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 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 at a 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 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 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 a 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:
$$ f"_x(g(x)) = f"_g \cdot g"_x $$
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 content of the article
DERIVATIVE– derivative of the function y = f(x), given on a certain interval ( a, b) at point x of this interval is called the limit to which the ratio of the increment of the function tends f at this point to the corresponding increment of the argument when the increment of the argument tends to zero.
The derivative is usually denoted as follows:
Other designations 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 there 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 found herself in a position M 1 - on distance s+D s from the initial position ( see pic.).
Thus, over a period of time D t distance s changed by the amount D s. In this case they 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 movement of a point M at a point in 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 the average speed will not be able to reflect the indicated features of the point’s movement 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 shorter period of time D t. 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 current speed:
Thus, the speed of movement at a given moment is called the limit of the path increment ratio D s to time increment D t, when the time increment tends to zero. Because
Geometric meaning of the derivative. Tangent to the graph of a function.
The construction of tangent lines is one of those problems that led to the birth of differential calculus. The first published work related to differential calculus, written by Leibniz, was entitled New method maxima and minima, as well as tangents, for which neither fractional nor irrational quantities, and a special type of calculus for this, serve as an obstacle.
Let the curve be the graph of the function y =f(x) in a rectangular coordinate system ( cm. rice.).
At some value x function matters y =f(x). These values x And y the point on the curve corresponds 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 function value 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 you draw a secant M 0M 1 and denoted by j the angle formed by a transversal with the positive direction of the axis Ox, it is immediately clear 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 D x. At Dx® 0 the angle j tends to a certain limit a and the straight line passing through the point M 0 and the component with the positive direction of the x-axis, angle a, will be the desired tangent. Its slope is:
Hence, f´( x) = tga
those. derivative value f´( x) for a given argument value 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 having 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, the function cannot have a derivative at discontinuity points. The opposite conclusion is incorrect, i.e. from the fact that at some point x = x 0 function y = f(x) is continuous does not mean that it is differentiable at this point. For example, the function y = |x| continuous for everyone x(–Ґ x x = 0 has no derivative. At this point there is no tangent to the graph. There is a right tangent and a left one, but they do not coincide.
Some theorems on differentiable functions. Theorem on the roots of the derivative (Rolle's theorem). If the function f(x) is continuous on the segment [a,b], is differentiable at all interior points of this segment and at the ends x = a And x = b goes to zero ( 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) goes to zero, i.e. fў( c) = 0.
Finite increment theorem (Lagrange's theorem). If the function f(x) is continuous on the interval [ 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 the increments of two functions (Cauchy's theorem). If f(x) And g(x) – 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 differentiating this function, we obtain the so-called second derivative of the function f(x), which is denoted f ўў ( x).
Derivative n- th order of function f(x) is called the (first order) derivative 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– independent variable, yes dy = f ў( x)dx, some function from x, but from x only the first factor can depend f ў( x), 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 differential or second-order differential of this function and is denoted d 2y:
d(dx) = d 2y = f ўў( x)(dx) 2 .
Differential n- of the first order is called the first differential of the differential n- 1- th order:
d n y = d(d n–1y) = f(n)(x)dx(n).
Partial derivative.
If a function depends not on one, but on several arguments x i(i varies from 1 to n,i= 1, 2,… n),f(x 1,x 2,… x n), then in differential calculus the concept of 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. 1st order partial derivative with respect to x i is defined as an ordinary derivative, and it is assumed that all arguments except x i, keep constant values. For partial derivatives, the notation is introduced
The 1st order partial derivatives defined in this way (as functions of the same arguments) can, in turn, also have partial derivatives, these are second order partial derivatives, etc. Such derivatives taken from different arguments 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
Date: 11/20/2014
What is a derivative?
Table of derivatives.
Derivative is one of the main concepts higher mathematics. In this lesson we will introduce this concept. Let's get to know each other, without strict mathematical formulations and proofs.
This acquaintance will allow you to:
Understand the essence of simple tasks with derivatives;
Successfully solving these very problems difficult tasks;
Prepare for more serious lessons on derivatives.
