What is the derivative of sin x. Sine derivative: (sin x)′. Derivatives of higher orders

When deriving the very first formula of the table, we will proceed from the definition of the derivative of a function at a point. Let's take where x- any real number, that is, x– any number from the function definition area . Let us write the limit of the ratio of the function increment to the argument increment at :

It should be noted that under the sign of the limit, an expression is obtained, which is not the uncertainty of zero divided by zero, since the numerator contains not an infinitesimal value, but precisely zero. In other words, the increment of a constant function is always zero.

In this way, derivative of a constant functionis equal to zero on the entire domain of definition.

Derivative of a power function.

The formula for the derivative of a power function has the form , where the exponent p is any real number.

Let us first prove the formula for the natural exponent, that is, for p = 1, 2, 3, ...

We will use the definition of a derivative. Let us 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 Newton's binomial formula:

Consequently,

This proves the formula for the derivative of a power function for a natural exponent.

Derivative of exponential function.

We derive the derivative formula based on the definition:

Came to uncertainty. To expand it, we introduce a new variable , and for . Then . In the last transition, we used the formula for the transition to a new base of the logarithm.

Let's perform a substitution in the original limit:

If we recall the second remarkable limit, then we come to the formula for the derivative of the exponential function:

Derivative of a logarithmic function.

Let us prove the formula for the derivative of the logarithmic function for all x from the scope and all valid base values a logarithm. By definition of the derivative, we have:

As you noticed, in the proof, the transformations were carried out using the properties of the logarithm. Equality is valid due to the second remarkable limit.

Derivatives of trigonometric functions.

To derive formulas for derivatives of trigonometric functions, we will have to recall some trigonometry formulas, as well as the first remarkable limit.

By definition of the derivative for the sine function, we have .

We use the formula for the difference of sines:

It remains to turn to the first remarkable limit:

So the derivative of the function sin x there is cos x.

The formula for the cosine derivative is proved in exactly the same way.

Therefore, the derivative of the function cos x there is –sin x.

The derivation of formulas for the table of derivatives for the tangent and cotangent will be carried out using the proven rules of differentiation (derivative of a fraction).

Derivatives of hyperbolic functions.

The rules of differentiation and the formula for the derivative of the exponential function from the table of derivatives allow us to derive formulas for the derivatives of the hyperbolic sine, cosine, tangent and cotangent.

Derivative of the inverse function.

So that there is no confusion in the presentation, let's denote in the lower index the argument of the function by which differentiation is performed, that is, it is the derivative of the function f(x) on x.

Now we formulate rule for finding the derivative of the inverse function.

Let the functions y = f(x) and x = g(y) mutually inverse, defined on the intervals and respectively. If at a point there exists a finite non-zero derivative of the function f(x), then at the point there exists a finite derivative of the inverse function g(y), and . In another entry .

This rule can be reformulated for any x from the interval , then we get .

Let's check the validity of these formulas.

Let's find the inverse function for the natural logarithm (here y is a function, and x- argument). Solving this equation for x, we get (here x is a function, and y her argument). That is, and mutually inverse functions.

From the table of derivatives, we see that and .

Let's make sure that the formulas for finding derivatives of the inverse function lead us to the same results:

As you can see, we got the same results as in the table of derivatives.

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

Let's start with the derivative of the arcsine.

. Then, by the formula for the derivative of the inverse function, we obtain

It remains to carry out the transformation.

Since the range of the arcsine is the interval , then (see the section on basic elementary functions, their properties and graphs). Therefore, we do not consider.

Consequently, . The domain of definition of the derivative of the arcsine is the interval (-1; 1) .

For the arccosine, everything is done in exactly the same way:

Find the derivative of the arc tangent.

For the inverse function is .

We express the arc tangent through the arc cosine to simplify the resulting expression.

Let arctanx = z, then

Consequently,

Similarly, the derivative of the inverse tangent is found:

The proof and derivation of the formula for the derivative of the sine - sin(x) is presented. Examples of calculating derivatives of sin 2x, sine squared and cubed. Derivation of the formula for the derivative of the sine of the nth order.

Content

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

The derivative with respect to the variable x of the sine of x is equal to the cosine of x:
(sin x)′ = cos x.

