When the first formula itself is derived, we will proceed from the definition of derivatives in the point. Take where x. - any valid number, that is, x. - Any number from the function of determining the function. We write the limit of the relationship of the increment function to the increment of the argument at:

It should be noted that under the sign of the limit it turns out an expression that is not a single zero to divide on zero, since in the numerator there is not an infinitely small value, namely zero. In other words, the increment of a constant function is always zero.

In this way, derivative permanent functionequal to zero throughout the field of definition.

The derivative of the power function.

The formula of the derivative of the power function has the form where the indicator of the degree p. - Any valid number.

We first prove the formula for the natural indicator, that is, for p \u003d 1, 2, 3, ...

We will use the definition of the derivative. We write the limit of the ratio of the increment of the power function to the increment of the argument:

To simplify the expression in the numerator, we turn to the Newton Binoma formula:

Hence,

This proved the formula for the derivative of the power function for the natural indicator.

Derivative indicative function.

The derivative of the derivative formula is based on the definition:

Came to uncertainty. For its disclosure, we introduce a new variable, and with. Then. In the last transition, we used the transition formula to the new base of the logarithm.

Perform a substitution to the initial limit:

If we recall the second wonderful limit, then we will come to the formula of the derivative of the indicative function:

Derivative logarithmic function.

We prove the formula of the derivative logarithmic function for all x. from the definition area and all allowable base values a. Logarithm. By definition, we have:

As you noticed, when proofing the transformation was carried out using the properties of the logarithm. Equality fairly due to the second remarkable limit.

Derived trigonometric functions.

To display the formulas of derivative trigonometric functions, we will have to recall some formulas trigonometry, as well as the first wonderful limit.

By definition of the derivative for the sinus function we have .

We use the sinus difference formula:

It remains to contact the first wonderful limit:

Thus, derivative function sIN X. there is cOS X..

Absolutely similarly proved the formula of the cosine derivative.

Consequently, derived function cOS X. there is -Sin X..

The output of the formulas of the tables of derivatives for Tangent and Kotangens will carry out using proven differentiation rules (derivative of the fraction).

Derivatives of hyperbolic functions.

The rules of differentiation and the formula of the derivative indicative function from the derivative table allow us to derive the formula of derivatives of hyperbolic sinus, cosine, tangent and catangent.

Derived reverse function.

In order not to be confused when exposing, let's refer to the lower index the argument of the function to which differentiation is performed, that is, it is derived f (x) by x..

Now formulate the rule of finding a derivative of the feedback.

Let functions y \u003d f (x) and x \u003d g (y) Mutually reverse, determined at intervals and, accordingly. If at the point there is a finite different derivative function from zero f (x), then at the point there is a finite derivative of the feedback g (Y), and . In another record .

You can reformulate this rule for any x. From the gap, then we get .

Let's check the validity of these formulas.

Find a reverse function for a natural logarithm (here y. - function, and x.- argument). Resolving this equation relative x., get here (here x. - function, and y. - its argument). I.e, and mutually reverse functions.

From the table derivatives we see that and .

Correct that formulas for finding derivative feedback leads us to the same results:

We give a consolidated table for convenience and visibility when studying the topic.

We analyze how the formulas of the specified table were obtained or, in other words, we prove the output of the derivative formulas for each type of functions.

Derivative constant

In order to derive this formula, take as a basis the definition of the derivative function at the point. Using x 0 \u003d x, where X. takes the meaning of any actual number, or, in other words, X. It is in any number from the function of determining the function f (x) \u003d c. We will make a record of the limit of the relationship of the function of the function to the increment of the argument at Δ x → 0:

lim Δ x → 0 Δ f (x) Δ x \u003d Lim Δ x → 0 C - C Δ x \u003d Lim Δ x → 0 0 Δ x \u003d 0

Note that the expression 0 Δ x is entering the limit. It does not eat the uncertainty "zero to divide on zero", since the numerator does not contain an infinitely small value, namely zero. In other words, the increment of a constant function is always zero.

So, the derivative of the constant function f (x) \u003d c is zero on the entire definition area.

Example 1.

