Find the derivative of the function x 2 3. Derivative of the sum and difference

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. Next, 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 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 familiarization 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 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.

Definition. Let the function $y = f(x)$ be defined in a certain interval containing the point $x_0$. 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 a new function, but naturally related to 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. Slope coefficient such a line does not have, which means that $f"(0) $ does not exist either

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) $ $

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

Everything seems to be clear. But try using this formula to calculate, 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 from the entire variety of functions we can distinguish the so-called elementary functions. It's relative simple expressions, the derivatives of which have long been calculated and listed in the table. Such functions are quite easy to remember - along with their derivatives.

Derivatives of elementary functions

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

So, derivatives of elementary functions:

Name	Function	Derivative
Constant	f(x) = C, C ∈ R	0 (yes, zero!)
Power 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/sin 2 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 particularly elementary, but also differentiated according to certain rules. These rules are discussed below.

Derivative of sum and difference

Let the functions be given f(x) And g(x), the derivatives of which 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:

(f + g)’ = f ’ + g ’
(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 of “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 + sin x; g(x) = x 4 + 2x 2 − 3.

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

f ’(x) = (x 2 + sin x)’ = (x 2)’ + (sin x)’ = 2x+ cos x;

We reason 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+ cos x;
g ’(x) = 4x · ( x 2 + 1).

Derivative of the product

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

(f · g) ’ = f ’ · g + f · g ’

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

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

Function f(x) is the 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 factor is a little more complicated, but general scheme this doesn't change. Obviously, the first factor 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 .

Please note that in the last step the derivative is factorized. Formally, this does not need to be done, but most derivatives are not calculated on their own, but to examine the function. This means that further the derivative will be equated to zero, its signs will be determined, and so on. For such a case, it is better to have an expression factorized.

If there are two functions f(x) And g(x), and g(x) ≠ 0 on the set we are interested in, we can define new feature 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 specific examples.

Task. Find derivatives of functions:

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

According to tradition, let's factorize the numerator - this will greatly simplify the answer:

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

What should I do? In such cases, replacing a variable and formula for the derivative of a complex function helps:

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

As a rule, the situation with understanding 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 out elementary function f(x) = e x. Therefore, we make a replacement: let 2 x + 3 = t, f(x) = f(t) = e t. We look for the derivative of a complex function using the formula:

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

And now - attention! We perform the reverse replacement: 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 it 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 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 “prime.” For example, a prime from the amount equal to the sum strokes. Is that clearer? Well, that's good.

Thus, calculating the derivative comes down to getting rid of these same 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 people know that in the role n may well perform a fractional number. For example, the root is x 0.5. What if there is something fancy under the root? Again, the result will be a complex function - they like to give such constructions to tests and exams.