The operation of finding a 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 and precisely defined rules of differentiation appeared. Isaac Newton (1643-1727) and Gottfried Wilhelm Leibniz (1646-1716) were the first to work in the field of finding derivatives.
Therefore, in our time, in order to find the derivative of any function, it is not necessary to calculate the above-mentioned limit of the ratio of the increment of the function to the increment of the argument, but 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 stroke sign break down simple functions and determine what actions (product, sum, quotient) these functions are related. Further, we find the derivatives of elementary functions in the table of derivatives, and the formulas for the derivatives of the product, sum and quotient - in the rules of differentiation. The table of derivatives 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 the 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 the sine is cosine. We substitute these values in 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 the sum, in which the second term with a constant factor can be taken out of the sign of the derivative:
If there are still questions about where something comes from, they, as a rule, become clear after reading the table of derivatives and the simplest rules of differentiation. We are going 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 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 | |
| 3. Derivative of degree. When solving problems, you need to convert non-square roots to a power. | |
| 4. Derivative of a variable to the power of -1 | |
| 5. Derivative of the square root | |
| 6. Sine derivative | |
| 7. Cosine derivative |  |
| 8. Tangent derivative |  |
| 9. Derivative of cotangent |  |
| 10. Derivative of the arcsine |  |
| 11. Derivative of arc cosine |  |
| 12. Derivative of arc tangent |  |
| 13. Derivative of the inverse tangent |  |
| 14. Derivative of natural logarithm | |
| 15. Derivative of a logarithmic function |  |
| 16. Derivative of the exponent | |
| 17. Derivative of exponential function |
Differentiation rules
| 1. Derivative of the sum or difference |  |
| 2. Derivative of a product |  |
| 2a. Derivative of an expression multiplied by a constant factor | |
| 3. Derivative of the quotient |  |
| 4. Derivative of a complex function |  |
Rule 1If functions
are differentiable at some point , then at the same point the functions
and
those. the derivative of the algebraic sum of functions is equal to the algebraic sum of the derivatives of these functions.
Consequence. If two differentiable functions differ by a constant, then their derivatives are, i.e.
Rule 2If functions
are differentiable at some point, then their product is also 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.
Consequence 1. The constant factor can be taken out of the sign of the derivative:
Consequence 2. The derivative of the product of several differentiable functions is equal to the sum of the products of the derivative of each of the factors and all the others.
For example, for three multipliers:
Rule 3If functions
differentiable at some point and , then at this point their quotient is also differentiable.u/v , and
those. the derivative of a quotient of two functions is equal to a fraction whose numerator 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 on other pages
When finding the derivative of the product and the quotient in real problems, it is always necessary to apply several differentiation rules at once, so more examples on these derivatives are in the article."The derivative of a product and a quotient".
Comment. You should not confuse a constant (that is, a number) as a term in the sum and as a constant factor! In the case of a term, its derivative is equal to zero, and in the case of a constant factor, it is taken out of the sign of the derivatives. This is a typical mistake that occurs at the initial stage of studying derivatives, but as the average student solves several one-two-component examples, this mistake no longer makes.
And if, when differentiating a product or a quotient, you have a term u"v, wherein 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 (such a case is analyzed in example 10).
Another common mistake is the mechanical solution of the derivative of a complex function as the derivative of a simple function. That's why derivative of a complex function devoted to a separate article. But first we will learn to find derivatives of simple functions.
Along the way, you can not do without transformations of expressions. To do this, you may need to open in new windows manuals Actions with powers and roots and Actions with fractions .
If you are looking for solutions to derivatives with powers and roots, that is, when the function looks like 
, then follow the lesson " Derivative of the sum of fractions with powers and roots".
If you have a task like 
, then you are in 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 determine the parts of the function expression: the entire expression represents the 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 and 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 with 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 - 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 the following values of 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 a quotient: the derivative of a quotient of two functions is equal to a fraction whose numerator 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's 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 such problems in which you need to find the derivative of a function, where there is a continuous pile of roots and degrees, such as, for example, 
then welcome to class "The derivative of the sum of fractions with powers and roots" .
If you need to learn more about the derivatives of sines, cosines, tangents and other trigonometric functions, that is, when the function looks like , then you have a lesson "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, with the derivative of which we familiarized ourselves in the table of derivatives. According to the product differentiation rule and the tabular value of the derivative of the square root, we get:
Example 6 Find the derivative of a function
Solution. In this function, we see the quotient, the dividend of which is the square root of the independent variable. According to the rule of differentiation of the quotient, which we repeated and applied in example 4, and the tabular value of the derivative of the square root, we get:
To get rid of the fraction in the numerator, multiply the numerator and denominator by .
Hello dear readers. After reading the article, you will probably have a logical question: “Why, in fact, is this necessary?”. Because of this, I first consider it necessary to inform in advance that the desired method for solving quadratic equations is presented more from the moral and aesthetic side of mathematics than from the side of practical dry application. I also apologize in advance to those readers who find my amateurish sayings unacceptable. So, let's start hammering nails with a microscope.
