Meaning of the first derivative. Function derivative

Here is 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 = 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 = log x (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 formulas for derivatives for each type of function.

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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Can be taken out of the sign derivative:

(af(x)"=af" (x).

For example:

Derivative of an algebraic sum several functions (taken in a constant number) is equal to the algebraic sum of their derivatives:

(f 1 (x) + f 2 (x) - f 3 (x))" = f 1 "(x) + f 2 "(x) - f 3 "(x).

For example:

(0.3 x 2 - 2 x + 0.8) "= (0.3 x 2)" - (2 x) "+ (0.8)" = 0.6 x - 2 ( derivative last term equation is zero).

If a function derivative g is nonzero, then the ratio f/g also has final derivative. This property can be written as:

.

Let functions y = f(x) and y = g(x) have finite derivatives at the point x 0 . Then functions f ± g and f g also have final derivatives in this point. Then we get:

(f ± g) ′ = f ′ ± g ′,

(f g) ′ = f ′ g + f g ′.

Derivative of a complex function.

Let function y = f(x) has final derivative at a point x 0 , the function z = s(y) has a finite derivative at the point y 0 = f(x 0).

Then complex function z = s (f(x)) also has a finite derivative at this point. This can be written in the form:

.

Derivative of the inverse function.

Let the function y = f(x) have inverse function x = g(y) on some interval(a, b) and there exists a nonzero final derivative this function at the point x 0 , which belongs to domains, i.e. x 0 ∈ (a, b).

Then inverse function It has derivative at the point y 0 = f(x 0):

.

Derivative of an implicit function.

If a function y = f(x) is implicitly defined equation F(x, y(x)) = 0, then its derivative is found from the condition:

.

They say that function y = f(x) set implicitly, If she identically satisfies the relation:

where F(x, y) is some function of two arguments.

Derivative of a function given parametrically.

If a function y = f(x) is given parametrically using the considered

The derivative of a function is one of the most difficult topics in the school curriculum. Not every graduate will answer the question of what a derivative is.

This article simply and clearly explains what a derivative is and why it is needed.. We will not now strive for mathematical rigor of presentation. The most important thing is to understand the meaning.

Let's remember the definition:

The derivative is the rate of change of the function.

The figure shows graphs of three functions. Which one do you think grows the fastest?

The answer is obvious - the third. It has the highest rate of change, that is, the largest derivative.

Here is another example.

Kostya, Grisha and Matvey got jobs at the same time. Let's see how their income changed during the year:

You can see everything on the chart right away, right? Kostya's income has more than doubled in six months. And Grisha's income also increased, but just a little bit. And Matthew's income decreased to zero. The starting conditions are the same, but the rate of change of the function, i.e. derivative, - different. As for Matvey, the derivative of his income is generally negative.

Intuitively, we can easily estimate the rate of change of a function. But how do we do it?

What we are really looking at is how steeply the graph of the function goes up (or down). In other words, how fast y changes with x. Obviously, the same function at different points can have a different value of the derivative - that is, it can change faster or slower.

The derivative of a function is denoted by .

Let's show how to find using the graph.

A graph of some function is drawn. Take a point on it with an abscissa. Draw a tangent to the graph of the function at this point. We want to evaluate how steeply the graph of the function goes up. A handy value for this is tangent of the slope of the tangent.

The derivative of a function at a point is equal to the tangent of the slope of the tangent drawn to the graph of the function at that point.

Please note - as the angle of inclination of the tangent, we take the angle between the tangent and the positive direction of the axis.

Sometimes students ask what is the tangent to the graph of a function. This is a straight line that has the only common point with the graph in this section, moreover, as shown in our figure. It looks like a tangent to a circle.

Let's find . We remember that the tangent of an acute angle in a right triangle is equal to the ratio of the opposite leg to the adjacent one. From triangle:

We found the derivative using the graph without even knowing the formula of the function. Such tasks are often found in the exam in mathematics under the number.

There is another important correlation. Recall that the straight line is given by the equation

The quantity in this equation is called slope of a straight line. It is equal to the tangent of the angle of inclination of the straight line to the axis.

.

We get that

Let's remember this formula. It expresses the geometric meaning of the derivative.

