1 - 1 2 e (- 1) 2 + 1 = 1 2 e (- 1) 2 + 1 = 1 2 e 2 (e 3 - 1)
Answer:∫ - 1 2 x e x 2 + 1 d x = 1 2 e 2 (e 3 - 1)
Example 3
Calculate the integrals ∫ - 4 - 1 2 4 x 3 + 2 x 2 d x and ∫ - 1 1 4 x 3 + 2 x 2 d x .
Solution
Segment - 4; - 1 2 says that the function under the integral sign is continuous, which means that it is integrable. From here we find the set of antiderivatives of the function y = 4 x 3 + 2 x 2 . We get that
∫ 4 x 3 + 2 x 2 d x = 4 ∫ x d x + 2 ∫ x - 2 d x = 2 x 2 - 2 x + C
It is necessary to take the antiderivative F (x) \u003d 2 x 2 - 2 x, then, applying the Newton-Leibniz formula, we obtain the integral, which we calculate:
∫ - 4 - 1 2 4 x 3 + 2 x 2 dx = 2 x 2 - 2 x - 4 - 1 2 = 2 - 1 2 2 - 2 - 1 2 - 2 - 4 2 - 2 - 4 = 1 2 + 4 - 32 - 1 2 = - 28
We make the transition to the calculation of the second integral.
From the segment [ - 1 ; 1 ] we have that the integrand is considered unbounded, because lim x → 0 4 x 3 + 2 x 2 = + ∞ , then it follows that a necessary condition for integrability from the segment. Then F (x) = 2 x 2 - 2 x is not an antiderivative for y = 4 x 3 + 2 x 2 from the interval [ - 1 ; 1 ] , since the point O belongs to the segment, but is not included in the domain of definition. This means that there is a definite integral of Riemann and Newton-Leibniz for the function y = 4 x 3 + 2 x 2 from the interval [ - 1 ; one ] .
Answer: ∫ - 4 - 1 2 4 x 3 + 2 x 2 d x \u003d - 28, there is a definite integral of Riemann and Newton-Leibniz for the function y = 4 x 3 + 2 x 2 from the interval [ - 1 ; one ] .
Before using the Newton-Leibniz formula, you need to know exactly about the existence of a definite integral.
Change of variable in a definite integral
When the function y = f (x) is defined and continuous from the segment [ a ; b ] , then the existing set [ a ; b ] is considered to be the range of the function x = g (z) defined on the interval α ; β with the existing continuous derivative, where g (α) = a and g β = b , hence we get that ∫ a b f (x) d x = ∫ α β f (g (z)) g " (z) d z .
This formula is used when it is necessary to calculate the integral ∫ a b f (x) d x , where the indefinite integral has the form ∫ f (x) d x , we calculate using the substitution method.
Example 4
Calculate a definite integral of the form ∫ 9 18 1 x 2 x - 9 d x .
Solution
The integrand is considered continuous on the integration interval, which means that the definite integral exists. Let's give the notation that 2 x - 9 = z ⇒ x = g (z) = z 2 + 9 2 . The value x \u003d 9 means that z \u003d 2 9 - 9 \u003d 9 \u003d 3, and for x \u003d 18 we get that z \u003d 2 18 - 9 \u003d 27 \u003d 3 3, then g α \u003d g (3) \u003d 9 , g β = g 3 3 = 18 . Substituting the obtained values into the formula ∫ a b f (x) d x = ∫ α β f (g (z)) g "(z) d z, we obtain that
∫ 9 18 1 x 2 x - 9 dx = ∫ 3 3 3 1 z 2 + 9 2 z z 2 + 9 2 "dz = = ∫ 3 3 3 1 z 2 + 9 2 z zdz = ∫ 3 3 3 2 z 2 + 9 dz
According to the table of indefinite integrals, we have that one of the antiderivatives of the function 2 z 2 + 9 takes the value 2 3 a r c t g z 3 . Then, applying the Newton-Leibniz formula, we obtain that
∫ 3 3 3 2 z 2 + 9 d z = 2 3 a r c t g z 3 3 3 3 = 2 3 a r c t g 3 3 3 - 2 3 a r c t g 3 3 = 2 3 a r c t g 3 - a r c t g 1 = 2 3 π 3 - π 4 = π 18
The finding could be done without using the formula ∫ a b f (x) d x = ∫ α β f (g (z)) g " (z) d z .