We have an algebraic equation of the second degree (it is also quadratic) in general form:
Let's move from a quadratic equation to a quadratic function:
Where, obviously, it is necessary to find such values of the function argument in which it would return zero.
It seems to just solve the quadratic equation using Vieta's theorem, or through the discriminant. But that's not what we're here for. Let's take the derivative! 
Based on the definition of the physical meaning of the first-order derivative, it is clear that by substituting the argument into the function obtained above, we (in particular) get speed function changes at the point given by this argument.
This time we got the "rate of speed" of the function change (i.e. acceleration) at a particular point. After analyzing the result a little, we can conclude that the "acceleration" is a constant that does not depend on the function argument - remember this.
Now let's remember a little physics and uniformly accelerated motion (RUD). What do we have in our arsenal? That's right, there is a formula for determining the coordinate of movement along the axis during the desired movement:
Where - time, - initial speed, - acceleration.
It is easy to see that our original function is just an RUD.
Isn't the displacement formula for throttles a consequence of solving a quadratic equation?
No. The formula for the throttle above is in fact the result of taking the integral of the speed formula for the PORD. Or from the graph you can find the area of \u200b\u200bthe figure. There will come out a trapezoid.
The displacement formula for throttle does not follow from the solution of any quadratic equations. This is very important, otherwise there would be no point in the article.
Now it remains to figure out what is what, and what we are missing.
We already have "acceleration" - it is the second order derivative, derived above. But to get the initial speed , we need to take, in general, any (let's denote it as ) and substitute it into the derivative of now the first order - because it will be the desired one.
In this case, the question arises, which one should be taken? Obviously, such that the initial speed is equal to zero, so that the formula for "displacement at the throttle" becomes:
In this case, we make an equation for the search:
[substituted in the derivative of the first order]
The root of such an equation with respect to will be:
And the value of the original function with such an argument will be: 
Now it becomes obvious that:
Putting all the pieces of the puzzle together:
Here we have the final solution to the problem. In general, we did not discover America - we simply came to the formula for solving a quadratic equation through the discriminant in a roundabout way. This does not carry practical meaning (in approximately the same way, equations of the first / second degree of any (not necessarily general) form can be solved).
The purpose of this article is, in particular, to stir up interest in the analysis of mat. functions and mathematics in general.
Peter was with you, thank you for your attention!
In this lesson, we will learn how to apply formulas and rules of differentiation.
Examples. Find derivatives of functions.
1. y=x 7 +x 5 -x 4 +x 3 -x 2 +x-9. Applying the Rule I, formulas 4, 2 and 1. We get:
y'=7x 6 +5x 4 -4x 3 +3x 2 -2x+1.
2. y=3x6 -2x+5. We solve similarly, using the same formulas and the formula 3.
y’=3∙6x 5 -2=18x 5 -2.
Applying the Rule I, formulas 3, 5
and 6
and 1.
Applying the Rule IV, formulas 5 and 1 .
In the fifth example, according to the rule I the derivative of the sum is equal to the sum of the derivatives, and we just found the derivative of the 1st term (example 4 ), therefore, we will find derivatives 2nd and 3rd terms, and for 1st term, we can immediately write the result.
Differentiating 2nd and 3rd terms according to the formula 4
. To do this, we transform the roots of the third and fourth degrees in denominators to powers with negative exponents, and then, according to 4
formula, we find the derivatives of the powers.
Look at this example and the result. Did you catch the pattern? Good. This means that we have a new formula and can add it to our derivatives table.
Let's solve the sixth example and derive one more formula.
We use the rule IV and formula 4
. We reduce the resulting fractions.
We look at this function and its derivative. You, of course, understood the pattern and are ready to name the formula:
Learning new formulas!
Examples.
1. Find argument increment and function increment y= x2 if the initial value of the argument was 4 , and the new 4,01 .
Solution.
New argument value x \u003d x 0 + Δx. Substitute the data: 4.01=4+Δx, hence the increment of the argument Δх=4.01-4=0.01. The increment of a function, by definition, is equal to the difference between the new and previous values of the function, i.e. Δy \u003d f (x 0 + Δx) - f (x 0). Since we have a function y=x2, then Δу\u003d (x 0 + Δx) 2 - (x 0) 2 \u003d (x 0) 2 + 2x 0 · Δx+(Δx) 2 - (x 0) 2 \u003d 2x 0 · ∆x+(∆x) 2 =
2 · 4 · 0,01+(0,01) 2 =0,08+0,0001=0,0801.
Answer: argument increment Δх=0.01; function increment Δу=0,0801.
It was possible to find the function increment in another way: Δy\u003d y (x 0 + Δx) -y (x 0) \u003d y (4.01) -y (4) \u003d 4.01 2 -4 2 \u003d 16.0801-16 \u003d 0.0801.