Permanent functions are given:

f 1 (x) \u003d 3, F 2 (x) \u003d a, a ∈ R, F 3 (x) \u003d 4. 13 7 22, F 4 (x) \u003d 0, F 5 (x) \u003d - 8 7

Decision

We describe the specified conditions. In the first function, we see a derivative of a natural number 3. In the following example, it is necessary to take a derivative from butwhere but - Any valid number. The third example sets us a derivative of the irrational number 4. 13 7 22, fourth - zero derivative (zero - integer). Finally, in the fifth case, we have a derivative of rational fraction - 8 7.

Answer: Derivative features are zero with any valid X. (throughout the field of definition)

f 1 "(x) \u003d (3)" \u003d 0, F 2 "(x) \u003d (a)" \u003d 0, a ∈ R, F 3 "(x) \u003d 4. 13 7 22" \u003d 0, F 4 "(x) \u003d 0" \u003d 0, f 5 "(x) \u003d - 8 7" \u003d 0

The derivative of the power function

We turn to the power function and the formula for its derivative, having a view: (x p) "\u003d p · x p - 1, where the indicator of the degree P. is any actual number.

Proof 2.

We present the proof of the formula, when the indicator of the degree - the natural number: P \u003d 1, 2, 3, ...

We rely on the definition of the derivative. We will record the limit of the ratio of the increment of the power function to the increment of the argument:

(x p) "\u003d Lim Δ x → 0 \u003d δ (x p) Δ x \u003d Lim Δ x → 0 (x + δ x) p - x p Δ x

To simplify the expression in the numerator, we use the Newton Binoma formula:

(x + δ x) p - x p \u003d c p 0 + x P + C p 1 · x p - 1 · Δ x + c p 2 · x p - 2 · (δ x) 2 +. . . + + C pp - 1 · x · (Δ x) p - 1 + c pp · (δ x) p - xp \u003d c p 1 · xp - 1 · Δ x + c p 2 · xp - 2 · (Δ x) 2 +. . . + C P P - 1 · X · (Δ x) p - 1 + C P p · (Δ x) p

In this way:

(xp) "\u003d Lim Δ x → 0 Δ (xp) Δ x \u003d Lim Δ x → 0 (x + Δ x) p - xp Δ x \u003d \u003d Lim Δ x → 0 (C p 1 · xp - 1 · Δ X + C P 2 · XP - 2 · (Δ x) 2 +... + C pp - 1 · x · (Δ x) p - 1 + C pp · (Δ x) p) Δ x \u003d \u003d Lim Δ X → 0 (C p 1 · Xp - 1 + C p 2 · Xp - 2 · Δ x +... + C pp - 1 · x · (Δ x) p - 2 + C pp · (Δ x) p - 1) \u003d \u003d c p 1 · xp - 1 + 0 + 0 +... + 0 \u003d p!1! · (P - 1)! · Xp - 1 \u003d p · xp - 1

So, we proved the formula for the derivative of the power function, when the indicator of the degree is a natural number.

Proof 3.

To bring proof for the case when P -any valid number, different from zero, use the logarithmic derivative (here should be understood the difference from the derivative of the logarithmic function). To have a more complete understanding, it is desirable to study the derivative of the logarithmic function and further understand the derivative of an implicitly specified function and a derivative of a complex function.

Consider two cases: when X. Positive and when X. Negative.

So, x\u003e 0. Then: x p\u003e 0. Logarithming Equality y \u003d x P based on E and applies the logarithm property:

y \u003d x p ln y \u003d ln x p ln y \u003d p · ln x

At this stage, an implicitly specified function was obtained. Determine its derivative:

(ln y) "\u003d (p · ln x) 1 y · y" \u003d p · 1 x ⇒ y "\u003d p · y x \u003d p · x p x \u003d p · x p - 1

Now we consider the case when x -a negative number.

If an indicator P. There is an even number, then the power function is determined and when x< 0 , причем является четной: y (x) = - y ((- x) p) " = - p · (- x) p - 1 · (- x) " = = p · (- x) p - 1 = p · x p - 1

Then X P.< 0 и возможно составить доказательство, используя логарифмическую производную.

