Rules of differentiation THEOREM 1. Differentiation of sum, product and quotient. If the functions f and g are differentiable at a point x, then f + g, f g, f /g are differentiable at this point (if g(x) 0) and, moreover, Let y = f g. 1) (f (x) + g (x)) "= f" (x) + g "(x); 2) (f (x) g (x))" = f "(x) g (x) + f(x)g "(x); Proof. We present the proof of property 2. f = f (x + x) – f(x) f (x + x) = f(x) + f ; g \u003d g (x + x) - g (x) g (x + x) \u003d g (x) + g. g "(x) f" (x) 0 at x 0 (Due to the implicit differential function.)

THEOREM 2. Differentiation complex function Let the function y = f(u) be differentiable at the point u 0, y 0 = f(u 0), and the function u = (x) be differentiable at the point x 0, u 0 = (x 0). Then the complex function y \u003d f ((x)) is differentiable at the point x 0 and f "((x 0)) \u003d f" (u 0) "(x 0) or NOTE. The rule for calculating the derivative of a complex function extends to the composition of any finite number of functions. For example: (f ((g (x))))" = f "((g (x))) "(g (x)) g" (x). Corollary. If f (x) is differentiable at the point x and C \u003d const, then (C f (x))" \u003d C f "(x); (f (x) / C)" \u003d f "(x) / C.

Example 1. y \u003d cosx, x R. (cosx) \u003d (sin (/ 2 - x)) \u003d cos (/ 2 - x) (/ 2 - x) \u003d - sinx. y = tgx, x /2 + k, k Z. Using Theorems 1 and 2, we find the derivatives trigonometric functions y = ctgx, x + k, k Z.

THEOREM 3. Differentiation of the inverse function. If y \u003d f (x) is continuous and strictly monotone on the segment and has a derivative f "(x 0), then the function inverse to it x \u003d g (y) is differentiable at the point y 0 \u003d f (x 0), and g "( y 0) \u003d 1/ f "(x 0). x0x0 x 0 - x 0 + y0y0 x y x y \u003d f (x) x \u003d g (y) Let y be such that y 0 + y (,). Denote x = g(y 0 + y) - g(y 0) It is necessary to prove that 0 exists Proof Let f(x) strictly increase by .Let = f(x 0 -), = f(x 0 +) Then on [,] the inverse function x = g(y) is defined, continuous and strictly increasing, and f(x 0) (,). y, then so does x, since x = g(y) is continuous at y 0.

Example 2. Find the derivatives of inverse trigonometric functions

0; (x n)´ = n x n-1, n N, x R; 3)(a x)´ = a x lna, a > 0, a 1, x R; (e x)´ = e x, x R; four). 5)(sin x) = cos x, x R; 6)(cos x) = - sin x, x R; 7)(tg x) = 1/ cos 2 " title="(!LANG: Table of derivatives elementary functions 1)(C)´= 0, C = const; 2)(x)´ = x -1, R, x > 0; (x n)´ = n x n-1, n N, x R; 3)(a x)´ = a x lna, a > 0, a 1, x R; (e x)´ = e x, x R; four). 5)(sin x) = cos x, x R; 6)(cos x) = - sin x, x R; 7)(tg x) = 1/ cos 2 "class="link_thumb"> 8 Table of derivatives of elementary functions 1)(С)´= 0, C = const; 2)(x)´ = x -1, R, x > 0; (x n)´ = n x n-1, n N, x R; 3)(a x)´ = a x lna, a > 0, a 1, x R; (e x)´ = e x, x R; four). 5)(sin x) = cos x, x R; 6)(cos x) = - sin x, x R; 7)(tg x) \u003d 1 / cos 2 x, x π / 2 + πn, n; 8) (ctg x) \u003d - 1 / sin 2 x, x πn, n; 9)10) 11)12) 0; (x n)´ = n x n-1, n N, x R; 3)(a x)´ = a x lna, a > 0, a 1, x R; (e x)´ = e x, x R; four). 5)(sin x) = cos x, x R; 6)(cos x) = - sin x, x R; 7)(tg x) = 1/ cos 2 "> 0; (x n)´ = n x n-1, n N, x R; 3)(a x)´ = a x lna, a > 0, a 1, x R ; (e x)´ = e x, x R; 4) 5)(sin x) = cos x, x R; 6)(cos x) = - sin x, x R; 7)(tg x) = 1/ cos 2 x, x π / 2 + πn, n; 8) (ctg x) \u003d - 1 / sin 2 x, x πn, n; 9) 10) 11) 12) "\u003e 0; (x n)´ = n x n-1, n N, x R; 3)(a x)´ = a x lna, a > 0, a 1, x R; (e x)´ = e x, x R; four). 5)(sin x) = cos x, x R; 6)(cos x) = - sin x, x R; 7)(tg x) = 1/ cos 2 "title="(!LANG: Table of derivatives of elementary functions 1)(С)´= 0, C = const; 2)(x)´ = x -1, R, x > 0; (x n)´ = n x n-1, n N, x R; 3)(a x)´ = a x lna, a > 0, a 1, x R; (e x)´ = e x, x R; four). 5)(sin x) = cos x, x R; 6)(cos x) = - sin x, x R; 7)(tg x) = 1/ cos 2"> title="Table of derivatives of elementary functions 1)(С)´= 0, C = const; 2)(x)´ = x -1, R, x > 0; (x n)´ = n x n-1, n N, x R; 3)(a x)´ = a x lna, a > 0, a 1, x R; (e x)´ = e x, x R; four). 5)(sin x) = cos x, x R; 6)(cos x) = - sin x, x R; 7)(tg x) = 1/ cos 2"> !}

