What is the derivative of the argument. What is a derivative? Definition and meaning of a derivative of a function

Definition. Let the function \(y = f(x) \) be defined in some interval containing the point \(x_0 \) inside. Let's increment \(\Delta x \) to the argument so as not to leave this interval. Find the corresponding increment of the function \(\Delta y \) (when passing from the point \(x_0 \) to the point \(x_0 + \Delta x \)) and compose the relation \(\frac(\Delta y)(\Delta x) \). If there is a limit of this relation at \(\Delta x \rightarrow 0 \), then the specified limit is called derivative function\(y=f(x) \) at the point \(x_0 \) and denote \(f"(x_0) \).

$$ \lim_(\Delta x \to 0) \frac(\Delta y)(\Delta x) = f"(x_0) $$

The symbol y is often used to denote the derivative. Note that y" = f(x) is a new function, but naturally associated with the function y = f(x), defined at all points x at which the above limit exists . This function is called like this: derivative of the function y \u003d f (x).

geometric sense derivative consists of the following. If a tangent that is not parallel to the y axis can be drawn to the graph of the function y \u003d f (x) at a point with the abscissa x \u003d a, then f (a) expresses the slope of the tangent:
\(k = f"(a)\)

Since \(k = tg(a) \), the equality \(f"(a) = tg(a) \) is true.

And now we interpret the definition of the derivative in terms of approximate equalities. Let the function \(y = f(x) \) have a derivative at a particular point \(x \):
$$ \lim_(\Delta x \to 0) \frac(\Delta y)(\Delta x) = f"(x) $$
This means that near the point x, the approximate equality \(\frac(\Delta y)(\Delta x) \approx f"(x) \), i.e. \(\Delta y \approx f"(x) \cdot\Deltax\). The meaningful meaning of the obtained approximate equality is as follows: the increment of the function is “almost proportional” to the increment of the argument, and the coefficient of proportionality is the value of the derivative in given point X. For example, for the function \(y = x^2 \) the approximate equality \(\Delta y \approx 2x \cdot \Delta x \) is true. If we carefully analyze the definition of the derivative, we will find that it contains an algorithm for finding it.

Let's formulate it.

How to find the derivative of the function y \u003d f (x) ?

1. Fix value \(x \), find \(f(x) \)
2. Increment \(x \) argument \(\Delta x \), go to new point\(x+ \Delta x \), find \(f(x+ \Delta x) \)
3. Find the function increment: \(\Delta y = f(x + \Delta x) - f(x) \)
4. Compose the relation \(\frac(\Delta y)(\Delta x) \)
5. Calculate $$ \lim_(\Delta x \to 0) \frac(\Delta y)(\Delta x) $$
This limit is the derivative of the function at x.

If the function y = f(x) has a derivative at the point x, then it is called differentiable at the point x. The procedure for finding the derivative of the function y \u003d f (x) is called differentiation functions y = f(x).

Let us discuss the following question: how are the continuity and differentiability of a function at a point related?

Let the function y = f(x) be differentiable at the point x. Then a tangent can be drawn to the graph of the function at the point M (x; f (x)) and, recall, the slope of the tangent is equal to f "(x). Such a graph cannot "break" at the point M, i.e., the function must be continuous at x.

It was reasoning "on the fingers". Let us present a more rigorous argument. If the function y = f(x) is differentiable at the point x, then the approximate equality \(\Delta y \approx f"(x) \cdot \Delta x \) holds. zero, then \(\Delta y \) will also tend to zero, and this is the condition for the continuity of the function at a point.

So, if a function is differentiable at a point x, then it is also continuous at that point.

The converse is not true. For example: function y = |x| is continuous everywhere, in particular at the point x = 0, but the tangent to the graph of the function at the “joint point” (0; 0) does not exist. If at some point it is impossible to draw a tangent to the function graph, then there is no derivative at this point.

One more example. The function \(y=\sqrt(x) \) is continuous on the entire number line, including at the point x = 0. And the tangent to the graph of the function exists at any point, including at the point x = 0. But at this point the tangent coincides with the y-axis, that is, it is perpendicular to the abscissa axis, its equation has the form x \u003d 0. There is no slope for such a straight line, which means that \ (f "(0) \) does not exist either

So, we got acquainted with a new property of a function - differentiability. How can you tell if a function is differentiable from the graph of a function?