If a P. There is an odd number, then the power function is determined and when x< 0 , причем является нечетной: y (x) = - y (- x) = - (- x) p . Тогда x p < 0 , а значит логарифмическую производную задействовать нельзя. В такой ситуации возможно взять за основу доказательства правила дифференцирования и правило нахождения производной сложной функции:

y "(x) \u003d (- (- x) p)" \u003d - ((- x) p) "\u003d - p · (- x) p - 1 · (- x)" \u003d \u003d p · (- x) P - 1 \u003d P · XP - 1

The last transition is possible due to the fact that if P. - odd number, then P - 1. either even number or zero (at p \u003d 1), therefore, with negative X. It is true equality (x) p - 1 \u003d x p - 1.

So, we have proved the formula for the derivative of the power function at any valid p.

Example 2.

Dana functions:

f 1 (x) \u003d 1 x 2 3, F 2 (x) \u003d x 2 - 1 4, F 3 (x) \u003d 1 x log 7 12

Determine their derivatives.

Decision

Some of the specified functions are converting into a table form y \u003d x p, based on the properties of the degree, and then use the formula:

f 1 (x) \u003d 1 x 2 3 \u003d x - 2 3 ⇒ F 1 "(x) \u003d - 2 3 · x - 2 3 - 1 \u003d - 2 3 · x - 5 3 f 2" (x) \u003d x 2 - 1 4 \u003d 2 - 1 4 · x 2 - 1 4 - 1 \u003d 2 - 1 4 · x 2 - 5 4 F 3 (x) \u003d 1 x log 7 12 \u003d x - Log 7 12 ⇒ F 3 "( x) \u003d - log 7 12 · x - log 7 12 - 1 \u003d - log 7 12 · x - log 7 12 - log 7 7 \u003d - log 7 12 · x - log 7 84

Derivative indicative function

We derive the derivative formula by taking the definition as a basis:

(AX) "\u003d LIM Δ X → 0 AX + Δ X - AX Δ x \u003d Lim Δ x Δ x \u003d Lim Δ x → 0 AX (A Δ x - 1) Δ x \u003d Ax · Lim Δ x → 0 A Δ x - 1 Δ x \u003d 0 0

We received uncertainty. To reveal it, write down the new variable z \u003d a Δ x - 1 (z → 0 at δ x → 0). In this case, A Δ x \u003d z + 1 ⇒ Δ x \u003d log A (z + 1) \u003d ln (z + 1) Ln a. For the latter transition, the formula for the transition to a new base of the logarithm is used.

Implement substitution to the initial limit:

(AX) "\u003d AX · LIM Δ x → 0 A Δ x - 1 Δ x \u003d Ax · Ln A · Lim Δ x → 0 1 1 z · ln (z + 1) \u003d \u003d AX · Ln A · Lim Δ x → 0 1 ln (z + 1) 1 z \u003d ax · ln a · 1 ln lim δ x → 0 (z + 1) 1 z

Recall the second wonderful limit and then we obtain a derivative formula of the derivative function:

(a x) "\u003d a x · ln a · 1 ln lim z → 0 (z + 1) 1 z \u003d a x · ln a · 1 ln e \u003d a x · ln a

Example 3.

Dame indicative functions:

f 1 (x) \u003d 2 3 x, f 2 (x) \u003d 5 3 x, f 3 (x) \u003d 1 (E) x

It is necessary to find their derivatives.

Decision

We use the formula of the derivative of the indicative function and the properties of the logarithm:

f 1 "(x) \u003d 2 3 x" \u003d 2 3 x · ln 2 3 \u003d 2 3 x · (ln 2 - ln 3) f 2 "(x) \u003d 5 3 x" \u003d 5 3 x · Ln 5 1 3 \u003d 1 3 · 5 3 x · ln 5 f 3 "(x) \u003d 1 (E) x" \u003d 1 ex "\u003d 1 ex · ln 1 e \u003d 1 ex · ln e - 1 \u003d - 1 ex