Derivative of the n-th order DEFINITION. Let f(x) be defined in U (x 0) and have a derivative f (x) at every point of this interval. If at the point x 0 there is a derivative of f (x), then it is called the second derivative of the function f (x) at this point and is denoted. Similarly, the derivative f (n) (x) of any order n \u003d 1, 2, ... If in U (x 0) exists f (n-1) (x) (in this case, the zero-order derivative means the function itself), then n = 1, 2, 3, .... A function that has derivatives up to the nth order inclusive at each point of the set X is called n times differentiable on the set X.

Let the functions f(x) and g(x) have nth order derivatives at the point x. Then the function Аf(x) + Вg(x), where А and В are constants, also has a derivative at the point x, and (Аf(x) + Вg(x)) (n) = Аf (n) (x) + Вg (n)(x). When calculating derivatives of any order, the following basic formulas are often used. y=x; y (n) = (-1)… (- (n-1)) x - n. y = x -1, y = (-1)x -2, y = (-1)(-2) x -3 ... In particular, if = m N, then y = a x ; y (n) = a x (lna) n. y \u003d a x lna, y \u003d a x (lna) 2, y \u003d a x (lna) 3, ... In particular, (e x) (n) \u003d e x. y "= ((x + a) - 1)" = - (x+a) - 2, y "" = 2 (x + a) - 3, y """ = (x + a) - 4, …

Y = ln(x+a); y (n) \u003d (-1) n-1 (n-1)! (x + a) -n. y \u003d (x + a) -1, y \u003d - (x + a) -2, y \u003d 2 (x + a) -3, y (4) \u003d - 2 3 (x + a) - 4, ... y = sinαx; y (n) = α n sin(αx+n /2) y = α cos αx = α sin(αx+ /2), y = α 2 cos(αx+ /2) = α 2 sin(αx+2 / 2), y = α 3 cos(αx + 2 /2) = α 3 sin(αx+3 /2), … y = cos αx; y (n) = α n cos(αx+n /2) y = – α sin αx = α cos(αx+ /2), y = – α 2 sin(αx+ /2) = α 2 cos(αx + 2 /2), y = – α 3 sin(αx+2 /2) = α 3 cos(αx + 3 /2),...

