Calculation of a definite integral. Newton-Leibniz formula. Start in science Derivative by Leibniz formula online

Higher order derivatives

In this lesson we will learn how to find derivatives of higher orders, as well as write the general formula for the “nth” derivative. In addition, Leibniz’s formula for such a derivative and, by popular demand, higher-order derivatives of implicit function. I suggest you take a mini-test right away:

Here's the function: and here is its first derivative:

In case you have any difficulties/confusion about this example, please start with the two basic articles of my course: How to find the derivative? And Derivative of a complex function. After mastering elementary derivatives, I recommend that you read the lesson The simplest problems with derivatives, on which we dealt, in particular with second derivative.

It’s not difficult to even guess that the second derivative is the derivative of the 1st derivative:

In principle, the second derivative is already considered a higher order derivative.

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 numbering, the following notations are often used in practice:
, the derivative of the “nth” order is denoted by . In this case, the superscript must be enclosed in parentheses– to distinguish the derivative from the “y” in degree.

Sometimes you see something like this: – third, fourth, fifth, ..., “nth” derivatives, respectively.

Forward without fear and doubt:

Example 1

The function is given. Find .

Solution: what can you say... - go ahead for the fourth derivative :)

It is no longer customary to put four strokes, so we switch to numerical indices:

Answer:

Okay, now let’s think about this question: what to do if the condition requires finding not the 4th, but for example, the 20th derivative? If for the derivative 3-4-5th (maximum 6-7th) order of magnitude, the solution is formalized quite quickly, then we will not “get to” derivatives of higher orders very soon. In fact, don’t write down 20 lines! In such a situation, you need to analyze several derivatives found, see the pattern and create a formula for the “nth” derivative. So, in Example No. 1 it is easy to understand that with each subsequent differentiation an additional “three” will “pop up” in front of the exponent, and at any step the degree of the “three” 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-long sheets”!

Warming up on our own:

Example 2

Find functions. Write the order derivative

The solution and answer are 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 managed to get acquainted with the lesson Sequence limit, it will be a little easier:

Example 3

Find for function.

Solution: to clarify the situation, let’s find several derivatives:

We are in no hurry to multiply the resulting numbers! ;-)

Perhaps that's enough. ...I even went a little overboard.

The next step is best to create the formula for the “nth” derivative (if 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 subsequent derivative is obtained.

Firstly, they alternate. Alignment ensures "flashing light", and since the 1st derivative is positive, the following factor will enter the general formula: . An equivalent option would also work, but personally, as an optimist, I love the plus sign =)

Secondly, in the numerator “winds up” factorial, and it “lags” behind the derivative number by one unit:

And thirdly, the power of “two” in the numerator increases, which is equal to the number of the derivative. The same can be said about the degree of the denominator. Finally:

To check, let’s substitute a couple of “en” values, for example, and :

Great, now making a mistake is simply a sin:

Answer:

A simpler function to solve on your own:

Example 4

Find functions.

And a more interesting problem:

Example 5

Find functions.

Let's repeat the procedure once again:

1) First we find several derivatives. To catch patterns, three or four are usually enough.

2) Then I strongly recommend making (at least in draft form) The “nth” derivative – it is guaranteed to protect you from errors. But you can do without it, 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 in question orally. However, you should remember that “quick” methods are fraught, and it is better to be safe.

3) At the final stage, we check the “nth” derivative - take a pair of “nth” values (preferably neighboring ones) and perform the substitution. And it’s even more reliable to check all previously found derivatives. Then we substitute it into the desired value, for example, or and carefully comb the result.

A short solution to examples 4 and 5 at the end of the lesson.

In some tasks, in order to avoid problems, you need to work a little magic on the function:

Example 6

Solution: I don’t want to differentiate the proposed function at all, since it will result in a “bad” fraction, which will greatly complicate finding subsequent derivatives.