Derivative logarithmic function

We present proof of the formula for the logarithmic function for any X. In the field of definition and any permissible values \u200b\u200bof the base and logarithm. Relying on the definition of the derivative, we get:

(log ax) "\u003d Lim Δ x → 0 log a (x + Δ x) - log ax Δ x \u003d Lim Δ x → 0 log ax + Δ xx Δ x \u003d \u003d Lim Δ x Δ x \u003d 1 Δ x · Log A 1 + Δ xx \u003d Lim Δ x → 0 log a 1 + δ xx 1 Δ x \u003d \u003d Lim Δ x → 0 Log A 1 + Δ xx 1 Δ x · xx \u003d Lim Δ x → 0 1 x · Log A 1 + Δ xxx Δ x \u003d \u003d 1 x · log a lim Δ x → 0 1 + Δ xxx Δ x \u003d 1 x · log ae \u003d 1 x · ln e ln a \u003d 1 x · ln a

From the specified chain of equations it can be seen that the transformation was based on the logarithm properties. Equality Lim Δ x → 0 1 + Δ x x x δ x \u003d e is correct in accordance with the second wonderful limit.

Example 4.

Logarithmic functions are set:

f 1 (x) \u003d log ln 3 x, f 2 (x) \u003d ln x

It is necessary to calculate their derivatives.

Decision

Apply the derived formula:

f 1 "(x) \u003d (log ln 3 x)" \u003d 1 x · ln (ln 3); f 2 "(x) \u003d (ln x)" \u003d 1 x · ln e \u003d 1 x

So, the derivative of the natural logarithm is the unit divided by X..

Derived trigonometric functions

We use some trigonometric formulas and the first wonderful limit to derive the formula of the derivative trigonometric function.

According to the definition of the derivative of the sinus function, we get:

(SIN X) "\u003d LIM Δ x → 0 sin (x + Δ x) - SIN x Δ x

The sinus difference formula will allow us to make the following actions:

(SIN X) "\u003d LIM Δ x → 0 sin (x + Δ x) - sin x Δ x \u003d \u003d Lim Δ x → 0 2 · sin x + δ x - x 2 · cos x + δ x + x 2 δ x \u003d \u003d Lim Δ x → 0 sin Δ x 2 · cos x + δ x 2 δ x 2 \u003d \u003d cos x + 0 2 · Lim Δ x → 0 sin Δ x 2 Δ x 2

Finally, we use the first wonderful limit:

sin "x \u003d cos x + 0 2 · Lim Δ x → 0 sin Δ x 2 Δ x 2 \u003d cos x

So derived function SIN X. will be COS X..

Let us also prove the formula of the cosine derivative:

cos "x \u003d Lim Δ x → 0 cos (x + Δ x) - cos x Δ x \u003d \u003d Lim Δ x → 0 - 2 · sin x + δ x - x 2 · sin x + δ x + x 2 Δ x \u003d \u003d - Lim Δ x → 0 sin Δ x 2 · sin x + δ x 2 Δ x 2 \u003d \u003d - sin x + 0 2 · Lim Δ x → 0 sin Δ x 2 Δ x 2 \u003d - sin x

Those. COS X derivative will be - SIN X..

The formulas of Tangent and Kotangen derivatives withdraw on the basis of differentiation rules:

tG "x \u003d sin x cos x" \u003d sin "x · cos x - sin x · cos" x cos 2 x \u003d cos x · cos x - sin x · (- sin x) cos 2 x \u003d sin 2 x + COS 2 X COS 2 X \u003d 1 COS 2 XCTG "X \u003d COS X SIN X" \u003d COS "X · SIN X - COS X · SIN" X SIN 2 X \u003d - SIN X · SIN X - COS X · COS X SIN 2 X \u003d - SIN 2 X + COS 2 X SIN 2 X \u003d - 1 SIN 2 x

Derivatives of inverse trigonometric functions

The section on the derivative of inverse functions provides comprehensive information on the proof of the formulas of the derivatives of Arksinus, Arkkosinus, Arctanens and Arkotanens, so we will not duplicate the material here.