N-th derivative of the product of two functions (Leibniz formula) where This formula is called the Leibniz formula. It can be written in the form where Let the functions f(x) and g(x) have nth order derivatives at the point x. By induction, we can prove that (f(x) g(x)) (n) = ?
Example 5. y \u003d (x 2 + 3x + 5) sin x, y (13) \u003d? = sin(x +13π /2) (x 2 +3x+5) + 13 sin (x +12π /2) (2x+3) + 78 sin (x +11π /2) 2 = = cos x (x 2 +3x+5) + 13 sin x (2x+3) + 78 (- cos x) 2 = = (x 2 +3x -151) cos x + 13 (2x+3) sin x. We apply the Leibniz formula, putting in it f (x) \u003d sin x, g (x) \u003d (x 2 + 3x + 5). Then

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Derivatives of elementary functions. Generalizing repetition lesson Grade 11 Kruglova A.N., mathematics teacher, secondary school No. 186

Lesson objectives 1. Generalize and consolidate the concept of a derivative. 2. Repeat the concept of the limit of a function and its continuity, the concept of a derivative. 3. Repeat the rules of differentiation, derivatives of power and some elementary functions. 4. Apply this knowledge in differentiation. 5. Implementation of an individual mode of operation.

History reference. The term "function" was first used in 1692 by the German mathematician G. Leibniz. In 1748 L. Euler defined the function and introduced the symbol f(x). In 1834, N.I. Lobachevsky gave a definition of a function based on the idea of ​​correspondence between two numerical sets. In 1837, the German mathematician P. Dirichlet formulated the generalized concept of a function: “y is a function of the variable x on the segment if each value of x corresponds to a certain value of y, and it doesn’t matter how this correspondence is established - by a formula, graph, table or verbal description ". The first definition of the limit was given by the English mathematician D. Vallis (1616-1703). The method of limits was developed in the works of the English scientist I. Newton (1643-1727), he also introduced the symbol lim. A significant contribution to the development of differential calculus was made by the French scientists P. Fermat (1601-1665) and R. Descartes (1596-1650). Newton came to the concept of a derivative by solving problems in mechanics related to finding the instantaneous velocity. The term "derivative" was introduced in 1800 by the French mathematician L. Arbogast (1759-1803). The notation for the derivative y’ and f(x)’ was introduced by the French mathematician J. Lagrange (1736-1813). An essential approximation of the theory of differential calculus to its modern presentation began the work of the French mathematician O. Cauchy (1789-1857).

Function limit. Construct graphs of functions 1) y \u003d x + 1 2) x ² - 1 x - 1 for x 1 y \u003d 3 for x \u003d 1 3) y \u003d (x ² - 1): (x - 1) Answer questions a) What are function graphs? Straight lines b) Through what points on the coordinate axes do the graphs pass? (0;1) and (-1;0) c) What is the difference between graphs? The second and third graphs with a "punched" point (1; 2), but on the second graph for x = 1, the value of the function is 3.

Function graphs. y y y x x x 1 2 3

Conclusion General property functions for x values ​​close to 1? The values ​​of each of the functions differ little from 2. Therefore, each of these functions has a limit equal to 2 at the point x = 1. How to write this? However, for the first function lim y(x) = y(1) = 2 For the second function lim y(x) ≠ y(1) , for the third function y(1) does not exist. The first function is called continuous, and the second and third functions are called discontinuous at the point x = 1. lim y(x) = 2 x 1

Definition of the derivative The derivative of the function f (x) at the point x 0 is called the limit of the difference relation f (x 0 + h) - f (x 0) h for h → 0: ƒ‘(x 0) = lim The operation of finding the derivative is called differentiation. 0h

Derivative of power and some elementary functions. (Find the continuation of formulas on the right side) (x ⁿ) " = 1 2 3 4 5 6 () ' = 1 2 3 4 5 6 3. (ln x)' = 1 2 3 4 5 6 4. (sin x) ' = 1 2 3 4 5 6 (cos x)' = 1 2 3 4 5 6 Continue = cos x = - sin x = = tg x = 1/x = nx ⁿˉ¹

Solve examples 1) (x ³)' = 2) (2 x)' = 3) ()' = 4) (lnx)' = 5) (-4 lnx)' = 6) (3)' = 7) ( 5 cosx)' = 8) (0.3 sinx)' = 3x ² 2 - 10 x ˉ ³ 1 / x - 4 / x 3 e - 5 sinx 0.3 cosx