In this regard, it is advisable to perform preliminary transformations: we use square difference formula And property of logarithm :

It's a completely different matter:

And old friends:

I think everything is being looked at. Please note that the 2nd fraction alternates sign, but the 1st fraction does not. We construct the order derivative:

Control:

Well, for beauty’s sake, let’s take the factorial out of brackets:

Answer:

An interesting task to solve on your own:

Example 7

Write down the order derivative formula for the function

And now about the unshakable mutual guarantee that even the Italian mafia would envy:

Example 8

The function is given. Find

The eighteenth derivative at the point. Just.

Solution: first, obviously, you need to find . Go:

We started with the sinus and ended up with the sinus. It is clear that with further differentiation this cycle will continue indefinitely, and the following question arises: what is the best way to “get” to the eighteenth derivative?

The “amateur” method: quickly write down the numbers of subsequent derivatives in the column on the right:

Thus:

But this 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 in 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 for the “nth” derivative of sine , into which the desired number is simply substituted. For example:
(reduction formula ) ;
(reduction formula )

In our case:

(1) Since sine is a periodic function with a period, the argument can be painlessly “unscrewed” 4 periods (i.e.).

The order derivative of the product of two functions can be found using the formula:

In particular:

There is no need to remember anything specifically, because the more formulas you know, the less you understand. It is much more useful to familiarize yourself with Newton's binomial, since Leibniz's formula is very, very similar to it. Well, those lucky ones who will get a derivative of the 7th or higher orders (which is really unlikely), will be forced to do this. However, when the turn comes combinatorics– then you still have to =)

Let's find the third derivative of the function. We use Leibniz's formula:

In this case: . The derivatives are easy to recite orally:

Now carefully and CAREFULLY perform the substitution and simplify the result:

Answer:

A similar task for independent solution:

Example 11

Find features

If in the previous example the “head-on” solution still competed with Leibniz’s formula, then here it will be really unpleasant. And even more unpleasant - in the case of a higher order derivative:

Example 12

Find the derivative of the specified order

Solution: the first and significant remark is that you probably don’t need to decide like this =) =)

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 “living” derivatives quickly “ended” and this is very good - three terms in Leibniz’s formula will be reset to zero:

Let me again dwell on the dilemma that appeared in the article about complex derivatives: Should I simplify the result? In principle, you can leave it this way - it will be even easier for the teacher to check. But he may demand that the decision be finalized. On the other hand, simplification on one's own initiative is fraught with algebraic errors. However, we have an answer obtained in a “primitive” way =) (see link at the beginning) and I hope it's correct:

Great, everything came together.

Answer:

Happy task for independent solution:

Example 13

For function:
a) find by direct differentiation;
b) find using Leibniz’s 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 a “right to life” - in some cases its complexity is comparable to the complexity of applying the Leibniz formula. Use if you deem it appropriate - this is unlikely to be a reason for failing the assignment.

A 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 functions specified implicitly

Many of us have spent long hours, days and weeks of our lives studying circles, parabolas, 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

The 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

The equation is given. Find .

I repeat that the 2nd derivative and the result should be expressed only through “x”/“y”!

A short solution and answer at the end of the lesson.

After children's pranks, let's look at German pornography, let's look at more adult examples, from which we will learn another important solution:

Example 16

Ellipse himself.

Solution: let's find the 1st derivative:

Now let’s stop and analyze the next point: now we have to differentiate the fraction, which is not at all pleasing. In this case, it is, of course, simple, but in real-life problems such gifts are too few and far between. 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 sides:

The second derivative must be expressed only in terms of and , so now (right now) It is convenient to get rid of the 1st derivative. To do this, substitute into the resulting equation:

To avoid unnecessary technical difficulties, let's multiply both parts by:

And only at the final stage do we formulate the 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 any point (which, of course, belongs to the ellipse), for example, at the point ? Very easy! This motive has already been encountered in the lesson about normal equation: you need to substitute the 2nd derivative into the expression :

Of course, in all three cases it is possible to obtain explicitly defined functions and differentiate them, but then be mentally prepared to work with two functions that contain roots. In my opinion, it is more convenient to carry out the solution in an “implicit way”.