The answer is actually given above. If at some point a tangent can be drawn to the graph of a function that is not perpendicular to the x-axis, then at this point the function is differentiable. If at some point the tangent to the graph of the function does not exist or it is perpendicular to the x-axis, then at this point the function is not differentiable.

Differentiation rules

The operation of finding the derivative is called differentiation. When performing this operation, you often have to work with quotients, sums, products of functions, as well as with "functions of functions", that is, complex functions. Based on the definition of the derivative, we can derive differentiation rules that facilitate this work. If C- constant number and f=f(x), g=g(x) are some differentiable functions, then the following are true differentiation rules:

$$ C"=0 $$ $$ x"=1 $$ $$ (f+g)"=f"+g" $$ $$ (fg)"=f"g + fg" $$ $$ ( Cf)"=Cf" $$ $$ \left(\frac(f)(g) \right) " = \frac(f"g-fg")(g^2) $$ $$ \left(\frac (C)(g) \right) " = -\frac(Cg")(g^2) $$ Compound function derivative:
$$ f"_x(g(x)) = f"_g \cdot g"_x $$

Table of derivatives of some functions

$$ \left(\frac(1)(x) \right) " = -\frac(1)(x^2) $$ $$ (\sqrt(x)) " = \frac(1)(2\ sqrt(x)) $$ $$ \left(x^a \right) " = a x^(a-1) $$ $$ \left(a^x \right) " = a^x \cdot \ln a $$ $$ \left(e^x \right) " = e^x $$ $$ (\ln x)" = \frac(1)(x) $$ $$ (\log_a x)" = \frac (1)(x\ln a) $$ $$ (\sin x)" = \cos x $$ $$ (\cos x)" = -\sin x $$ $$ (\text(tg) x) " = \frac(1)(\cos^2 x) $$ $$ (\text(ctg) x)" = -\frac(1)(\sin^2 x) $$ $$ (\arcsin x) " = \frac(1)(\sqrt(1-x^2)) $$ $$ (\arccos x)" = \frac(-1)(\sqrt(1-x^2)) $$ $$ (\text(arctg) x)" = \frac(1)(1+x^2) $$ $$ (\text(arctg) x)" = \frac(-1)(1+x^2) $ $

When a person took the first independent steps in the study mathematical analysis and starts asking uncomfortable questions, then it is no longer so easy to get rid of the phrase that "differential calculus is found in cabbage." Therefore, it is time to be determined and solve the mystery of the birth of tables of derivatives and differentiation rules. Started in the article about the meaning of the derivative, which I highly recommend for study, because there we just considered the concept of a derivative and started clicking tasks on the topic. The same lesson has a pronounced practical orientation, moreover,

the examples considered below, in principle, can be mastered purely formally (for example, when there is no time / desire to delve into the essence of the derivative). It is also highly desirable (but again not necessary) to be able to find derivatives using the "usual" method - at least at the level of two basic classes: How to find the derivative? and Derivative of a complex function.

But without something, which is now definitely indispensable, it is without function limits. You must UNDERSTAND what a limit is and be able to solve them, at least at an intermediate level. And all because the derivative

function at a point is defined by the formula:

I remind you of the designations and terms: they call argument increment;

– function increment;

- these are SINGLE symbols (“delta” cannot be “torn off” from “X” or “Y”).

Obviously, is a "dynamic" variable, is a constant and the result of calculating the limit - number (sometimes - "plus" or "minus" infinity).

As a point, you can consider ANY value belonging to domains a function that has a derivative.

Note: the clause "in which the derivative exists" - V general case significant! So, for example, the point, although it enters the domain of the function, but the derivative

does not exist there. Therefore the formula

not applicable at the point

and a shortened wording without a reservation would be incorrect. Similar facts are also valid for other functions with "breaks" in the graph, in particular, for the arcsine and arccosine.

Thus, after replacing , we obtain the second working formula:

Pay attention to an insidious circumstance that can confuse the teapot: in this limit, "x", being itself an independent variable, plays the role of an extra, and "dynamics" is again set by the increment. The result of limit calculation

is the derivative function.

Based on the foregoing, we formulate the conditions of two typical problems:

- Find derivative at a point using the definition of a derivative.

- Find derivative function using the definition of a derivative. This version, according to my observations, occurs much more often and will be given the main attention.

