Derivative f x 1 2. The derivative of the difference is equal to the difference of the derivatives

Application

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Definition. Let the function $y = f(x) $ be defined in a certain interval containing the point $x_0$ within itself. Let's give the argument an increment $\Delta x $ such that it does not leave this interval. Let's find the corresponding increment of the function $\Delta y $ (when moving 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 to this ratio at $\Delta x \rightarrow 0$, then the specified limit is called derivative of a 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 new feature, 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 = f(x).

Geometric meaning of derivative is as follows. If it is possible to draw a tangent to the graph of the function y = f(x) at the point with abscissa x=a, which is not parallel to the y-axis, then f(a) expresses the slope of the tangent:
$k = f"(a)$

Since $k = tg(a) $, then the equality $f"(a) = tan(a) $ is true.

Now let’s interpret the definition of derivative from the point of view of approximate equalities. Let the function $y = f(x)$ have a derivative in specific 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. $\Delta y \approx f"(x) \cdot\Delta x$. The meaningful meaning of the resulting 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 valid. If we carefully analyze the definition of a 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 = f(x)?

1. Fix the value of $x$, find $f(x)$
2. Give the argument $x$ an increment $\Delta x$, go to new point$x+ \Delta x $, find $f(x+ \Delta x) $
3. Find the increment of the function: $\Delta y = f(x + \Delta x) - f(x) $
4. Create 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 point x.

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

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

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 point M(x; f(x)), and, recall, the angular coefficient of the tangent is equal to f "(x). Such a graph cannot “break” at point M, i.e. the function must be continuous at point x.

These were “hands-on” arguments. Let us give a more rigorous reasoning. 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. If in this equality $\Delta x $ tends to zero, then $\Delta y $ will 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 continuous at that point.

The reverse statement 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 “junction point” (0; 0) does not exist. If at some point a tangent cannot be drawn to the graph of a function, then the derivative does not exist at that 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, i.e., it is perpendicular to the abscissa axis, its equation has the form x = 0. Such a straight line does not have an angle coefficient, which means that $f"(0)$ does not exist.

So, we got acquainted with a new property of a function - differentiability. How can one conclude from the graph of a function that it is differentiable?

The answer is actually given above. If at some point it is possible to draw a tangent to the graph of a function that is not perpendicular to the abscissa axis, then at this point the function is differentiable. If at some point the tangent to the graph of a function does not exist or it is perpendicular to the abscissa axis, then at this point the function is not differentiable.

Rules of differentiation

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 “functions of functions,” that is, complex functions. Based on the definition of derivative, we can derive differentiation rules that make this work easier. If C - constant number and f=f(x), g=g(x) are some differentiable functions, then the following are true differentiation rules:

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

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(arcctg) x)" = \frac(-1)(1+x^2) $ $

The operation of finding the derivative is called differentiation.

As a result of solving problems of finding derivatives of the simplest (and not very simple) functions by defining the derivative as the limit of the ratio of the increment to the increment of the argument, a table of derivatives appeared and exactly certain rules differentiation. The first to work in the field of finding derivatives were Isaac Newton (1643-1727) and Gottfried Wilhelm Leibniz (1646-1716).

Therefore, in our time, to find the derivative of any function, you do not need to calculate the above-mentioned limit of the ratio of the increment of the function to the increment of the argument, but you only need to use the table of derivatives and the rules of differentiation. The following algorithm is suitable for finding the derivative.

To find the derivative, you need an expression under the prime sign break down simple functions into components and determine what actions (product, sum, quotient) these functions are related. Further derivatives elementary functions we find in the table of derivatives, and the formulas for the derivatives of the product, sum and quotient are in the rules of differentiation. The derivative table and differentiation rules are given after the first two examples.

Example 1. Find the derivative of a function

Solution. From the rules of differentiation we find out that the derivative of a sum of functions is the sum of derivatives of functions, i.e.

From the table of derivatives we find out that the derivative of "x" is equal to one, and the derivative of sine is equal to cosine. We substitute these values into the sum of derivatives and find the derivative required by the condition of the problem:

Example 2. Find the derivative of a function

Solution. We differentiate as a derivative of a sum in which the second term has a constant factor; it can be taken out of the sign of the derivative:

If questions still arise about where something comes from, they are usually cleared up after familiarizing yourself with the table of derivatives and the simplest rules of differentiation. We are moving on to them right now.