A final example to solve on your own:

Example 17

Find an implicitly specified function

The text of the work is posted without images and formulas.
The full version of the work is available in the "Work Files" tab in PDF format

"Me too, Newton's binomial!»

from the novel "The Master and Margarita"

“Pascal’s triangle is so simple that even a ten-year-old child can write it down. At the same time, it conceals inexhaustible treasures and connects 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 diagrams in all of mathematics.”

Martin Gardner.

Goal of the work: generalize abbreviated multiplication formulas and show their application to problem solving.

Tasks:

1) study and systematize information on this issue;

2) analyze examples of problems using Newton’s binomial and formulas for the sum and difference of powers.

Objects of study: Newton's binomial, formulas for sums and differences of powers.

Research methods:

Work with educational and popular science literature, Internet resources.

Calculations, comparison, analysis, analogy.

Relevance. A person often has to deal with problems in which he needs to count the number of all possible ways of placing some objects or the number of all possible ways of performing some action. The 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 searching for answers to the questions: how many combinations are there in a given case?

Representatives of many specialties have to deal with combinatorial quantities: chemical scientist, biologist, designer, dispatcher, etc. Increased interest in combinatorics has recently been caused by the rapid development of cybernetics and computer technology.

Introduction

When they want to emphasize that the interlocutor is exaggerating the complexity of the problems he is faced with, they say: “I also like Newton’s binomial!” They say, here’s Newton’s binomial, it’s complicated, 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 binomial, i.e. the sum of two terms. The so-called abbreviated multiplication formulas are known from the school course:

( 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. Formulas for factoring differences of squares, sums and differences of cubes are also used in school. Do they generalize to other degrees? Yes, there are such formulas, they are often used in solving various problems: proving divisibility, reducing fractions, approximate calculations.

Studying generalizing formulas develops deductive-mathematical thinking and general thinking abilities.

SECTION 1. NEWTON'S BINOMAL FORMULA

Combinations and their properties

Let X be a set consisting of n elements. Any subset Y of a set X containing k elements is called a combination of k elements from n, with k ≤ n.

The number of different combinations of k elements from n is denoted by 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 is correct: С n k = С n - k n , (3)

The meaning of formula (3) is that there is a one-to-one correspondence between the set of all k-member subsets of X and the set of all (n - k)-member subsets of X: to establish this correspondence, it is sufficient for each k-member subset of Y compare its complement in the set X.

The correct formula is С 0 n + С 1 n + С 2 n + … + С n n = 2 n (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 one-member subsets, etc.).

For any k, 1≤ k≤ n, the equality is true

C k n = 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

Let us note the following patterns:

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 degree of the binomial in the condition;

Each monomial is the product of the first and second expressions in various powers and a certain number - a 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.

Let us derive formula (6). First of all, let's write down:

(a + b ) n = (a + b )(a + b ) ... (a + b ), (7)

where the number of brackets to be multiplied is equal to n. From the usual rule of 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, to any term of the third sum, etc.

From the above it is clear that the term in the expression for (a + b ) n correspond (one-to-one) to 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 . This means that the sum of all terms containing the letter a with a factor of exactly k times is equal to C n k a n - k b k . Since k can take values 0, 1, 2, ..., n-1, n, then formula (6) follows from our reasoning. Note that (6) can be written shorter: (8)

Although formula (6) is called after Newton, in fact 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 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 you put A= 1, b = -1, then we will have:

0 = C 0 n - C 1 n + C 2 n - C 3 n + ... + (-1) n C n n

or C 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 terms equidistant from the ends of the expansion are equal. These properties follow from the relationship: C n k = C 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 select 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 bringing similar terms is equal to the number of ways in which such a choice can be made. The first step of the sequence of elections can be carried out in ways, the second step - in, the third - etc., the th step - in ways. The required coefficient is equal to the product

SECTION 2. Higher order derivatives.

The concept of higher order derivatives.

Let the function be differentiable in some interval. Then its derivative, generally speaking, depends on X, that is, is a function of X. Consequently, in relation to it, the question of the existence of a derivative can again be raised.

Definition . The derivative of the first derivative is called second order derivative or second derivative and is denoted by the symbol or, that is

Definition . The derivative of the second derivative is called the third order derivative or third derivative and is denoted by the symbol or.