The fundamental difference between the tasks is that in the first case it is required to find the number (optionally infinity), and in the second

function . In addition, the derivative may not exist at all.

How ?

Make a ratio and calculate the limit.

Where did table of derivatives and differentiation rules ? With a single limit

Seems like magic, but

reality - sleight of hand and no fraud. At the lesson What is a derivative? I began to consider specific examples, where, using the definition, I found the derivatives of the linear and quadratic function. For the purpose of cognitive warm-up, we will continue to disturb derivative table, honing the algorithm and technical solutions:

Essentially, we need to prove special case derivative power function, which usually appears in the table: .

The solution is technically formalized in two ways. Let's start with the first, already familiar approach: the ladder starts with a plank, and the derivative function starts with a derivative at a point.

Consider some (concrete) point belonging to domains a function that has a derivative. Set the increment at this point (of course, not beyond o / o - z) and compose the corresponding increment of the function:

Let's calculate the limit:

Uncertainty 0:0 is eliminated by a standard technique considered as far back as the first century BC. multiply

numerator and denominator per adjoint expression :

The technique for solving such a limit is discussed in detail in the introductory lesson. about the limits of functions.

Since ANY point of the interval can be chosen as

Then, by substituting, we get:

Once again, let's rejoice at the logarithms:

Find the derivative of the function using the definition of the derivative

Solution: Let's consider a different approach to spinning up the same task. It is exactly the same, but more rational in terms of design. The idea is to get rid of the

subscript and use a letter instead of a letter.

Consider an arbitrary point belonging to domains function (interval), and set the increment in it. And here, by the way, as in most cases, you can do without any reservations, since logarithmic function is differentiable at any point of the domain of definition.

Then the corresponding function increment is:

Let's find the derivative:

Simplicity of design is balanced by confusion, which can

arise in beginners (and not only). After all, we are used to the fact that the letter “X” changes in the limit! But here everything is different: - an antique statue, and - a living visitor, briskly walking along the corridor of the museum. That is, “x” is “like a constant”.

I will comment on the elimination of uncertainty step by step:

(1) Using the property of the logarithm.

(2) Divide the numerator by the denominator in parentheses.

(3) In the denominator we artificially multiply and divide by "x" so that

take advantage of the wonderful , while as infinitesimal performs.

Answer: By definition of derivative:

Or in short:

I propose to independently construct two more tabular formulas:

Find derivative by definition

IN this case the compounded increment is immediately conveniently reduced to common denominator. An approximate sample of the assignment at the end of the lesson (the first method).

Find derivative by definition

And here everything must be reduced to a remarkable limit. The solution is framed in the second way.

Similarly, a number of other tabular derivatives. Full list can be found in a school textbook, or, for example, the 1st volume of Fichtenholtz. I don’t see much point in rewriting from books and proofs of the rules of differentiation - they are also generated

formula .

Let's move on to real-life tasks: Example 5

Find the derivative of a function , using the definition of the derivative

Solution: use the first style. Let's consider some point that belongs to, and set the increment of the argument in it. Then the corresponding function increment is:

Perhaps some readers have not yet fully understood the principle by which an increment should be made. We take a point (number) and find the value of the function in it: , that is, into the function

instead of "x" should be substituted. Now we take

Composed Function Increment it is beneficial to immediately simplify. For what? Facilitate and shorten the solution of the further limit.

We use formulas, open brackets and reduce everything that can be reduced:

The turkey is gutted, no problem with the roast:

Eventually:

Since you can choose any real number, then we make the substitution and get .

Answer : a-priory.

For verification purposes, we find the derivative using the rules

differentiations and tables:

It is always useful and pleasant to know the correct answer in advance, so it is better to mentally or on a draft differentiate the proposed function in a “quick” way at the very beginning of the solution.

Find the derivative of a function by the definition of the derivative

This is an example for independent decision. The result lies on the surface:

Back to Style #2: Example 7

Let's find out immediately what should happen. By the rule of differentiation of a complex function:

Solution: consider arbitrary point belonging, we set the increment of the argument in it and make the increment

Let's find the derivative:

(1) We use the trigonometric formula

(2) Under the sine we open the brackets, under the cosine we give like terms.

(3) Under the sine we reduce the terms, under the cosine we divide the numerator by the denominator term by term.