Table of derivatives of simple functions

1. Derivative of a constant (number). Any number (1, 2, 5, 200...) that is in the function expression. Always equal to zero. This is very important to remember, as it is required very often
2. Derivative of the independent variable. Most often "X". Always equal to one. This is also important to remember for a long time
3. Derivative of degree. When solving problems, you need to convert non-square roots into powers.
4. Derivative of a variable to the power -1
5. Derivative of square root
6. Derivative of sine
7. Derivative of cosine
8. Derivative of tangent
9. Derivative of cotangent
10. Derivative of arcsine
11. Derivative of arccosine
12. Derivative of arctangent
13. Derivative of arc cotangent
14. Derivative of the natural logarithm
15. Derivative of a logarithmic function
16. Derivative of the exponent
17. Derivative of an exponential function

Rules of differentiation

1. Derivative of a sum or difference
2. Derivative of the product
2a. Derivative of an expression multiplied by a constant factor
3. Derivative of the quotient
4. Derivative of a complex function

Rule 1.If the functions

are differentiable at some point, then the functions are differentiable at the same point

and

those. the derivative of the algebraic sum of functions is equal to algebraic sum derivatives of these functions.

Consequence. If two differentiable functions differ by a constant term, then their derivatives are equal, i.e.

Rule 2.If the functions

are differentiable at some point, then their product is differentiable at the same point

and

those. The derivative of the product of two functions is equal to the sum of the products of each of these functions and the derivative of the other.

Corollary 1. The constant factor can be taken out of the sign of the derivative:

Corollary 2. The derivative of the product of several differentiable functions is equal to the sum of the products of the derivative of each factor and all the others.

For example, for three multipliers:

Rule 3.If the functions

differentiable at some point And , then at this point their quotient is also differentiableu/v , and

those. the derivative of the quotient of two functions is equal to a fraction, the numerator of which is the difference between the products of the denominator and the derivative of the numerator and the numerator and the derivative of the denominator, and the denominator is the square of the former numerator.

Where to look for things on other pages

When finding the derivative of a product and the quotient in real problems It is always necessary to apply several differentiation rules at once, therefore more examples for these derivatives - in the article"Derivative of the product and quotient of functions".

Comment. You should not confuse a constant (that is, a number) as a term in a sum and as a constant factor! In the case of a term, its derivative is equal to zero, and in the case constant factor it is taken out of the derivative sign. This typical mistake, which occurs on initial stage studying derivatives, but as they solve several one- and two-part examples, the average student no longer makes this mistake.

And if, when differentiating a product or quotient, you have a term u"v, in which u- a number, for example, 2 or 5, that is, a constant, then the derivative of this number will be equal to zero and, therefore, the entire term will be equal to zero (this case is discussed in example 10).

Other common mistake - mechanical solution derivative of a complex function as a derivative of a simple function. That's why derivative of a complex function a separate article is devoted. But first we will learn to find derivatives simple functions.

Along the way, you can’t do without transforming expressions. To do this, you may need to open the manual in new windows. Actions with powers and roots And Operations with fractions .

If you are looking for solutions to derivatives of fractions with powers and roots, that is, when the function looks like , then follow the lesson “Derivative of sums of fractions with powers and roots.”

If you have a task like , then you will take the lesson “Derivatives of simple trigonometric functions”.

Step-by-step examples - how to find the derivative

Example 3. Find the derivative of a function

Solution. We define the parts of the function expression: the entire expression represents a product, and its factors are sums, in the second of which one of the terms contains a constant factor. We apply the product differentiation rule: the derivative of the product of two functions is equal to the sum of the products of each of these functions by the derivative of the other:

Next, we apply the rule of differentiation of the sum: the derivative of the algebraic sum of functions is equal to the algebraic sum of the derivatives of these functions. In our case, in each sum the second term has a minus sign. In each sum we see both an independent variable, the derivative of which is equal to one, and a constant (number), the derivative of which is equal to zero. So, “X” turns into one, and minus 5 turns into zero. In the second expression, "x" is multiplied by 2, so we multiply two by the same unit as the derivative of "x". We get following values derivatives:

We substitute the found derivatives into the sum of products and obtain the derivative of the entire function required by the condition of the problem:

Example 4. Find the derivative of a function

Solution. We are required to find the derivative of the quotient. We apply the formula for differentiating the quotient: the derivative of the quotient of two functions is equal to a fraction, the numerator of which is the difference between the products of the denominator and the derivative of the numerator and the numerator and the derivative of the denominator, and the denominator is the square of the former numerator. We get:

We have already found the derivative of the factors in the numerator in example 2. Let us also not forget that the product, which is the second factor in the numerator in the current example, is taken with a minus sign:

If you are looking for solutions to problems in which you need to find the derivative of a function, where there is a continuous pile of roots and powers, such as, for example, , then welcome to class "Derivative of sums of fractions with powers and roots" .

If you need to learn more about the derivatives of sines, cosines, tangents and others trigonometric functions, that is, when the function looks like , then a lesson for you "Derivatives of simple trigonometric functions" .