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 first are called higher derivatives.

Comment. Similarly, we can obtain the formula n-th derivative of the function:

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 the independent variable.

Since then

and taking into account that,

We get it, that is.

The third derivative can be found similarly.

Differential of sum, product and quotient.

Since the differential is obtained from the derivative by multiplying it by the differential of the 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 the constant is zero.

2 0 . The differential of an 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 by the differential of the second and the second function by the differential of the first .

Consequence. The constant multiplier can be taken out of the differential sign.

2.3. Functions defined parametrically, their differentiation.

Definition . A function is said to be specified parametrically if both variables X And y are defined each separately as single-valued functions of the same auxiliary variable - parametert :

Wheret varies within.

Comment . Let us present the parametric equations of a circle and an ellipse.

a) Circle with center at the origin and radius r has parametric equations:

b) Let us 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 equations where and are differentiable with respect tot functions and then.

2.4. Leibniz formula

To find the derivative n the th order of the product of two functions, Leibniz’s formula is of great practical importance.

Let u And v- some functions from a variable X, having derivatives of any order and y = uv. Let's 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 exponents there are numbers that determine the order of the derivative, and the functions themselves can be considered as “zero order derivatives”. Taking this into account, we obtain Leibniz’s formula:

This formula can be proven by mathematical induction.

SECTION 3. APPLICATION OF LEIBNITZ 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, use Leibniz formula.

Using this formula, we will consider examples of calculating the nth-order derivative of the product of two functions.

Example 1.

Find the second order derivative of a function

According to the definition, the second derivative is the first derivative of the first derivative, that is

Therefore, we first find the first-order derivative of the given function according to differentiation rules and using table of derivatives:

Now let's 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 a 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 order derivative is the first derivative of the first derivative, that is

Third order derivative:

Fourth derivative:

Note the pattern: in the numerator there is a factorial of a number that is equal to the order of the derivative, and in the denominator the expression to the power is one greater than the order of the derivative, that is

Answer.

Example 3.

Find the value of the third derivative of the function at a point.

Solution.

According to table of higher order derivatives, we have:

In the example under consideration, that is, we get

Note that a similar result could be obtained by sequentially finding the derivatives.

At a given point the third derivative is equal to:

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, to find the fourth-order derivative it would be advisable to apply the Leibniz formula:

Let's find all the derivatives and calculate the coefficients of the terms.

1) Let's calculate the coefficients of the terms:

2) Find the derivatives of the function:

3) Find the derivatives of the function:

Answer:

Example 6.

Given the function y=x 2 cos3x. Find the third order derivative.

Let u=cos3x , v=x 2 . Then, using Leibniz’s formula, we find:

The derivatives in this expression have the form:

(cos3x)′=−3sin3x,

(cos3x)′′=(−3sin3x)′=−9cos3x,

(cos3x)′′′=(−9cos3x)′=27sin3x,

(x2)′=2x,

(x2)′′=2,

(x2)′′′=0.

Therefore, the third derivative of the given function is equal to

1 ⋅ 27sin3x ⋅ x2+3 ⋅ (−9cos3x) ⋅ 2x+3 ⋅ (−3sin3x) ⋅ 2+1 ⋅ cos3x ⋅ 0

27x2sin3x−54xcos3x−18sin3x=(27x2−18)sin3x−54xcos3x.

Example 7.

Find the derivative n th order function y=x 2 cosx.

Let us use Leibniz's formula, assumingu=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 of the cosine function:

Therefore, the derivative of our function is equal to

CONCLUSION

At school, the so-called abbreviated multiplication formulas are studied and used: squares and cubes of the sum and difference of two expressions and formulas for factoring the difference of squares, sum and difference of cubes of two expressions. A generalization of these formulas is the formula called Newton's binomial formula and the formula 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 work systematizes information on the topic, provides examples of problems using Newton's binomial and formulas for the sum and difference of powers. The work can be used in the work of a mathematical circle, as well as for self-study those who are interested in mathematics.