Differentiation rules. Derivative of the sum (f(x) + g(x))' = f'(x) – g'(x) = f'(x) + g'(x) = f'(x) * g'(x ) Constant factor (cf(x))' = = c + f'(x) = f'(x) – c = cf'(x) Derivative of the product (f(x) g(x))' = f' (x) g(x) + f(x) g'(x) = f'(x) g'(x) = f'(x) g(x) g(x))' = f'(x)/g'(x) = (f'(x) g(x) - f(x) g'(x)) / g²(x) = f' (x) g(x) – f(x) g'(x) Let's continue the lesson.

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DERIVATIVE

MOU Srednesantimirskaya secondary school

Done by a math teacher

Singatullova G.Sh.

• Definition of a derivative.
• physical meaning derivative.
• .

• Basic rules of differentiation.
• Derivative of a complex function.
• Examples of solving problems on the topic derivative.

Derivative Definition

Let on some interval (a, b) the function y= f(x). Let's take any point x 0 from this interval and set the argument x at the point x 0 to an arbitrary increment ∆ x such that the point x 0 + ∆ x belongs to this interval. The function will be incremented

derivative functions y= f(x) at the point x \u003d x 0 the limit of the ratio of the increment of the function ∆y at this point to the increment of the argument ∆x is called, as the increment of the argument tends to zero.

The geometric meaning of the derivative

Let the function y= f(x) is defined on some interval (a, b). Then the tangent of the slope of the secant MP to the graph of the function.

Where  is the slope of the tangent function f(x) at the point (x 0 , f(x 0)).

The angle between curves can be defined as the angle between tangents drawn to these curves at some point.

Equation of a tangent to a curve:

The physical meaning of the derivative 1. The problem of determining the speed of motion of a material particle

Let a point move along some straight line according to the law s= s(t), where s is the distance traveled, t is the time, and it is necessary to find the speed of the point at the moment t 0 .

By the time t 0, the distance traveled is equal to s 0 = s(t 0), and by the time (t 0 + ∆t) - the path s 0 + ∆s=s(t 0 + ∆t).

Then, over the interval ∆t, the average speed will be

The smaller ∆t, the better the average speed characterizes the movement of a point at the moment t 0 . Therefore, under the speed of the point at the moment t 0 one should understand the limit of the average speed for the interval from t 0 to t 0 +∆t, when ∆t⇾0 , i.e.

2. THE PROBLEM OF THE VELOCITY OF THE CHEMICAL REACTIONS

Let some substance enter into a chemical reaction. The amount of this substance Q changes during the reaction depending on the time t and is a function of time. Let the amount of substance change by ∆Q during the time ∆t, then the ratio will express average speed chemical reaction over time ∆t, and the limit of this ratio

Current rate of a chemical reaction

time t.

3. A TASK DETERMINATIONS OF THE RADIOACTIVE DECAY RATE

If m is the mass of the radioactive substance and t is the time, then the phenomenon of radioactive decay at time t, provided that the mass of the radioactive substance decreases over time, is characterized by the function m = m(t).

The average decay rate over time ∆t is expressed by the ratio

and the instantaneous decay rate at time t

ALGORITHM for calculating the derivative

The derivative of the function y= f(x) can be found as follows:

1. Let's increment ∆x≠0 to the argument x and find the accumulated value of the function y+∆y= f(x+∆x).

2. Find the increment of the function ∆y= f(x+∆x) - f(x).

3. We make a relationship

4. Find the limit of this ratio at ∆x⇾0, i.e.

(if this limit exists).

Basic rules of differentiation

Let u=u(x) and v=v(x) - differentiable functions at the point x.

1) (u  v)  =u   v 

2) (uv)  =u  v+uv 

(cu)  = cu 

3) , if v  0

Derivative of a compound function

Theorem. If the function is differentiable at a point x and the function

is differentiable at the corresponding point, then the complex function is differentiable at the point x, and:

those. the derivative of a complex function is equal to the product of the derivative of the function with respect to the intermediate argument by the derivative of the intermediate argument with respect to x.

Task 1.

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Task 8 .