The derivative of a function at a point is equal to the slope of the tangent drawn to the graph of the function at that point.

In other words, the derivative is equal to the tangent of the slope of the tangent.

We have already said that the same function can have different derivatives at different points. Let's see how the derivative is related to the behavior of the function.

Let's draw a graph of some function. Let this function increase in some areas, and decrease in others, and at different rates. And let this function have maximum and minimum points.

At a point, the function is increasing. The tangent to the graph, drawn at the point, forms an acute angle with the positive direction of the axis. So the derivative is positive at the point.

At the point, our function is decreasing. The tangent at this point forms an obtuse angle with the positive direction of the axis. Since the tangent of an obtuse angle is negative, the derivative at the point is negative.

Here's what happens:

If a function is increasing, its derivative is positive.

If it decreases, its derivative is negative.

And what will happen at the maximum and minimum points? We see that at (maximum point) and (minimum point) the tangent is horizontal. Therefore, the tangent of the slope of the tangent at these points is zero, and the derivative is also zero.

The point is the maximum point. At this point, the increase of the function is replaced by a decrease. Consequently, the sign of the derivative changes at the point from "plus" to "minus".

At the point - the minimum point - the derivative is also equal to zero, but its sign changes from "minus" to "plus".

Conclusion: with the help of the derivative, you can find out everything that interests us about the behavior of the function.

If the derivative is positive, then the function is increasing.

If the derivative is negative, then the function is decreasing.

At the maximum point, the derivative is zero and changes sign from plus to minus.

At the minimum point, the derivative is also zero and changes sign from minus to plus.

We write these findings in the form of a table:

	increases	maximum point	decreasing	minimum point	increases
+	0	-	0	+

Let's make two small clarifications. You will need one of them when solving exam problems. Another - in the first year, with a more serious study of functions and derivatives.

A case is possible when the derivative of a function at some point is equal to zero, but the function has neither a maximum nor a minimum at this point. This so-called :

At a point, the tangent to the graph is horizontal and the derivative is zero. However, before the point the function increased - and after the point it continues to increase. The sign of the derivative does not change - it has remained positive as it was.

It also happens that at the point of maximum or minimum, the derivative does not exist. On the graph, this corresponds to a sharp break, when it is impossible to draw a tangent at a given point.

But how to find the derivative if the function is given not by a graph, but by a formula? In this case, it applies

It is absolutely impossible to solve physical problems or examples in mathematics without knowledge about the derivative and methods for calculating it. The derivative is one of the most important concepts of 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 the derivative

Let there be a function f(x) , given in some interval (a,b) . The points x and x0 belong to this interval. When x changes, the function itself changes. Argument change - difference of its values x-x0 . This difference is written as delta x and is called argument increment. The change or increment of a function is the difference between the values ​​of the function at two points. Derivative definition:

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 is the point in finding such a limit? But which one:

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.

The physical meaning of the derivative: the time derivative of the path is equal to the speed of the rectilinear motion.

Indeed, since school days, everyone knows that speed is a private path. x=f(t) and time t . Average speed over a certain period of time:

To find out the speed of movement at a time t0 you need to calculate the limit:

Rule one: take out the constant

The constant can be taken out of the sign of the derivative. Moreover, it must be done. When solving examples in mathematics, take as a rule - if you can simplify the expression, be sure to simplify .

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 a function:

Rule three: the 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:

Here it is important to say about the calculation of derivatives of complex functions. The derivative of a complex function is equal to the product of the derivative of this function with respect to the intermediate argument by the derivative of the intermediate argument with respect to the independent variable.

In the above example, we encounter the expression:

In this case, the intermediate argument is 8x to the fifth power. In order to calculate the derivative of such an expression, we first consider 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: The derivative of the quotient of two functions

Formula for determining the derivative of a quotient of two functions:

We tried to talk about derivatives for dummies from scratch. This topic is not as simple as it sounds, so be warned: there are often pitfalls in the examples, so be careful when calculating derivatives.

With any question on this and other topics, you can contact the student service. In a short time, we will help you solve the most difficult control and deal with tasks, even if you have never dealt with the calculation of derivatives before.