LIST OF SOURCES USED

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. 10th grade: textbook. for general education organizations basic and advanced levels - M.: Prosveshchenie, 2014. - 431 p.

3. Solving problems in statistics, combinatorics and probability theory. 7-9 grades / author - compiler V.N. Studenetskaya. - ed. 2nd, revised, - Volgograd: Teacher, 2009.

4. Savushkina I.A., Khugaev K.D., Tishkin S.B. Algebraic equations higher degrees/methodological manual for students of the interuniversity preparatory department. - St. Petersburg, 2001.

5. Sharygin I.F. Optional course in mathematics: Problem solving. Textbook for 10th grade. high school. - M.: Education, 1989.

6.Science and life, Newton's binomial and Pascal's triangle[Electronic resource]. - Access mode: http://www.nkj.ru/archive/articles/13598/

Leibniz's formula is given for nth calculations derivative of the product of two functions. Its proof is given in two ways. An example of calculating the nth order derivative is considered.

Content

Leibniz formula

Using Leibniz's formula, you can calculate the nth order derivative of the product of two functions. It looks like this:
(1) ,
Where
- binomial coefficients.

Binomial coefficients are the coefficients of the expansion of a binomial in powers and:
.
Also the number is the number of combinations of n through k.

Proof of Leibniz's formula

Applicable 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 variable x, and the other on the variable y. At the end of the calculation we assume . Then the previous formula can be written as follows:
(3) .
Since the derivative is equal to the sum of the terms, and each term is the product of two functions, then to calculate derivatives of higher orders, rule (3) can be consistently applied.

Then for the nth order derivative we have:

.
Considering that and , we obtain Leibniz’s formula:
(1) .

Proof by induction

Let us present a proof of Leibniz's formula using the method of mathematical induction.

Let us write out Leibniz’s formula once again:
(4) .
For n = 1 we have:
.
This is the formula for the derivative of the product of two functions. She's 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.

Let's differentiate (4):
;

.
So we found:
(5) .

Let's 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 holds for some number n = m it follows that it holds for n = m + 1 .
Leibniz's formula has been proven.

Example

Calculate the nth derivative of a function
.

Let's apply Leibniz's formula
(2) .
In our case
;
.

By derivative table we have:
.
We apply properties of trigonometric functions :
.
Then
.
This shows that differentiation of the sine function leads to its shift by . Then
.

Finding derivatives of the function.
;
;
;
, .

Since for , then in Leibniz’s formula only the first three terms are nonzero. Finding binomial coefficients.
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According to Leibniz's formula we have:

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Newton-Leibniz formula

Definition 1

When the function y = y (x) is continuous from the interval [ 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 produce a proof of this formula, it is necessary to use the concept of an integral with an available variable upper limit.

When the function y = f (x) is continuous from the interval [ 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 take the notation of the function will take the form ∫ a x f (t) d t = Φ (x) , it is continuous, and an inequality of the form ∫ a x f (t) d t " = Φ " (x) = f (x) is valid for it.

Let us 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 the definite integral and we obtain

Φ (x + ∆ x) - Φ x = ∫ a x + ∆ x f (t) d t - ∫ a x f (t) d t = = ∫ a x + ∆ x f (t) d t = f (c) x + ∆ x - x = f (c) ∆ x

where value c ∈ x; x + ∆ x .

Let us 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 go to the limit as ∆ x → 0, then we obtain a formula of the form Φ " (x) = f (x). We find that Φ (x) is one of the antiderivatives for a function of the form y = f (x), 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 we get that C = F (a). The result is applicable when calculating F (b) and we get:

F (b) = Φ (b) + C = ∫ a b f (t) d t + C = ∫ a b f (t) d t + F (a), in other words, F (b) = ∫ a b f (t) d t + F ( a) . The equality is proved by the Newton-Leibniz formula ∫ a b f (x) d x + F (b) - F (a) .