(4) Due to the oddness of the sine, we take out the "minus". Under cosine

indicate that the term .

(5) We artificially multiply the denominator to use first wonderful limit . Thus, the uncertainty is eliminated, we comb the result.

Answer: by definition As you can see, the main difficulty of the problem under consideration rests on

the complexity of the limit itself + a slight originality of the packaging. In practice, both methods of design are encountered, so I describe both approaches in as much detail as possible. They are equivalent, but still, in my subjective impression, it is more expedient for dummies to stick to the 1st option with “X zero”.

Using the definition, find the derivative of the function

This is a task for independent decision. The sample is formatted in the same spirit as the previous example.

Let's analyze a rarer version of the problem:

Find the derivative of a function at a point using the definition of a derivative.

First, what should be the bottom line? Number Calculate the answer in the standard way:

Decision: from the point of view of clarity, this task is much simpler, since in the formula instead of

considered a specific value.

We set an increment at the point and compose the corresponding increment of the function:

Calculate the derivative at a point:

We use very rare formula tangent differences and for the umpteenth time we reduce the solution to the first

amazing limit:

Answer: by definition of the derivative at a point.

The problem is not so difficult to solve and general view”- it is enough to replace nail or simple, depending on the design method. In this case, of course, you get not a number, but a derivative function.

Example 10 Using the definition, find the derivative of a function at the point

This is a do-it-yourself example.

The final bonus task is intended primarily for students with in-depth study mathematical analysis, but it will not hurt everyone else either:

Will the function be differentiable at the point?

Solution: It is obvious that a piecewise given function is continuous at a point, but will it be differentiable there?

The solution algorithm, and not only for piecewise functions, is as follows:

1) Find the left-hand derivative at a given point: .

2) Find the right-hand derivative at the given point: .

3) If one-sided derivatives are finite and coincide:

, then the function is differentiable at the point and

geometrically, there is a common tangent here (see the theoretical part of the lesson Definition and meaning of derivative).

If received two different meanings: (one of which may be infinite), then the function is not differentiable at a point.

If both one-sided derivatives are equal to infinity

(even if they have different signs), then the function does not

is differentiable at a point, but there exists an infinite derivative and a common vertical tangent to the graph (see Example 5 of lessonNormal Equation) .

The content of the article

DERIVATIVE-derivative of the function y = f(x) defined on some interval ( a, b) at the point x this interval is called the limit to which the ratio of the increment of the function tends f at that point to the corresponding increment of the argument as the increment of the argument approaches zero.

The derivative is usually denoted as follows:

Other notations are also widely used:

Instant speed.

Let the point M moves in a straight line. Distance s moving point, counted from some initial position M 0 , depends on time t, i.e. s is a function of time t: s= f(t). Let at some point in time t moving point M was at a distance s from the starting position M 0, and at some next moment t+ D t was in a position M 1 - on distance s+ D s from the initial position ( see pic.).

Thus, for a period of time D t distance s changed by the value D s. In this case, we say that during the time interval D t magnitude s received increment D s.

The average speed cannot in all cases accurately characterize the speed of moving a point. M at the time t. If, for example, the body at the beginning of the interval D t moved very quickly, and at the end very slowly, then average speed will not be able to reflect the indicated features of the movement of the point and give an idea of ​​the true speed of its movement at the moment t. To more accurately express the true speed using the average speed, you need to take a smaller period of time D t. It most fully characterizes the speed of movement of a point at the moment t the limit to which the average speed tends at D t® 0. This limit is called the speed of movement at a given moment:

Thus, the speed of movement at a given moment is the limit of the ratio of the increment of the path D s to the time increment D t when the time increment tends to zero. Because

The geometric value of the derivative. Tangent to the graph of a function.

The construction of tangents is one of those problems that led to the birth of differential calculus. The first published work relating to differential calculus and feathered Leibniz, had the name New method maxima and minima, as well as tangents, for which neither fractional nor irrational quantities are an obstacle, and a special kind of calculus for this.

Let the curve be the graph of the function y =f(x) V rectangular system coordinates ( cm. rice.).

For some value x function matters y =f(x). These values x And y point on the curve M 0(x, y). If the argument x give increment D x, then the new value of the argument x+ D x corresponds to the new value of the function y+ D y = f(x + D x). The corresponding point of the curve will be the point M 1(x+ D x,y+ D y). If we draw a secant M 0M 1 and denote by j angle formed by a secant with a positive axis direction Ox, it is directly seen from the figure that .