Proof

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

To find this limit, we need to transform the expression in such a way as to reduce it to known laws, properties, and rules. To do this, we need to know four properties.
1) Meaning of the first remarkable limit:
(1) ;
2) Continuity of the cosine function:
(2) ;
3) Trigonometric Formulas. We need the following formula:
(3) ;
4) Arithmetic properties of the function limit:
If and then
(4) .

We apply these rules to our limit. First we transform the algebraic expression
.
For this we apply the formula
(3) .
In our case
; . Then
;
;
;
.

Now let's make a substitution. At , . Let us apply the first remarkable limit (1):
.

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

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

.

The formula for the derivative of the sine has been proven.

Examples

Consider simple examples of finding derivatives of functions containing a sine. We will find derivatives of the following functions:
y=sin2x; y= sin2x and y= sin3x.

Example 1

Find the derivative of sin 2x.

First we find the derivative of the simplest part:
(2x)′ = 2(x)′ = 2 1 = 2.
We apply.
.
Here .

(sin 2x)′ = 2 cos 2x.

Example 2

Find the derivative of the squared sine:
y= sin2x.

Let's rewrite the original function in a more understandable form:
.
Find the derivative of the simplest part:
.
We apply the formula for the derivative of a complex function.

.
Here .

One of the trigonometry formulas can be applied. Then
.

Example 3

Find the derivative of the sine cubed:
y= sin3x.

Derivatives of higher orders

Note that the derivative of sin x of the first order can be expressed in terms of the sine as follows:
.

Let's find the second order derivative using the formula for the derivative of a complex function:

.
Here .

Now we can see that the differentiation sin x causes its argument to be incremented by . Then the derivative of the nth order has the form:
(5) .

Let us prove this by applying the method of mathematical induction.

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

Let us assume that formula (5) is valid for some value of . Let us prove that it follows from this that formula (5) is valid for .

We write formula (5) for :
.
We differentiate this equation by applying the rule of differentiation of a complex function:

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

The formula has been proven.

Derivative of a constant

Proof 1

In order to derive this formula, we take as a basis the definition of the derivative of a function at a point. We use x 0 = x, where x takes on the value of any real number, or, in other words, x is any number from the domain of the function f (x) = C . Let's write the limit of the ratio of the increment of the function to the increment of the argument as ∆ x → 0:

lim ∆ x → 0 ∆ f (x) ∆ x = lim ∆ x → 0 C - C ∆ x = lim ∆ x → 0 0 ∆ x = 0

Please note that the expression 0 ∆ x falls under the limit sign. It is not the uncertainty of “zero divided by zero”, since the numerator contains not an infinitesimal value, but zero. In other words, the increment of a constant function is always zero.

So, the derivative of the constant function f (x) = C is equal to zero over the entire domain of definition.

Example 1

Given constant functions:

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

Solution

Let us describe the given conditions. In the first function we see the derivative of the natural number 3 . In the following example, you need to take the derivative of a, where a- any real number. The third example gives us the derivative of the irrational number 4 . 13 7 22 , the fourth - the derivative of zero (zero is an integer). Finally, in the fifth case, we have the derivative of the rational fraction - 8 7 .

Answer: the derivatives of the given functions are zero for any real x(over the entire domain of definition)

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

Power function derivative

We turn to the power function and the formula for its derivative, which has the form: (x p) " = p x p - 1, where the exponent p is any real number.

Proof 2

Here is the proof of the formula when the exponent is a natural number: p = 1 , 2 , 3 , …

Again, we rely on the definition of a derivative. Let's write the limit of the ratio of the increment of the power function to the increment of the argument:

(x p) " = lim ∆ x → 0 = ∆ (x p) ∆ x = lim ∆ x → 0 (x + ∆ x) p - x p ∆ x

To simplify the expression in the numerator, we use Newton's binomial formula:

(x + ∆ x) p - x p = C p 0 + x p + C p 1 x p - 1 ∆ x + C p 2 x p - 2 (∆ x) 2 + . . . + + C p p - 1 x (∆ x) p - 1 + C p p (∆ x) p - x p = = C p 1 x p - 1 ∆ x + C p 2 x p - 2 (∆ x) 2 + . . . + C p p - 1 x (∆ x) p - 1 + C p p (∆ x) p

In this way:

(x p) " = lim ∆ x → 0 ∆ (x p) ∆ x = lim ∆ x → 0 (x + ∆ x) p - x p ∆ x = = lim ∆ x → 0 (C p 1 x p - 1 ∆ x + C p 2 x p - 2 (∆ x) 2 + . . . + C p p - 1 x (∆ x) p - 1 + C p p (∆ x) p) ∆ x = = lim ∆ x → 0 (C p 1 x p - 1 + C p 2 x p - 2 ∆ x + . . . + C p p - 1 x (∆ x) p - 2 + C p p (∆ x) p - 1) = = C p 1 x p - 1 + 0 + 0 + . . . + 0 = p! 1! (p - 1)! x p - 1 = p x p - 1

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

Proof 3

To give proof for the case when p- any real number other than zero, we use the logarithmic derivative (here we should understand 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 additionally deal with the derivative of an implicitly given function and the derivative of a complex function.

Consider two cases: when x positive and when x are negative.

So x > 0 . Then: x p > 0 . We take the logarithm of the equality y \u003d x p to the base e and apply the property of the logarithm:

y = x p ln y = ln x p ln y = p ln x

At this stage, an implicitly defined function has been obtained. Let's define its derivative:

(ln y) " = (p ln x) 1 y y " = p 1 x ⇒ y " = p y x = p x p x = p x p - 1

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

If the indicator p is an even number, then the power function is also defined for x< 0 , причем является четной: y (x) = - y ((- x) p) " = - p · (- x) p - 1 · (- x) " = = p · (- x) p - 1 = p · x p - 1

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

If a p is an odd number, then the power function is defined for 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 p (- x) p - 1 = p x p - 1

The last transition is possible because if p is an odd number, then p - 1 either an even number or zero (for p = 1), therefore, for negative x the equality (- x) p - 1 = x p - 1 is true.

So, we have proved the formula for the derivative of a power function for any real p.

Example 2

Given functions:

f 1 (x) = 1 x 2 3 , f 2 (x) = x 2 - 1 4 , f 3 (x) = 1 x log 7 12

Determine their derivatives.

Solution

We transform part of the given functions into a tabular form y = 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 = 2 - 1 4 x 2 - 1 4 - 1 = 2 - 1 4 x 2 - 5 4 f 3 (x) = 1 x log 7 12 = x - log 7 12 ⇒ f 3 "( x) = - log 7 12 x - log 7 12 - 1 = - log 7 12 x - log 7 12 - log 7 7 = - log 7 12 x - log 7 84

Derivative of exponential function

Proof 4

We derive the formula for the derivative, based on the definition:

(a x) " = lim ∆ x → 0 a x + ∆ x - a x ∆ x = lim ∆ x → 0 a x (a ∆ x - 1) ∆ x = a x lim ∆ x → 0 a ∆ x - 1 ∆ x = 0 0

We got uncertainty. To expand it, we write a new variable z = a ∆ x - 1 (z → 0 as ∆ x → 0). In this case a ∆ x = z + 1 ⇒ ∆ x = log a (z + 1) = ln (z + 1) ln a . For the last transition, the formula for the transition to a new base of the logarithm is used.

Let's perform a substitution in the original limit:

(a x) " = a x lim ∆ x → 0 a ∆ x - 1 ∆ x = a x ln a lim ∆ x → 0 1 1 z ln (z + 1) = = a x ln a lim ∆ x → 0 1 ln (z + 1) 1 z = a x ln a 1 ln lim ∆ x → 0 (z + 1) 1 z

Recall the second wonderful limit and then we get the formula for the derivative of the exponential function:

(a x) " = a x ln a 1 ln lim z → 0 (z + 1) 1 z = a x ln a 1 ln e = a x ln a

Example 3

The exponential functions are given:

f 1 (x) = 2 3 x , f 2 (x) = 5 3 x , f 3 (x) = 1 (e) x

We need to find their derivatives.

