SIN AX derivative. Sinus derivative: (SIN X) '

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 when:

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:

As you can see, they obtained the same results as in the table of derivatives.

Now we have knowledge to prove the formulas of derivatives inverse trigonometric functions.

Let's start with the Arksinus derivative.

. Then according to the derivative formula, we get

It remains to be transformed.

Since the area of \u200b\u200bArksinus values \u200b\u200bis the interval T. (See the main elementary functions, properties and graphics). Therefore, but not consider.

Hence, . The area of \u200b\u200bdetermination of the derivative of Arksinus is the gap (-1; 1) .

For Arkkosinus, everything is done in absolutely similar:

Find a derivative of Arctangent.

For reverse function is .

Express Arctanens through Arkkosinus to simplify the resulting expression.

Let be arctgx \u003d Z., then

Hence,

The derivative of Arkkothangence is similar:

Derivative

The calculation of the derivative of the mathematical function (differentiation) is a very frequent task when solving the highest mathematics. For simple (elementary) mathematical functions, this is a fairly simple matter, since the tables of derivatives for elementary functions have been prepared and easily accessible. However, the finding of a derivative complex mathematical function is not a trivial task and often requires considerable effort and time costs.

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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

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The proof and output of the formula for the sinus derivative - sin (x) is presented. Examples of calculating derivatives from SIN 2X, sinus in a square and cube. The output of the formula for the sinus derivative of the N-th order.

See also: Sine and cosine - properties, graphs, formulas

The derivative of the variable x from the sinus x is equal to cosine X:
(SIN X) '\u003d COS X.

Evidence

To derive the formula of the sinus derivative, we will use the definition of the derivative:
.

To find this limit, we need to convert an expression in such a way as to reduce it to well-known laws, properties and rules. To do this, we need to know four properties.
1) Value the first wonderful limit :
(1) ;
2) Continuity of the function of cosine :
(2) ;
3) Trigonometric formulas . We will need the following formula:
(3) ;
4) Arithmetic properties of the limit function:
If and then
(4) .

Apply these rules to our limit. First we transform an algebraic expression
.
To do this, apply the formula
(3) .
In our case
; . Then
;
;
;
.

Now make a substitution. With ,. Apply the first wonderful limit (1):
.

We will make the same substitution and use the property of continuity (2):
.

Since the limits calculated above exist, we apply the property (4):

.

The formula of the sine derivative is proved.

Examples

Consider simple examples of finding derivatives from functions containing sinus. We will find derivatives from the following functions:
y \u003d sin 2x; y \u003d. sIN 2 X. and y \u003d. sIN 3 X..

Example 1.

Find a derivative OT sIN 2X..

First we find a derivative from the simplest part:
(2x) '\u003d 2 (x)' \u003d 2 · 1 \u003d 2.
Apply.
.
Here .

(SIN 2X) '\u003d 2 COS 2X.

Example 2.

Find a derivative of sine in a square:
y \u003d. sIN 2 X..

I rewrite the original function in a more understandable form:
.
Find a derivative from the simplest part:
.
Apply the formula of the derivative complex function.

.
Here .

You can apply one of the formulas of trigonometry. Then
.

Example 3.

Find a derivative of sinus in Cuba:
y \u003d. sIN 3 X..

Derivatives of higher orders

Note that the derivative of sIN X. First order can be expressed through sinus as follows:
.

Find a second order derivative using formula derivative complex function :

.
Here .

Now we can notice that differentiation sIN X. leads to an increase in its argument on. Then the derivative of the N-th order has the form:
(5) .

We prove it by applying the method of mathematical induction.

We have already checked that when, formula (5) is valid.

Suppose that formula (5) is valid for some meaning. We prove that from this it follows that formula (5) is performed for.

We drank formula (5) with:
.
Differentiating this equation, applying the differentiation rule of a complex function:

.
Here .
So, we found:
.
If we substitute, then this formula will take the form (5).

The formula is proved.