Derivatives of hyperbolic functions

The derivation of the formulas of the derivatives of hyperbolic sine, cosine, tangent and catangens will be carried out using the number of differentiation and the formula of the derivative of the indicative function:

sh "x \u003d ex - e - x 2" \u003d 1 2 ex "- e - x" \u003d \u003d 1 2 ex - e - x \u003d ex + e - x 2 \u003d chxch "x \u003d ex + e - x 2" \u003d 1 2 EX "+ E - X" \u003d 1 2 EX + - E - X \u003d EX - E - X 2 \u003d SHXTH "X \u003d SHXCHX" \u003d SH "X · CHX - SHX · CH" XCH 2 x \u003d CH 2 x - sh 2 xch 2 x \u003d 1 ch 2 xcth "x \u003d chxshx" \u003d ch "x · shx - chx · sh" xsh 2 x \u003d sh 2 x - ch 2 xsh 2 x \u003d - 1 sh 2 x

If you notice a mistake in the text, please select it and press Ctrl + Enter

Basic concepts

Before making a question about the derivative of the exhibit to the degree of $ x $, we recall the definitions

1. functions;
2. sequence limit;
3. derivative;
4. exhibitors.

This is necessary for a clear understanding of the derivative of the exhibit to the degree of $ x $.

Definition 1.

The function is called the relationship between two variables.

Take $ Y \u003d F (x) $, where $ x $ and $ y $ are variable values. Here $ x $ is called an argument, and $ y $ is a function. The argument can make arbitrary values. In turn, the $ y $ variable varies on a certain law depending on the argument. That is, an argument $ x $ is an independent variable, and the function $ y $ is a dependent variable. Any value of $ x $ corresponds to the only value of $ y $.

If each natural number is $ n \u003d 1, 2, 3, ... $ to put into line with a certain law of $ x_n $, then they say that the sequence of numbers $ x_1, x_2, ..., x_n $ is determined. Otherwise, this sequence is written as $ \\ (x_n \\) $. All numbers $ x_n $ are called members or elements of the sequence.

Definition 2.

The sequence limit is called the final or infinitely remote point of the numerical line. The limit is written as follows: $ \\ lim x_n \u003d \\ l \\ limits_ (n \\ to \\ infty) x_n \u003d a $. This record means that the $ x_n $ variable tends to $ a $ $ x_n \\ to a $.

The derivative function $ F $ at $ x_0 is the following limit:

$ \\ limits_ (x \\ to x_0) \\ FRAC (F (X) - F (X_O)) (x-x_o) $. It is denoted by $ f "(x_0) $.

The number $ E $ is the next limit:

$ E \u003d \\ Limits_ (X \\ To \\ infty) (1+ \\ FRAC (1) (N)) \\ APPROX2,718281828459045 ... $

In this limit, $ n $ is a natural or valid number.

Owning the concepts of the limit, derivative and exponential, we can proceed with the proof of the formula $ (E ^ x) "\u003d E ^ X $.

Derivative derivative from exponent to $ x $

We have $ E ^ x $, where $ x: - \\ infty

$ y "\u003d \\ l \\ limits _ (\\ deelta x \\ to 0) \\ FRAC (E ^ (X + \\ Delta X) -e ^ x) (\\ Delta X) $.

By the property of the exponent of $ E ^ (A + BX) \u003d E ^ A * E ^ b $ we can convert the limit number:

$ E ^ (x + \\ Delta x) -e ^ x \u003d E ^ x * E ^ (\\ Delta x) -e ^ x \u003d E ^ x (E ^ (\\ Delta x) -1) $.

That is, $ y "\u003d \\ l \\ limits _ (\\ lta x \\ to 0) \\ FRAC (E ^ (x + \\ Delta X) -e ^ x) (\\ Delta X) \u003d \\ LIM \\ Limits _ (\\ Delta X \\ To 0) \\ FRAC (E ^ X (E ^ (\\ Delta X) -1)) (\\ Delta X) $.

Denote by $ T \u003d E ^ (\\ Delta X) -1 $. We obtain $ E ^ (\\ Delta X) \u003d T + $ 1, and according to the logarithm property, it turns out that $ \\ delta x \u003d ln (t + 1) $.