If now D x tends to zero, then the point M 1 moves along the curve, approaching the point M 0 and angle j changes with change D x. At Dx® 0 the angle j tends to some limit a and the line passing through the point M 0 and the component with the positive direction of the abscissa axis, angle a, will be the desired tangent. Its slope:

Hence, f´( x) = tga

those. derivative value f´( x) at given value argument x equals the tangent of the angle formed by the tangent to the graph of the function f(x) at the corresponding point M 0(x,y) with positive axis direction Ox.

Differentiability of functions.

Definition. If the function y = f(x) has a derivative at the point x = x 0, then the function is differentiable at this point.

Continuity of a function that has a derivative. Theorem.

If the function y = f(x) is differentiable at some point x = x 0, then it is continuous at this point.

Thus, at discontinuity points, the function cannot have a derivative. The converse conclusion is false, i.e. from the fact that at some point x = x 0 function y = f(x) is continuous, it does not follow that it is differentiable at this point. For example, the function y = |x| continuous for all x(–Ґ x x = 0 has no derivative. There is no tangent to the graph at this point. There is a right tangent and a left tangent, but they do not coincide.

Some theorems on differentiable functions. Theorem on the roots of the derivative (Roll's theorem). If the function f(x) is continuous on the interval [a,b], is differentiable in all internal points this segment and at the ends x = a And x = b vanishes ( f(a) = f(b) = 0), then inside the segment [ a,b] there is at least one point x= With, a c b, in which the derivative fў( x) vanishes, i.e. fў( c) = 0.

Finite increment theorem (Lagrange's theorem). If the function f(x) is continuous on the segment [ a, b] and is differentiable at all interior points of this segment, then inside the segment [ a, b] there is at least one point With, a c b that

f(b) – f(a) = fў( c)(b– a).

Theorem on the ratio of increments of two functions (Cauchy's theorem). If f(x) And g(x) are two functions continuous on the segment [a, b] and differentiable at all interior points of this segment, and gў( x) does not vanish anywhere inside this segment, then inside the segment [ a, b] there is such a point x = With, a c b that

Derivatives of various orders.

Let the function y =f(x) is differentiable on some interval [ a, b]. Derivative values f ў( x), generally speaking, depend on x, i.e. derivative f ў( x) is also a function of x. When this function is differentiated, the so-called second derivative of the function is obtained f(x), which is denoted f ўў ( x).

derivative n- order of the function f(x) is called the derivative (of the first order) of the derivative n- 1- th and is denoted by the symbol y(n) = (y(n– 1))ў.

Differentials of various orders.

Function differential y = f(x), Where x is an independent variable, is dy = f ў( x)dx, some function from x, but from x only the first factor can depend f ў( x), while the second factor ( dx) is the increment of the independent variable x and does not depend on the value of this variable. Because dy there is a function from x, then we can determine the differential of this function. The differential of the differential of a function is called the second or second-order differential of this function and is denoted d 2y:

d(dx) = d 2y = f ўў( x)(dx) 2 .

Differential n- order is called the first differential of the differential n- 1- order:

d n y = d(d n–1y) = f(n)(x)dx(n).

Private derivative.

If the function depends not on one, but on several arguments x i(i changes from 1 to n,i= 1, 2,… n),f(x 1,x 2,… x n), then in differential calculus the concept of a partial derivative is introduced, which characterizes the rate of change of a function of several variables when only one argument changes, for example, x i. Partial derivative of the 1st order with respect to x i is defined as the ordinary derivative, it is assumed that all arguments except x i, keep constant values. For partial derivatives, we introduce the notation

Partial derivatives of the 1st order defined in this way (as functions of the same arguments) can, in turn, also have partial derivatives, these are partial derivatives of the second order, etc. Taken with respect to different arguments, such derivatives are called mixed. Continuous mixed derivatives of the same order do not depend on the order of differentiation and are equal to each other.