Example 5. Find the derivative of a function

Solution. In this function we see a product, one of the factors of which is the square root of the independent variable, the derivative of which we familiarized ourselves with in the table of derivatives. According to the rule of differentiation of the product and table value derivative of the square root we get:

Example 6. Find the derivative of a function

Solution. In this function we see a quotient whose dividend is the square root of the independent variable. Using the rule for differentiating quotients, which we repeated and applied in example 4, and the tabular value of the derivative of the square root, we obtain.

The problem of finding the derivative of given function is one of the main courses in mathematics high school and in higher education institutions. It is impossible to fully explore a function and construct its graph without taking its derivative. The derivative of a function can be easily found if you know the basic rules of differentiation, as well as the table of derivatives of basic functions. Let's figure out how to find the derivative of a function.

The derivative of a function is the limit of the ratio of the increment of the function to the increment of the argument when the increment of the argument tends to zero.

Understanding this definition is quite difficult, since the concept of a limit is not fully studied in school. But in order to find derivatives various functions, it is not necessary to understand the definition, let’s leave it to mathematicians and move straight to finding the derivative.

The process of finding the derivative is called differentiation. When we differentiate a function, we will obtain a new function.

To denote them we will use letters f, g, etc.

There are many different notations for derivatives. We will use a stroke. For example, writing g" means that we will find the derivative of the function g.

Derivatives table

In order to answer the question of how to find the derivative, it is necessary to provide a table of derivatives of the main functions. To calculate derivatives of elementary functions it is not necessary to perform complex calculations. It is enough just to look at its value in the table of derivatives.

(sin x)"=cos x
(cos x)"= –sin x
(x n)"=n x n-1
(e x)"=e x
(ln x)"=1/x
(a x)"=a x ln a
(log a x)"=1/x ln a
(tg x)"=1/cos 2 x
(ctg x)"= – 1/sin 2 x
(arcsin x)"= 1/√(1-x 2)
(arccos x)"= - 1/√(1-x 2)
(arctg x)"= 1/(1+x 2)
(arcctg x)"= - 1/(1+x 2)

Example 1. Find the derivative of the function y=500.

We see that this is a constant. From the table of derivatives it is known that the derivative of a constant is equal to zero (formula 1).

Example 2. Find the derivative of the function y=x 100.

This power function whose exponent is 100 and to find its derivative you need to multiply the function by the exponent and reduce it by 1 (formula 3).

(x 100)"=100 x 99

Example 3. Find the derivative of the function y=5 x

This exponential function, let's calculate its derivative using formula 4.

Example 4. Find the derivative of the function y= log 4 x

We find the derivative of the logarithm using formula 7.

(log 4 x)"=1/x ln 4

Rules of differentiation

Let's now figure out how to find the derivative of a function if it is not in the table. Most of the functions studied are not elementary, but are combinations of elementary functions using simple operations (addition, subtraction, multiplication, division, and multiplication by a number). To find their derivatives, you need to know the rules of differentiation. Below, the letters f and g denote functions, and C is a constant.

1. The constant coefficient can be taken out of the sign of the derivative

Example 5. Find the derivative of the function y= 6*x 8

We take it out constant coefficient 6 and differentiate only x 4 . This is a power function, the derivative of which is found using formula 3 of the table of derivatives.

(6*x 8)" = 6*(x 8)"=6*8*x 7 =48* x 7

2. The derivative of a sum is equal to the sum of the derivatives

(f + g)"=f" + g"

Example 6. Find the derivative of the function y= x 100 +sin x

A function is the sum of two functions, the derivatives of which we can find from the table. Since (x 100)"=100 x 99 and (sin x)"=cos x. The derivative of the sum will be equal to the sum of these derivatives:

(x 100 +sin x)"= 100 x 99 +cos x

3. The derivative of the difference is equal to the difference of the derivatives

(f – g)"=f" – g"

Example 7. Find the derivative of the function y= x 100 – cos x

This function is the difference of two functions, the derivatives of which we can also find from the table. Then the derivative of the difference is equal to the difference of the derivatives and don’t forget to change the sign, since (cos x)"= – sin x.

(x 100 – cos x)"= 100 x 99 + sin x

Example 8. Find the derivative of the function y=e x +tg x– x 2.

This function has both a sum and a difference; let’s find the derivatives of each term:

(e x)"=e x, (tg x)"=1/cos 2 x, (x 2)"=2 x. Then the derivative of the original function is equal to:

(e x +tg x– x 2)"= e x +1/cos 2 x –2 x

4. Derivative of the product

(f * g)"=f" * g + f * g"

Example 9. Find the derivative of the function y= cos x *e x

To do this, we first find the derivative of each factor (cos x)"=–sin x and (e x)"=e x. Now let's substitute everything into the product formula. We multiply the derivative of the first function by the second and add the product of the first function by the derivative of the second.