Since the exhibitor is continuous, we have $ \\ lim \\ limits _ (\\ delta x \\ to 0) E ^ (\\ Delta X) \u003d E ^ 0 \u003d 1. $ therefore if $ \\ delta x \\ to 0 $, then $ t \\ As a result, we show the conversion:

$ Y "\u003d \\ LIM \\ Limits _ (\\ Delta X \\ To 0) \\ FRAC (E ^ (\\ Delta X) -1) (\\ Delta X) \u003d E ^ X \\ LIM \\ Limits_ (t \\ to 0) \\ FRAC (t) (Ln (T + 1)) $.

Denote by $ n \u003d \\ FRAC (1) (T) $, then $ T \u003d \\ FRAC (1) (N) $. It turns out that if $ t \\ to 0 $, then $ n \\ to \\ infty $.

We transform our limit:

$ Y "\u003d E ^ X \\ LIM \\ Limits_ (T \\ To 0) \\ FRAC (T) (Ln (T + 1)) \u003d E ^ X \\ LIM \\ Limits_ (n \\ to \\ infty) \\ FRAC (1) (N \\ Cdot Ln (\\ FRAC (1) (N) +1) ^ n) $.

By the property of the logarithm $ B \\ Cdot LN C \u003d LN C ^ b $ we have

$ n \\ cdot ln (\\ FRAC (1) (N) +1) \u003d ln (\\ FRAC (1) (N) +1) ^ n \u003d ln (1+ \\ FRAC (1) (N)) ^ n $ .

The limit is converted as follows:

$ y "\u003d E ^ X \\ LIM \\ Limits_ (n \\ to \\ infty) \\ FRAC (1) (N \\ CDOT LN (\\ FRAC (1) (N) +1)) \u003d E ^ X \\ LIM \\ Limits_ ( N \\ To \\ infty) \\ FRAC (1) (Ln (\\ FRAC (1) (N) +1) ^ n) \u003d E ^ X \\ FRAC (1) (\\ Limits_ (N \\ To \\ infty) ln (\\ FRAC (1) (N) +1) ^ n) $.

According to the function of the continuity of the logarithm and the properties of the limits for the continuous function: $ \\ lim \\ limits_ (x \\ to x_0) ln (f (x)) \u003d ln (\\ lim \\ limits_f (x)) $, where $ F (x) $ has The positive limit is $ \\ l \\ limits_ (x \\ to x_0) f (x) $. So, due to the fact that the logarithm is continuous and there is a positive limit of $ \\ limits_ (n \\ to \\ infty) (\\ FRAC (1) (N) +1) ^ N $, we can derive:

$ \\ Lim \\ Limits_ (n \\ to \\ infty) ln (1+ \\ FRAC (1) (N)) ^ n \u003d ln \\ limits_ (n \\ to \\ infty) ln (1+ \\ FRAC (1) ( n)) ^ n \u003d ln e \u003d 1 $.

We use the value of the second remarkable limit of $ \\ Lim \\ Limits_ (n \\ to \\ infty) (1+ \\ FRAC (1) (N)) ^ n \u003d E $. We get:

$ Y "\u003d E ^ X \\ FRAC (1) (\\ Lim \\ Limits_ (n \\ to \\ infty) ln (\\ FRAC (1) (N) +1) ^ n) \u003d E ^ X \\ CDOT \\ FRAC (1 ) (Ln E) \u003d E ^ X \\ CDOT \\ FRAC (1) (1) \u003d E ^ x $.

Thus, we brought the formula for the derivative of the exhibitance and can argue that the derivative of the exhibitors to the degree of $ x $ is equivalent to the exhibitor to the degree $ x $:

There are also other ways to output this formula using other formulas and rules.

Consider an example of finding a derived function.

Example 1.

Condition

: Find the derivative function $ y \u003d 2 ^ x + 3 ^ x + 10 ^ x + e ^ x $.: To the term $ 2 ^ x, 3 ^ x $ and $ 10 ^ x $ we use the formula $ (a ^ x) "\u003d a ^ x \\ cdot ln a $. According to the derived formula $ (E ^ x)" \u003d E ^ x $ The fourth term $ e ^ x $ does not change.