We take the increment of the function as F x a b = F (b) - F (a) . Using the notation, the Newton-Leibniz formula takes the form ∫ 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 function y = f (x) from the segment [ a ; b ], calculate the increment of the antiderivative from this segment. Let's look at 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 it is integrable on this interval. From the table of indefinite integrals we see that the function y = x 2 has a set of antiderivatives for all real values of x, which means 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 = 0, then we obtain that F (x) = x 3 3.

We use the Newton-Leibniz formula and find that the calculation of the definite integral takes 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 it is integrable on it. It is necessary to find the value of the indefinite integral ∫ x · e x 2 + 1 d x using the method of subsuming under the differential sign, then we obtain ∫ x · e x 2 + 1 d x = 1 2 ∫ e x 2 + 1 d (x 2 + 1) = 1 2 e x 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 · e x 2 + 1 d x = 1 2 e x 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 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) = 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 d x = 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 proceed 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 antiderivative for y = 4 x 3 + 2 x 2 from the interval [ - 1 ; 1 ], since point O belongs to the segment, but is not included in the domain of definition. This means that there is a definite Riemann and Newton-Leibniz integral for the function y = 4 x 3 + 2 x 2 from the interval [ - 1 ; 1 ] .

Answer: ∫ - 4 - 1 2 4 x 3 + 2 x 2 d x = - 28 , there is a definite Riemann and Newton-Leibniz integral for the function y = 4 x 3 + 2 x 2 from the interval [ - 1 ; 1 ] .

Before using the Newton-Leibniz formula, you need to know exactly about the existence of a definite integral.

Changing a variable in a definite integral

When the function y = f (x) is defined and continuous from the interval [ a ; b], then the available set [a; b] is considered to be the range of values of the function x = g (z), defined on the segment α; β with the existing continuous derivative, where g (α) = a and g β = b, we obtain from this that ∫ a b f (x) d x = ∫ α β f (g (z)) g " (z) d z.

This formula is used when you need 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 function is considered continuous on the interval of integration, which means that a definite integral exists. Let's give the notation that 2 x - 9 = z ⇒ x = g (z) = z 2 + 9 2. The value x = 9 means that z = 2 9 - 9 = 9 = 3, and for x = 18 we get that z = 2 18 - 9 = 27 = 3 3, then g α = g (3) = 9, g β = g 3 3 = 18. When 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 d x = ∫ 3 3 3 1 z 2 + 9 2 · z · z 2 + 9 2 " d z = = ∫ 3 3 3 1 z 2 + 9 2 · z · z d z = ∫ 3 3 3 2 z 2 + 9 d z

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, when 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 using the replacement method we use an integral of the form ∫ 1 x 2 x - 9 d x, then we can come to 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 carry out calculations using the Newton-Leibniz formula and calculate the definite integral. We get that

∫ 9 18 2 z 2 + 9 d z = 2 3 a r c t g z 3 9 18 = = 2 3 a r c t g 2 18 - 9 3 - a r c t g 2 9 - 9 3 = = 2 3 a r c t g 3 - a r c t g 1 = 2 3 π 3 - π 4 = π 18

The results were the same.

Answer: ∫ 9 18 2 x 2 x - 9 d x = π 18

Integration by parts when calculating 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, thus from this segment for the integrable function u " (x) · v ( x) the equality ∫ a b v " (x) · u (x) d x = (u (x) · v (x)) a b - ∫ a b u " (x) · v (x) d x 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 look for 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 interval - π 2 ; 3 π 2, which means it is continuous.

Let u (x) = x, then d (v (x)) = v " (x) d x = sin x 3 + π 6 d x, and d (u (x)) = u " (x) d x = d x, 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 obtain that

∫ - π 2 3 π 2 x · sin x 3 + π 6 d x = - 3 x · cos x 3 + π 6 - π 2 3 π 2 - ∫ - π 2 3 π 2 - 3 cos x 3 + π 6 d x = = - 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 = 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 example can be solved 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 x x 3 + π 6 d x = u = x , d v = sin x 3 + π 6 d x ⇒ d u = d x , v = - 3 cos x 3 + π 6 = = - 3 cos x 3 + π 6 + 3 ∫ cos x 3 + π 6 d x = = - 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 d x = - 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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