First - a pleasant surprise.)
The strict definition of the derivative is based on the theory of limits and the thing is quite complicated. This is upsetting. But the practical application of derivatives, as a rule, does not require such extensive and deep knowledge!
To successfully complete most tasks at school and university, it is enough to know just a few terms- to understand the task, and just a few rules- to solve it. That's all. This makes me happy.
Let's start getting acquainted?)
Terms and designations.
There are many different mathematical operations in elementary mathematics. Addition, subtraction, multiplication, exponentiation, logarithm, etc. If you add one more operation to these operations, elementary mathematics becomes higher. This new operation called differentiation. The definition and meaning of this operation will be discussed in separate lessons.
It is important to understand here that differentiation is simply a mathematical operation on a function. We take any function and, according to certain rules, transform it. The result will be a new function. This new function is called: derivative.
Differentiation- action on a function.
Derivative- the result of this action.
Just like, for example, sum- the result of addition. Or private- the result of division.
Knowing the terms, you can at least understand the tasks.) The formulations are as follows: find the derivative of a function; take the derivative; differentiate the function; calculate derivative and so on. This is all same. Of course, there are also more complex tasks, where finding the derivative (differentiation) will be just one of the steps in solving the problem.
The derivative is indicated by a dash at the top right of the function. Like this: y" or f"(x) or S"(t) and so on.
Reading igrek stroke, ef stroke from x, es stroke from te, well, you understand...)
A prime can also indicate the derivative of a particular function, for example: (2x+3)", (x 3 )" , (sinx)" etc. Often derivatives are denoted using differentials, but we will not consider such notation in this lesson.
Let's assume that we have learned to understand the tasks. All that’s left is to learn how to solve them.) Let me remind you once again: finding the derivative is transformation of a function according to certain rules. Surprisingly, there are very few of these rules.
To find the derivative of a function, you need to know only three things. Three pillars on which all differentiation stands. Here they are these three pillars:
1. Table of derivatives (differentiation formulas).
3. Derivative of a complex function.
Let's start in order. In this lesson we will look at the table of derivatives.
Table of derivatives.
There are an infinite number of functions in the world. Among this variety, there are functions that are most important for practical application. These functions are found in all laws of nature. From these functions, like from bricks, you can construct all the others. This class of functions is called elementary functions. It is these functions that are studied at school - linear, quadratic, hyperbola, etc.
Differentiation of functions "from scratch", i.e. Based on the definition of derivative and the theory of limits, this is a rather labor-intensive thing. And mathematicians are people too, yes, yes!) So they simplified their (and us) life. They calculated the derivatives of elementary functions before us. The result is a table of derivatives, where everything is ready.)
Here it is, this plate for the most popular functions. On the left is an elementary function, on the right is its derivative.
| Function y |
Derivative of function y y" |
|
| 1 | C (constant value) | C" = 0 |
| 2 | x | x" = 1 |
| 3 | x n (n - any number) | (x n)" = nx n-1 |
| x 2 (n = 2) | (x 2)" = 2x | |
 |
||
 |
 |
|
| 4 | sin x | (sin x)" = cosx |
| cos x | (cos x)" = - sin x | |
| tg x |  |
|
| ctg x |  |
|
| 5 | arcsin x |  |
| arccos x |  |
|
| arctan x |  |
|
| arcctg x |  |
|
| 4 | a x |  |
| e x | ||
| 5 | log a x |  |
| ln x ( a = e) |
I recommend paying attention to the third group of functions in this table of derivatives. Derivative power function- one of the most common formulas, if not the most common! Do you get the hint?) Yes, it is advisable to know the table of derivatives by heart. By the way, this is not as difficult as it might seem. Try to solve more examples, the table itself will be remembered!)
Finding the table value of the derivative, as you understand, is not the most difficult task. Therefore, very often in such tasks there are additional chips. Either in the wording of the task, or in the original function, which doesn’t seem to be in the table...