DecisionAnswer

: $ y "\u003d 2 ^ x \\ cdot ln 2 + 3 ^ X \\ Cdot Ln 3 + 10 ^ x \\ Cdot Ln 10 + E ^ x $.{!LANG-f3b53ee8b29ed15168829a43d2d8bdec!}

Thus, we derived the formula $ (E ^ x) "\u003d E ^ X $, while giving the definition with basic concepts, disassembled an example of finding a derivative function with an exponent as one of the terms.

Many numbers have gained their magnitude and superstitious significance in antiquity. Nowadays, new myths are added to them. There are many legends on the number of pi, a little inferior to him in fond of the famous numbers of Fibonacci. But perhaps the most awesome is the number E, without which can not do modern mathematics, physics and even economy.

The arithmetic value of the number E is approximately 2.718. Why not exactly, and approximately? Because this number is irrational and transcendental, it cannot be expressed by a fraction with natural integers or polynomial with rational coefficients. For most calculations of the specified accuracy, the value of 2.718 is sufficient, although the current level of computing technology allows you to determine its value with an accuracy of more than a trillion after a comma.

The main feature of the number E is that the derivative of its indicative function f (x) \u003d e x is equal to the value of the function itself e x. There is no other mathematical dependence of such an unusual property. Let's tell about this a little more.

What is the limit

First we will deal with the concept of limit. Consider some mathematical expression, for example, i \u003d 1 / n. Can see, that with increasing "n ", The value" I "will decrease, and with the desire" n "to infinity (which is denoted by the ∞ icon)," I "will strive for the limit value (called more often the limit) equal to zero. The expression of the limit (denoted as Lim) for the case under consideration can be written as Lim N → ∞ (1 / N) \u003d 0.

There are various limits for different expressions. One of these limits included in Soviet and Russian textbooks as a second wonderful limit is the expression of Lim N → ∞ (1 + 1 / N) N. Already in the Middle Ages, it was found that the limit of this expression is the number E.

To the first wonderful limit include the expression Lim N → ∞ (SIN N / N) \u003d 1.

How to find a derivative E X - in this video.

What is derived function

For disclosure, the concept of the derivative should be reminded that such a function in mathematics. In order not to clutter the text with complex definitions, we will focus on the intuitive mathematical concept of the function in the fact that in it one or more values \u200b\u200bfully determine the value of another value if they are interrelated. For example, in the formula S \u003d π ∙ R 2 of the circle area, the value of the radius R completely and uniquely determines the area of \u200b\u200bthe circle S.

Depending on the species, functions can be algebraic, trigonometric, logarithmic, etc. Two, three or more arguments can be interconnected. For example, the distance S, which the object overcame with an equilibrium speed, is described by the function S \u003d 0.5 ∙ A ∙ T 2 + V ∙ T, where "T" - the time of movement, the argument "A" acceleration (may be both positive, so and a negative value) and "V" the initial speed of movement. Thus, the magnitude of the distance has passed depends on the values \u200b\u200bof the three arguments, two of which ("A" and "V") are constant.

We show the elementary concept of the derivative function on this example. It characterizes the speed of change of function at this point. In our example, it will be the speed of the object at a specific point in time. With constant "A" and "V", it depends only on the time "T", that is, in the scientific language, it is necessary to take the derivative of the function s in time "T".

This process is called differentiation, performed by calculating the limit of the ratio of the function of the function to the increase in its argument on a negligible value. Solutions of such tasks for individual functions are often not easy and not considered here. It should also be noted that some functions at certain points do not have such limits at all.

In our example, the derivative s By time, "T" will take the form S "\u003d ds / dt \u003d a ∙ t + v, from which it can be seen that the speed S" varies according to the linear law depending on "T".

Derivative exhibit

The exponent is called an indicative function, which is the basis of which the number e. It is usually displayed as f (x) \u003d e x, where the degree x is a variable value. This function has complete differentiability in the entire range of real numbers. With increasing x, it is constantly increasing and always more zero. The function is logarithm back to it.