2. Find the angle of inclination of the tangent to the function graph y=f(x) at the point x 0, if f "(x 0) \u003d 1.
Solution.
The value of the derivative at the point of contact x 0 and is the value of the tangent of the slope of the tangent (the geometric meaning of the derivative). We have: f "(x 0) \u003d tgα \u003d 1 → α \u003d 45 °, because tg45°=1.
Answer: the tangent to the graph of this function forms an angle with the positive direction of the Ox axis, equal to 45°.
3. Derive the formula for the derivative of a function y=xn.
Differentiation is the act of finding the derivative of a function.
When finding derivatives, formulas are used that were derived on the basis of the definition of the derivative, in the same way as we derived the formula for the derivative degree: (x n)" = nx n-1.
Here are the formulas.
Derivative table it will be easier to memorize by pronouncing verbal formulations:
1. The derivative of a constant value is zero.
2. X stroke is equal to one.
3. The constant factor can be taken out of the sign of the derivative.
4. The derivative of a degree is equal to the product of the exponent of this degree by the degree with the same base, but the exponent is one less.
5. The derivative of the root is equal to one divided by two of the same roots.
6. The derivative of unity divided by x is minus one divided by x squared.
7. The derivative of the sine is equal to the cosine.
8. The derivative of cosine is equal to minus sine.
9. The derivative of the tangent is equal to one divided by the square of the cosine.
10. The derivative of the cotangent is minus one divided by the square of the sine.
We teach differentiation rules.
1.
The derivative of the algebraic sum is equal to the algebraic sum of the derivative terms.
2. The derivative of the product is equal to the product of the derivative of the first factor by the second plus the product of the first factor by the derivative of the second.
3. The derivative of “y” divided by “ve” is equal to a fraction, in the numerator of which “y is a stroke multiplied by “ve” minus “y, multiplied by a stroke”, and in the denominator - “ve squared”.
4. A special case of the formula 3.
Let's learn together!
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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).
The geometric meaning of the 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 at a 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 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 is a constant number and f=f(x), g=g(x) are some differentiable functions, then the following are true differentiation rules:
$$ f"_x(g(x)) = f"_g \cdot g"_x $$
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) $ $It is absolutely impossible to solve physical problems or examples in mathematics without knowledge about the derivative and methods for calculating it. The derivative is one of the most important concepts of mathematical analysis. We decided to devote today's article to this fundamental topic. What is a derivative, what is its physical and geometric meaning, how to calculate the derivative of a function? All these questions can be combined into one: how to understand the derivative?
Geometric and physical meaning of the derivative
Let there be a function f(x) , given in some interval (a,b) . The points x and x0 belong to this interval. When x changes, the function itself changes. Argument change - difference of its values x-x0 . This difference is written as delta x and is called argument increment. The change or increment of a function is the difference between the values of the function at two points. Derivative definition:
The derivative of a function at a point is the limit of the ratio of the increment of the function at a given point to the increment of the argument when the latter tends to zero.
Otherwise it can be written like this:
What is the point in finding such a limit? But which one:
the derivative of a function at a point is equal to the tangent of the angle between the OX axis and the tangent to the graph of the function at a given point.
The physical meaning of the derivative: the time derivative of the path is equal to the speed of the rectilinear motion.
Indeed, since school days, everyone knows that speed is a private path. x=f(t) and time t . Average speed over a certain period of time:
To find out the speed of movement at a time t0 you need to calculate the limit:
Rule one: take out the constant
The constant can be taken out of the sign of the derivative. Moreover, it must be done. When solving examples in mathematics, take as a rule - if you can simplify the expression, be sure to simplify .
Example. Let's calculate the derivative:
Rule two: derivative of the sum of functions
The derivative of the sum of two functions is equal to the sum of the derivatives of these functions. The same is true for the derivative of the difference of functions.
We will not give a proof of this theorem, but rather consider a practical example.
Find the derivative of a function:
Rule three: the derivative of the product of functions
The derivative of the product of two differentiable functions is calculated by the formula:
Example: find the derivative of a function:
Solution: 
Here it is important to say about the calculation of derivatives of complex functions. The derivative of a complex function is equal to the product of the derivative of this function with respect to the intermediate argument by the derivative of the intermediate argument with respect to the independent variable.
In the above example, we encounter the expression:
In this case, the intermediate argument is 8x to the fifth power. In order to calculate the derivative of such an expression, we first consider the derivative of the external function with respect to the intermediate argument, and then multiply by the derivative of the intermediate argument itself with respect to the independent variable.
Rule Four: The derivative of the quotient of two functions
Formula for determining the derivative of a quotient of two functions:
We tried to talk about derivatives for dummies from scratch. This topic is not as simple as it sounds, so be warned: there are often pitfalls in the examples, so be careful when calculating derivatives.
With any question on this and other topics, you can contact the student service. In a short time, we will help you solve the most difficult control and deal with tasks, even if you have never dealt with the calculation of derivatives before.