(cos x* e x)"= e x cos x – e x *sin x

5. Derivative of the quotient

(f / g)"= f" * g – f * g"/ g 2

Example 10. Find the derivative of the function y= x 50 /sin x

To find the derivative of a quotient, we first find the derivative of the numerator and denominator separately: (x 50)"=50 x 49 and (sin x)"= cos x. Substituting the derivative of the quotient into the formula, we get:

(x 50 /sin x)"= 50x 49 *sin x – x 50 *cos x/sin 2 x

Derivative of a complex function

A complex function is a function represented by a composition of several functions. There is also a rule for finding the derivative of a complex function:

(u (v))"=u"(v)*v"

Let's figure out how to find the derivative of such a function. Let y= u(v(x)) - complex function. Let's call the function u external, and v - internal.

For example:

y=sin (x 3) is a complex function.

Then y=sin(t) is the outer function

t=x 3 - internal.

Let's try to calculate the derivative of this function. According to the formula, you need to multiply the derivatives of the internal and external functions.

(sin t)"=cos (t) - derivative of the external function (where t=x 3)

(x 3)"=3x 2 - derivative of the internal function

Then (sin (x 3))"= cos (x 3)* 3x 2 is the derivative of a complex function.

Derivative calculation- one of the most important operations in differential calculus. Below is a table for finding derivatives of simple functions. More complex rules differentiation, see other lessons:

Table of derivatives of exponential and logarithmic functions

Use the given formulas as reference values. They will help you decide differential equations and tasks. In the picture, in the table of derivatives of simple functions, there is a “cheat sheet” of the main cases of finding a derivative in a form that is understandable for use, next to it are explanations for each case.

Derivatives of simple functions

1. The derivative of a number is zero
с´ = 0
Example:
5´ = 0

Explanation:
The derivative shows the rate at which the value of a function changes when its argument changes. Since the number does not change in any way under any conditions, the rate of its change is always zero.

2. Derivative of a variable equal to one
x´ = 1

Explanation:
With each increment of the argument (x) by one, the value of the function (the result of the calculation) increases by the same amount. Thus, the rate of change in the value of the function y = x is exactly equal to the rate of change in the value of the argument.

3. The derivative of a variable and a factor is equal to this factor
сx´ = с
Example:
(3x)´ = 3
(2x)´ = 2
Explanation:
IN in this case, each time the function argument changes ( X) its value (y) increases in With once. Thus, the rate of change of the function value in relation to the rate of change of the argument is exactly equal to the value With.

Whence it follows that
(cx + b)" = c
that is, the differential linear function y=kx+b is equal slope slope of the straight line (k).

4. Modulo derivative of a variable equal to the quotient of this variable to its modulus
|x|"= x / |x| provided that x ≠ 0
Explanation:
Since the derivative of a variable (see formula 2) is equal to unity, the derivative of the module differs only in that the value of the rate of change of the function changes to the opposite when crossing the point of origin (try drawing a graph of the function y = |x| and see for yourself. This is exactly what value and returns the expression x / |x|< 0 оно равно (-1), а когда x >0 - one. That is, when negative values variable x, with each increase in argument, the value of the function decreases by exactly the same value, and for positive ones, on the contrary, it increases, but by exactly the same value.

5. Derivative of a variable to a power equal to the product of a number of this power and a variable to the power reduced by one
(x c)"= cx c-1, provided that x c and cx c-1 are defined and c ≠ 0
Example:
(x 2)" = 2x
(x 3)" = 3x 2
To remember the formula:
Move the degree of the variable down as a factor, and then reduce the degree itself by one. For example, for x 2 - the two was ahead of the x, and then the reduced power (2-1 = 1) simply gave us 2x. The same thing happened for x 3 - we “move down” the triple, reduce it by one and instead of a cube we have a square, that is, 3x 2. A little "unscientific" but very easy to remember.

6.Derivative of a fraction 1/x
(1/x)" = - 1 / x 2
Example:
Since a fraction can be represented by raising it to negative degree
(1/x)" = (x -1)", then you can apply the formula from rule 5 of the table of derivatives
(x -1)" = -1x -2 = - 1 / x 2

7. Derivative of a fraction with a variable of arbitrary degree in the denominator
(1 / x c)" = - c / x c+1
Example:
(1 / x 2)" = - 2 / x 3

8. Derivative of the root(derivative of the variable under square root)
(√x)" = 1 / (2√x) or 1/2 x -1/2
Example:
(√x)" = (x 1/2)" means you can apply the formula from rule 5
(x 1/2)" = 1/2 x -1/2 = 1 / (2√x)

9. Derivative of a variable under the root of an arbitrary degree
(n √x)" = 1 / (n n √x n-1)