Anna Chugainova

When solving various problems of geometry, mechanics, physics and other branches of knowledge, it became necessary to use the same analytical process from a given function y=f(x) receive new feature, which is called derivative function(or simply derivative) of this function f(x) and are symbolized

The process by which a given function f(x) get a new function f"(x), called differentiation and it consists of the following three steps: 1) we give the argument x increment  x and determine the corresponding increment of the function  y = f(x+ x)-f(x); 2) make up the relation

3) counting x permanent, and  x0, we find
, which is denoted by f"(x), as if emphasizing that the resulting function depends only on the value x, at which we pass to the limit. Definition: Derivative y "=f" (x) given function y=f(x) given x is called the limit of the ratio of the increment of the function to the increment of the argument, provided that the increment of the argument tends to zero, if, of course, this limit exists, i.e. finite. Thus,
, or

Note that if for some value x, for example when x=a, relation
at  x0 does not tend to a finite limit, then in this case we say that the function f(x) at x=a(or at the point x=a) has no derivative or is not differentiable at a point x=a.

2. The geometric meaning of the derivative.

Consider the graph of the function y \u003d f (x), differentiable in the vicinity of the point x 0

f(x)

Let's consider an arbitrary straight line passing through the point of the graph of the function - the point A (x 0, f (x 0)) and intersecting the graph at some point B (x; f (x)). Such a straight line (AB) is called a secant. From ∆ABC: ​​AC = ∆x; BC \u003d ∆y; tgβ=∆y/∆x .

Since AC || Ox, then ALO = BAC = β (as corresponding in parallel). But ALO is the angle of inclination of the secant AB to the positive direction of the Ox axis. Hence, tgβ = k is the slope of the straight line AB.

Now we will decrease ∆x, i.e. ∆x→ 0. In this case, point B will approach point A according to the graph, and the secant AB will rotate. The limiting position of the secant AB at ∆x → 0 will be the straight line (a), called the tangent to the graph of the function y \u003d f (x) at point A.

If we pass to the limit as ∆х → 0 in the equality tgβ =∆y/∆x, then we get
or tg \u003d f "(x 0), since
-angle of inclination of the tangent to the positive direction of the Ox axis
, by definition of a derivative. But tg \u003d k is the slope of the tangent, which means that k \u003d tg \u003d f "(x 0).

So, the geometric meaning of the derivative is as follows:

Derivative of a function at a point x 0 is equal to angular coefficient tangent to the graph of the function drawn at the point with the abscissa x 0 .

3. Physical meaning of the derivative.

Consider the movement of a point along a straight line. Let the coordinate of a point at any time x(t) be given. It is known (from the course of physics) that the average speed over a period of time is equal to the ratio of the distance traveled during this period of time to the time, i.e.

Vav = ∆x/∆t. Let us pass to the limit in the last equality as ∆t → 0.

lim Vav (t) = (t 0) - instantaneous speed at time t 0 , ∆t → 0.

and lim = ∆x/∆t = x "(t 0) (by the definition of a derivative).

So, (t) = x"(t).

The physical meaning of the derivative is as follows: the derivative of the functiony = f(x) at the pointx 0 is the rate of change of the functionf(x) at the pointx 0

The derivative is used in physics to find the speed from known function coordinates from time, acceleration from a known function of velocity from time.

 (t) \u003d x "(t) - speed,

a(f) = "(t) - acceleration, or

If the law of motion of a material point along a circle is known, then it is possible to find the angular velocity and angular acceleration in rotation:

φ = φ(t) - change in angle with time,

ω \u003d φ "(t) - angular velocity,

ε = φ"(t) - angular acceleration, or ε = φ"(t).

If the distribution law for the mass of an inhomogeneous rod is known, then the linear density of the inhomogeneous rod can be found:

m \u003d m (x) - mass,

x  , l - rod length,

p \u003d m "(x) - linear density.

With the help of the derivative, problems from the theory of elasticity and harmonic vibrations are solved. Yes, according to Hooke's law

F = -kx, x – variable coordinate, k – coefficient of elasticity of the spring. Putting ω 2 \u003d k / m, we obtain the differential equation of the spring pendulum x "(t) + ω 2 x (t) \u003d 0,

where ω = √k/√m is the oscillation frequency (l/c), k is the spring rate (H/m).

An equation of the form y "+ ω 2 y \u003d 0 is called the equation of harmonic oscillations (mechanical, electrical, electromagnetic). The solution to such equations is the function

y = Asin(ωt + φ 0) or y = Acos(ωt + φ 0), where

A - oscillation amplitude, ω - cyclic frequency,

φ 0 - initial phase.