If the replacement method uses an integral of the form ∫ 1 x 2 x - 9 d x , then we can arrive at the result ∫ 1 x 2 x - 9 d x = 2 3 a r c t g 2 x - 9 3 + C .
From here we will perform calculations using the Newton-Leibniz formula and calculate the definite integral. We get that
∫ 9 18 2 z 2 + 9 dz = 2 3 arctgz 3 9 18 = = 2 3 arctg 2 18 - 9 3 - arctg 2 9 - 9 3 = = 2 3 arctg 3 - arctg 1 = 2 3 π 3 - π 4 \u003d π 18
The results matched.
Answer: ∫ 9 18 2 x 2 x - 9 d x = π 18
Integration by parts in the calculation of a definite integral
If on the segment [ a ; b ] the functions u (x) and v (x) are defined and continuous, then their first-order derivatives v " (x) u (x) are integrable, so from this interval for the integrable function u " (x) v ( x) the equality ∫ abv " (x) u (x) dx = (u (x) v (x)) ab - ∫ abu " (x) v (x) dx is true.
The formula can be used then, it is necessary to calculate the integral ∫ a b f (x) d x , and ∫ f (x) d x it was necessary to find it using integration by parts.
Example 5
Calculate the definite integral ∫ - π 2 3 π 2 x · sin x 3 + π 6 d x .
Solution
The function x sin x 3 + π 6 is integrable on the segment - π 2; 3 π 2 , so it is continuous.
Let u (x) \u003d x, then d (v (x)) \u003d v "(x) dx \u003d sin x 3 + π 6 dx, and d (u (x)) \u003d u "(x) dx \u003d dx, and v (x) = - 3 cos π 3 + π 6 . From the formula ∫ a b v "(x) u (x) d x = (u (x) v (x)) a b - ∫ a b u " (x) v (x) d x we get that
∫ - π 2 3 π 2 x sin x 3 + π 6 dx = - 3 x cos x 3 + π 6 - π 2 3 π 2 - ∫ - π 2 3 π 2 - 3 cos x 3 + π 6 dx \u003d \u003d - 3 3 π 2 cos π 2 + π 6 - - 3 - π 2 cos - π 6 + π 6 + 9 sin x 3 + π 6 - π 2 3 π 2 \u003d 9 π 4 - 3 π 2 + 9 sin π 2 + π 6 - sin - π 6 + π 6 = 9 π 4 - 3 π 2 + 9 3 2 = 3 π 4 + 9 3 2
The solution of the example can be done in another way.
Find the set of antiderivatives of the function x sin x 3 + π 6 using integration by parts using the Newton-Leibniz formula:
∫ x sin xx 3 + π 6 dx = u = x, dv = sin x 3 + π 6 dx ⇒ du = dx, v = - 3 cos x 3 + π 6 = = - 3 cos x 3 + π 6 + 3 ∫ cos x 3 + π 6 dx = = - 3 x cos x 3 + π 6 + 9 sin x 3 + π 6 + C ⇒ ∫ - π 2 3 π 2 x sin x 3 + π 6 dx = - 3 cos x 3 + π 6 + 9 sincos x 3 + π 6 - - - 3 - π 2 cos - π 6 + π 6 + 9 sin - π 6 + π 6 = = 9 π 4 + 9 3 2 - 3 π 2 - 0 = 3 π 4 + 9 3 2
Answer: ∫ x sin x x 3 + π 6 d x = 3 π 4 + 9 3 2
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