Let's look at a few examples:
1. Find the derivative of the function y = x 3
There is no such function in the table. But there is a derivative of the power function in general view(third group). In our case n=3. So we substitute three instead of n and carefully write down the result:
(x 3) " = 3 x 3-1 = 3x 2
That's it.
Answer: y" = 3x 2
2. Find the value of the derivative of the function y = sinx at the point x = 0.
This task means that you must first find the derivative of the sine, and then substitute the value x = 0 into this very derivative. Exactly in that order! Otherwise, it happens that they immediately substitute zero into the original function... We are asked to find not the value of the original function, but the value its derivative. The derivative, let me remind you, is a new function.
Using the tablet we find the sine and the corresponding derivative:
y" = (sin x)" = cosx
We substitute zero into the derivative:
y"(0) = cos 0 = 1
This will be the answer.
3. Differentiate the function:
What, does it inspire?) There is no such function in the table of derivatives.
Let me remind you that to differentiate a function is simply to find the derivative of this function. If you forget elementary trigonometry, looking for the derivative of our function is quite troublesome. The table doesn't help...
But if we see that our function is double angle cosine, then everything gets better right away!
Yes Yes! Remember that transforming the original function before differentiation quite acceptable! And it happens to make life a lot easier. Using the double angle cosine formula:
Those. our tricky function is nothing more than y = cosx. And this is a table function. We immediately get:
Answer: y" = - sin x.
Example for advanced graduates and students:
4. Find the derivative of the function:
There is no such function in the derivatives table, of course. But if you remember elementary mathematics, operations with powers... Then it is quite possible to simplify this function. Like this:
And x to the power of one tenth is already a tabular function! Third group, n=1/10. We write directly according to the formula:
That's all. This will be the answer.
I hope that everything is clear with the first pillar of differentiation - the table of derivatives. It remains to deal with the two remaining whales. In the next lesson we will learn the rules of differentiation.
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. These are relatively simple expressions, the derivatives of which have long been calculated and tabulated. 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 differentiable with respect 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, huh? 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 at 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) - this is a 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 we get an 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 a final 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 be 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.
Task. Find the derivative of the 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 replacement: let x 2 + 8x − 7 = t. We find the derivative using the formula:
f ’(x) = f ’(t) · t ’ = (t 0.5)’ · t’ = 0.5 · t−0.5 · t ’.
Let's do the reverse replacement: 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:
Solving physical problems or examples in mathematics is completely impossible without knowledge of the derivative and methods for calculating it. Derivative is one of the most important concepts 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 derivative
Let there be a function f(x) , specified in a certain interval (a, b) . Points x and x0 belong to this interval. When x changes, the function itself changes. Changing the argument - the difference in its values x-x0 . This difference is written as delta x and is called argument increment. A change or increment of a function is the difference between the values of a function at two points. Definition of derivative:
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's the point of finding such a limit? And here's what it is:
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.
Physical meaning derivative: the derivative of the path with respect to time is equal to the speed of rectilinear motion.
Indeed, since school days everyone knows that speed is a particular path x=f(t) and time t . Average speed over a certain period of time:
To find out the speed of movement at a moment in time t0 you need to calculate the limit:
Rule one: set a constant
The constant can be taken out of the derivative sign. Moreover, this must be done. When solving examples in mathematics, take it as a rule - If you can simplify an expression, be sure to simplify it .
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 the function:
Rule three: 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: 
It is important to talk about calculating derivatives of complex functions here. The derivative of a complex function is equal to the product of the derivative of this function with respect to the intermediate argument and the derivative of the intermediate argument with respect to the independent variable.
In the above example we come across the expression:
IN in this case the intermediate argument is 8x to the fifth power. In order to calculate the derivative of such an expression, we first calculate 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: derivative of the quotient of two functions
Formula for determining the derivative of the quotient of two functions:
We tried to talk about derivatives for dummies from scratch. This topic is not as simple as it seems, so be warned: there are often pitfalls in the examples, so be careful when calculating derivatives.
With any questions on this and other topics, you can contact the student service. In a short time, we will help you solve the most difficult test and understand the tasks, even if you have never done derivative calculations before.