Solution

We use the formula for the derivative of the exponential function and the properties of the logarithm:

f 1 "(x) = 2 3 x" = 2 3 x ln 2 3 = 2 3 x (ln 2 - ln 3) f 2 "(x) = 5 3 x" = 5 3 x ln 5 1 3 = 1 3 5 3 x ln 5 f 3 "(x) = 1 (e) x" = 1 e x " = 1 e x ln 1 e = 1 e x ln e - 1 = - 1 e x

Derivative of a logarithmic function

Proof 5

We present the proof of the formula for the derivative of the logarithmic function for any x in the domain of definition and any valid values of the base a of the logarithm. Based on the definition of the derivative, we get:

(log a x) " = lim ∆ x → 0 log a (x + ∆ x) - log a x ∆ x = lim ∆ x → 0 log a x + ∆ x x ∆ x = = lim ∆ x → 0 1 ∆ x log a 1 + ∆ x x = lim ∆ x → 0 log a 1 + ∆ x x 1 ∆ x = = lim ∆ x → 0 log a 1 + ∆ x x 1 ∆ x x x = lim ∆ x → 0 1 x log a 1 + ∆ x x x ∆ x = = 1 x log a lim ∆ x → 0 1 + ∆ x x x ∆ x = 1 x log a e = 1 x ln e ln a = 1 x ln a

It can be seen from the specified chain of equalities that the transformations were built on the basis of the logarithm property. The equality lim ∆ x → 0 1 + ∆ x x x ∆ x = e is true in accordance with the second remarkable limit.

Example 4

Logarithmic functions are given:

f 1 (x) = log log 3 x , f 2 (x) = log x

It is necessary to calculate their derivatives.

Solution

Let's apply the derived formula:

f 1 "(x) = (log ln 3 x)" = 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 one divided by x.

Derivatives of trigonometric functions

Proof 6

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

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

(sin x) " = lim ∆ x → 0 sin (x + ∆ x) - sin x ∆ x

The formula for the difference of sines will allow us to perform the following actions:

(sin x) " = lim ∆ x → 0 sin (x + ∆ x) - sin x ∆ x = = lim ∆ x → 0 2 sin x + ∆ x - x 2 cos x + ∆ x + x 2 ∆ x = = lim ∆ x → 0 sin ∆ x 2 cos x + ∆ x 2 ∆ x 2 = = cos x + 0 2 lim ∆ x → 0 sin ∆ x 2 ∆ x 2

Finally, we use the first wonderful limit:

sin "x = cos x + 0 2 lim ∆ x → 0 sin ∆ x 2 ∆ x 2 = cos x

So the derivative of the function sin x will be cos x.

We will also prove the formula for the cosine derivative in the same way:

cos "x = lim ∆ x → 0 cos (x + ∆ x) - cos x ∆ x = = lim ∆ x → 0 - 2 sin x + ∆ x - x 2 sin x + ∆ x + x 2 ∆ x = = - lim ∆ x → 0 sin ∆ x 2 sin x + ∆ x 2 ∆ x 2 = = - sin x + 0 2 lim ∆ x → 0 sin ∆ x 2 ∆ x 2 = - sin x

Those. the derivative of the function cos x will be – sin x.

We derive the formulas for the derivatives of the tangent and cotangent based on the rules of differentiation:

t g "x = sin x cos x" = sin "x cos x - sin x cos "x cos 2 x = = cos x cos x - sin x (- sin x) cos 2 x = sin 2 x + cos 2 x cos 2 x = 1 cos 2 x c t g "x = cos x sin x" = cos "x sin x - cos x sin "x sin 2 x = = - sin x sin x - cos x cos x sin 2 x = - sin 2 x + cos 2 x sin 2 x = - 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 for the derivatives of the arcsine, arccosine, arctangent and arccotangent, so we will not duplicate the material here.

Derivatives of hyperbolic functions

Proof 7

We can derive formulas for the derivatives of the hyperbolic sine, cosine, tangent and cotangent using the differentiation rule and the formula for the derivative of the exponential function:

s h "x = e x - e - x 2" = 1 2 e x "- e - x" == 1 2 e x - - e - x = e x + e - x 2 = c h x c h "x = e x + e - x 2" = 1 2 e x "+ e - x" == 1 2 e x + - e - x = e x - e - x 2 = s h x t h "x = s h x c h x" = s h "x c h x - s h x c h "x c h 2 x = c h 2 x - s h 2 x c h 2 x = 1 c h 2 x c t h "x = c h x s h x" = c h "x s h x - c h x s h "x s h 2 x = s h 2 x - c h 2 x s h 2 x = - 1 s h 2 x

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