If we follow the definition, then the derivative of a function at a point is the limit of the increment ratio of the function Δ y to the increment of the argument Δ x:

Everything seems to be clear. But try to calculate by this formula, say, the derivative of the function f(x) = x 2 + (2x+ 3) · e x sin x. If you do everything by definition, then after a couple of pages of calculations you will simply fall asleep. Therefore, there are simpler and more effective ways.

To begin with, we note that the so-called elementary functions can be distinguished from the whole variety of functions. It's relative simple expressions, whose derivatives have long been calculated and entered in the table. Such functions are easy enough to remember, along with their derivatives.

Derivatives of elementary functions

Elementary functions are everything listed below. The derivatives of these functions must be known by heart. Moreover, it is not difficult to memorize them - that's why they are elementary.

So, the derivatives of elementary functions:

Name	Function	Derivative
Constant	f(x) = C, C ∈ R	0 (yes, yes, zero!)
Degree with rational exponent	f(x) = x n	n · x n − 1
Sinus	f(x) = sin x	cos x
Cosine	f(x) = cos x	− sin x(minus sine)
Tangent	f(x) = tg x	1/cos 2 x
Cotangent	f(x) = ctg x	− 1/sin2 x
natural logarithm	f(x) = log x	1/x
Arbitrary logarithm	f(x) = log a x	1/(x ln a)
Exponential function	f(x) = e x	e x(nothing changed)

If an elementary function is multiplied by an arbitrary constant, then the derivative of the new function is also easily calculated:

(C · f)’ = C · f ’.

In general, constants can be taken out of the sign of the derivative. For example:

(2x 3)' = 2 ( x 3)' = 2 3 x 2 = 6x 2 .

Obviously, elementary functions can be added to each other, multiplied, divided, and much more. This is how new functions will appear, no longer very elementary, but also differentiable according to certain rules. These rules are discussed below.

Derivative of sum and difference

Let the functions f(x) And g(x), whose derivatives are known to us. For example, you can take the elementary functions discussed above. Then you can find the derivative of the sum and difference of these functions:

1. (f + g)’ = f ’ + g ’
2. (f − g)’ = f ’ − g ’

So, the derivative of the sum (difference) of two functions is equal to the sum (difference) of the derivatives. There may be more terms. For example, ( f + g + h)’ = f ’ + g ’ + h ’.

Strictly speaking, there is no concept of "subtraction" in algebra. There is a concept negative element". Therefore, the difference f − g can be rewritten as a sum f+ (−1) g, and then only one formula remains - the derivative of the sum.

f(x) = x 2 + sinx; g(x) = x 4 + 2x 2 − 3.

Function f(x) is the sum of two elementary functions, so:

f ’(x) = (x 2+ sin x)’ = (x 2)' + (sin x)’ = 2x+ cosx;

We argue similarly for the function g(x). Only there are already three terms (from the point of view of algebra):

g ’(x) = (x 4 + 2x 2 − 3)’ = (x 4 + 2x 2 + (−3))’ = (x 4)’ + (2x 2)’ + (−3)’ = 4x 3 + 4x + 0 = 4x · ( x 2 + 1).

Answer:
f ’(x) = 2x+ cosx;
g ’(x) = 4x · ( x 2 + 1).

Derivative of a product

Mathematics is a logical science, so many people believe that if the derivative of the sum is equal to the sum of the derivatives, then the derivative of the product strike"\u003e equal to the product of derivatives. But figs to you! The derivative of the product is calculated using a completely different formula. Namely:

(f · g) ’ = f ’ · g + f · g ’

The formula is simple, but often forgotten. And not only schoolchildren, but also students. The result is incorrectly solved problems.

Task. Find derivatives of functions: f(x) = x 3 cosx; g(x) = (x 2 + 7x− 7) · e x .

Function f(x) is a product of two elementary functions, so everything is simple:

f ’(x) = (x 3 cos x)’ = (x 3)' cos x + x 3 (cos x)’ = 3x 2 cos x + x 3 (−sin x) = x 2 (3cos x − x sin x)

Function g(x) the first multiplier is a bit more complicated, but general scheme this does not change. Obviously, the first multiplier of the function g(x) is a polynomial, and its derivative is the derivative of the sum. We have:

g ’(x) = ((x 2 + 7x− 7) · e x)’ = (x 2 + 7x− 7)' · e x + (x 2 + 7x− 7) ( e x)’ = (2x+ 7) · e x + (x 2 + 7x− 7) · e x = e x(2 x + 7 + x 2 + 7x −7) = (x 2 + 9x) · e x = x(x+ 9) · e x .

