Derivative

Calculating the derivative of a mathematical function (differentiation) is a very common problem when solving higher mathematics. For simple (elementary) mathematical functions, this is a fairly simple matter, since tables of derivatives for elementary functions have long been compiled and are easily accessible. However, finding the derivative of a complex mathematical function is not a trivial task and often requires significant effort and time.

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

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

Thus, derivative of a constant functionis equal to zero throughout 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– 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 derivative. Let us write down the limit of the ratio of the increment of a power function to the increment of the argument:

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

Hence,

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

Derivative of an exponential function.

We present the derivation of the derivative formula based on the definition:

We have arrived at uncertainty. To expand it, we introduce a new variable, and at . Then . In the last transition, we used the formula for transitioning to a new logarithmic base.

Let's substitute into the original limit:

If we recall the second remarkable limit, we arrive at the formula for the derivative of the exponential function:

Derivative of a logarithmic function.

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

As you noticed, during the proof the transformations were carried out using the properties of the logarithm. Equality is true 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 .

Let's use the difference of sines formula:

It remains to turn to the first remarkable limit:

Thus, the derivative of the function sin x There is cos x.

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

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

We will derive formulas for the table of derivatives for tangent and cotangent using 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.

To avoid confusion during presentation, let's denote in subscript the argument of the function by which differentiation is performed, that is, it is the derivative of the function f(x) By x.

Now let's formulate rule for finding the derivative of an 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 is a finite non-zero derivative of the function f(x), then at the point there is a finite derivative of the inverse function g(y), and . In another post .

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). Having resolved 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 the derivatives of the inverse function lead us to the same results:

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

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

Let's start with the derivative of the arcsine.

. Then, using the formula for the derivative of the inverse function, we get

All that remains is to carry out the transformations.

Since the arcsine range is the interval , That (see the section on basic elementary functions, their properties and graphs). Therefore, we are not considering it.

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

For the arc cosine, everything is done in exactly the same way:

Let's find the derivative of the arctangent.

For the inverse function is .

Let's express the arctangent in terms of arccosine to simplify the resulting expression.

Let arctgx = z, Then

Hence,

The derivative of the arc cotangent is found in a similar way:

We present a summary table for convenience and clarity when studying the topic.

Constanty = C Power function y = x p (x p) " = p x p - 1	Exponential functiony = a x (a x) " = a x ln a In particular, whena = ewe have y = e x (e x) " = e x
Logarithmic function (log a x) " = 1 x ln a In particular, whena = ewe have y = logx (ln x) " = 1 x	Trigonometric functions (sin x) " = cos x (cos x) " = - sin x (t g x) " = 1 cos 2 x (c t g x) " = - 1 sin 2 x
Inverse trigonometric functions (a r c sin x) " = 1 1 - x 2 (a r c cos x) " = - 1 1 - x 2 (a r c t g x) " = 1 1 + x 2 (a r c c t g x) " = - 1 1 + x 2	Hyperbolic functions (s h x) " = c h x (c h x) " = s h x (t h x) " = 1 c h 2 x (c t h x) " = - 1 s h 2 x

Let us analyze how the formulas of the specified table were obtained or, in other words, we will prove the derivation of derivative formulas for each type of function.

Derivative of a constant

Evidence 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 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 down the limit of the ratio of the increment of a 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 “zero divided by zero,” since the numerator does not contain an infinitesimal value, but precisely 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 throughout the entire domain of definition.

Example 1

The constant functions are given:

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 is the derivative of zero (zero is an integer). Finally, in the fifth case we have the derivative of the rational fraction - 8 7.

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

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

Derivative of a power function

Let's move on 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.

Evidence 2

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

We again rely on the definition of a derivative. Let's write down the limit of the ratio of the increment of a 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

Thus:

(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

Thus, we have proven the formula for the derivative of a power function when the exponent is a natural number.

Evidence 3

To provide evidence 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 a logarithmic function). To have a more complete understanding, it is advisable to study the derivative of a logarithmic function and additionally understand the derivative of an implicit function and the derivative of a complex function.

Let's consider two cases: when x positive and when x negative.

So x > 0. Then: x p > 0 . Let us logarithm the equality y = x p to 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, we have obtained an implicitly specified function. 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 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 x p< 0 и возможно составить доказательство, используя логарифмическую производную.

If 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) = (- (- x) p) " = - ((- x) p) " = - p · (- x) p - 1 · (- x) " = = p · (- x) p - 1 = p x p - 1

The last transition is possible due to the fact that 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 proven the formula for the derivative of a power function for any real p.

Example 2

Functions given:

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 some of the given functions into tabular form y = x p , based on the properties of the degree, and then use the formula:

f 1 (x) = 1 x 2 3 = x - 2 3 ⇒ f 1 " (x) = - 2 3 x - 2 3 - 1 = - 2 3 x - 5 3 f 2 " (x) = 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 an exponential function

Proof 4

Let us derive the derivative formula using the definition as a basis:

(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, let's 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 transition to a new logarithm base was used.

Let us substitute into 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

Let us remember the second remarkable limit and then we obtain 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

It is necessary 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

Evidence 5

Let us provide a proof of the formula for the derivative of a logarithmic function for any x in the domain of definition and any permissible values ​​of the base a of the logarithm. Based on the definition of 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

From the indicated chain of equalities it is clear that the transformations were based on the property of the logarithm. 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 ln 3 x , f 2 (x) = ln 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) = (ln x) " = 1 x ln e = 1 x

So, the derivative of the natural logarithm is one divided by x.

Derivatives of trigonometric functions

Proof 6

Let's 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 cos x.

We will also prove the formula for the derivative of the cosine:

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 cos x function will be – sin x.

We derive the formulas for the derivatives of 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 arcsine, arccosine, arctangent and arccotangent, so we will not duplicate the material here.

Derivatives of hyperbolic functions

Evidence 7

We can derive the 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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