The famous Mathematics Taylor managed to decompose this function in a row, called it named E x \u003d 1 + x / 1! + x 2/2! + x 3/3! + ... in the range x from - ∞ to + ∞.

Law based on this function, called exponential. He describes:

• increasing complex bank interest;
• an increase in the population of animals and the population of the planet;
• the time of the inquition of the corpse and much more.

We repeat once again a wonderful property of this dependence - the value of its derivative at any point is always equal to the value of the function at this point, that is, (E x) "\u003d E x.

We give derivatives for the most common exhibitors:

• (E AX) "\u003d A ∙ E AX;
• (E F (x)) "\u003d F" (x) ∙ e F (x).

Using these dependencies, it is easy to find derivatives for other private species of this function.

Some interesting facts about

With this number, the names of such scientists, as a forever, Otraditz, Guygens, Bernoulli, Leibniz, Newton, Euler, and others are connected with this number. The latter actually introduced the designation E for this number, and also found the first 18 characters, using the OPERATION OPERATE E \u003d 1 + 1/1 for the calculation! + 2/2! + 3/3! ...

The number E is found in the most unexpected places. For example, it enters the chain line equation, which describes the rope provisis under the action of its own weight when its ends are fixed on supports.

Video

The topic of the video language is a derivative of the indicative function.

Proof and output of the formulas derivative derivative (E to degree x) and an indicative function (A to degree x). Examples of calculating derivatives from E ^ 2x, E ^ 3x and E ^ nx. Formulas derivatives of higher orders.

See also: Indicative function - properties, formulas, graph
Exhibitor, E to degree x - properties, formulas, schedule

Basic formulas

The derivative of the exhibit is equal to the exhibitor itself (the derivative E to the degree x is equal to e to the degree x):
(1) (E X) '\u003d E X.

The derivative of the indicative function with the base of the degree A is equal to the function itself multiplied by the natural logarithm from A:
(2) .

The exhibitor is an indicative function in which the degree base is equal to the number e, which is the following limit:
.
Here can be both natural and actual number. Next, we derive the formula (1) of the derivative of the exhibit.

Output of the formula derivative exhibit

Consider the exhibitor, E to the degree x:
y \u003d E X.
This feature is defined for all. Find its derivative in the variable x. By definition, the derivative is the following limit:
(3) .

We transform this expression to reduce it to well-known mathematical properties and rules. For this, we will need the following facts:
BUT) Property Exhibitors:
(4) ;
B) Logarithm property:
(5) ;
IN) Continuity of logarithm and property of limits for continuous function:
(6) .
Here is some function that has a limit and this limit is positive.
D) The value of the second remarkable limit:
(7) .

We use these facts to our limit (3). We use property (4):
;
.

Make a substitution. Then; .
Due to continuity of exhibitors,
.
Therefore, when,. As a result, we get:
.

Make a substitution. Then. With ,. And we have:
.

Applicate Logarithm property (5):
. Then
.

Apply Property (6). Since there is a positive limit and logarithm continuously, then:
.
Here we also took advantage of the second wonderful limit (7). Then
.

Thus, we obtained formula (1) of the derivative of the exhibit.

Output of the formula of the derivative of the indicative function

Now we will derive the formula (2) of the derivative of the indicative function with the basis of degree a. We believe that. Then the indicative function
(8)
Defined for all.

We transform formula (8). To do this, we use the properties of the indicative function and logarithm.
;
.
So, we transformed formula (8) to the following form:
.

Derivatives of higher orders from E to degree x

Now we find derivatives of higher orders. First consider the exhibitor:
(14) .
(1) .

We see that the derivative of the function (14) is equal to the function itself (14). Differentiating (1), we obtain derivatives of the second and third order:
;
.

It can be seen that the derivative of the N-th order is also equal to the source function:
.

Derivatives of higher orders of indicative function

Now consider an indicative function with the basis of degree A:
.
We found her first order derivative:
(15) .

Differentiating (15), we obtain derivatives of the second and third order:
;
.

We see that each differentiation leads to multiplication of the original function on. Therefore, the derivative of the N-th order has the following form:
.