Answer:
f ’(x) = x 2 (3cos x − x sin x);
g ’(x) = x(x+ 9) · e x .

Note that in the last step, the derivative is factorized. Formally, this is not necessary, but most derivatives are not calculated on their own, but to explore the function. This means that further the derivative will be equated to zero, its signs will be found out, and so on. For such a case, it is better to have an expression decomposed into factors.

If there are two functions f(x) And g(x), and g(x) ≠ 0 on the set of interest to us, we can define a new function h(x) = f(x)/g(x). For such a function, you can also find the derivative:

Not weak, right? Where did the minus come from? Why g 2? And like this! This is one of the most complex formulas You can't figure it out without a bottle. Therefore, it is better to study it on concrete examples.

Task. Find derivatives of functions:

There are elementary functions in the numerator and denominator of each fraction, so all we need is the formula for the derivative of the quotient:

By tradition, we factor the numerator into factors - this will greatly simplify the answer:

A complex function is not necessarily a formula half a kilometer long. For example, it suffices to take the function f(x) = sin x and replace the variable x, say, on x 2+ln x. It turns out f(x) = sin ( x 2+ln x) - That's what it is complex function. She also has a derivative, but it will not work to find it according to the rules discussed above.

How to be? In such cases, the replacement of a variable and the formula for the derivative of a complex function help:

f ’(x) = f ’(t) · t', If x is replaced by t(x).

As a rule, the situation with the understanding of this formula is even more sad than with the derivative of the quotient. Therefore, it is also better to explain it with specific examples, with detailed description every step.

Task. Find derivatives of functions: f(x) = e 2x + 3 ; g(x) = sin ( x 2+ln x)

Note that if in the function f(x) instead of expression 2 x+ 3 will be easy x, then it will work elementary function f(x) = e x. Therefore, we make a substitution: let 2 x + 3 = t, f(x) = f(t) = e t. We are looking for the derivative of a complex function by the formula:

f ’(x) = f ’(t) · t ’ = (e t)’ · t ’ = e t · t ’

And now - attention! Performing a reverse substitution: t = 2x+ 3. We get:

f ’(x) = e t · t ’ = e 2x+ 3 (2 x + 3)’ = e 2x+ 3 2 = 2 e 2x + 3

Now let's look at the function g(x). Obviously needs to be replaced. x 2+ln x = t. We have:

g ’(x) = g ’(t) · t' = (sin t)’ · t' = cos t · t ’

Reverse replacement: t = x 2+ln x. Then:

g ’(x) = cos( x 2+ln x) · ( x 2+ln x)' = cos ( x 2+ln x) · (2 x + 1/x).

That's all! As can be seen from the last expression, the whole problem has been reduced to calculating the derivative of the sum.

Answer:
f ’(x) = 2 e 2x + 3 ;
g ’(x) = (2x + 1/x) cos ( x 2+ln x).

Very often in my lessons, instead of the term “derivative”, I use the word “stroke”. For example, a stroke from the sum is equal to the sum strokes. Is that clearer? Well, that's good.

Thus, the calculation of the derivative comes down to getting rid of these very strokes according to the rules discussed above. As last example Let's return to the derivative power with a rational exponent:

(x n)’ = n · x n − 1

Few know that in the role n may well act a fractional number. For example, the root is x 0.5 . But what if there is something tricky under the root? Again, a complex function will turn out - they like to give such constructions on control work and exams.

Task. Find the derivative of a function:

First, let's rewrite the root as a power with a rational exponent:

f(x) = (x 2 + 8x − 7) 0,5 .

Now we make a substitution: let x 2 + 8x − 7 = t. We find the derivative by the formula:

f ’(x) = f ’(t) · t ’ = (t 0.5)' t' = 0.5 t−0.5 t ’.

We make a reverse substitution: t = x 2 + 8x− 7. We have:

f ’(x) = 0.5 ( x 2 + 8x− 7) −0.5 ( x 2 + 8x− 7)' = 0.5 (2 x+ 8) ( x 2 + 8x − 7) −0,5 .

Finally, back to the roots: