When the derivative of a function is positive. Research of functions. You need to know this
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First derivative If the derivative of a function is positive (negative) in a certain interval, then the function in this interval monotonically increases (monotonically decreases). If the derivative of a function is positive (negative) in a certain interval, then the function monotonically increases (monotonically decreases) in this interval. Further
Definition A curve is called convex at a point if in some neighborhood of this point it is located under its tangent at a point. A curve is called convex at a point if in some neighborhood of this point it is located under its tangent at a point. A curve is called concave at a point if in some neighborhood of this point it is located above its tangent at a point A curve is called concave at a point if in some neighborhood of this point it is located above its tangent at a point Next
Sign of concavity and convexity If the second derivative of a function in a given interval is positive, then the curve is concave in this interval, and if it is negative, it is convex in this interval. If the second derivative of a function in a given interval is positive, then the curve is concave in this interval, and if it is negative, it is convex in this interval. Definition
At a given interval, the function has 2 maximums and 2 minimums, for a total of 4 extrema. Assignment The figure shows a graph of the derivative of a function defined on an interval. Solution On a given interval, the derivative of a function is positive, so the function increases on this interval. Solution If the derivative at a certain point is equal to zero, and in its vicinity changes sign, then this is an extremum point.
Calculation of the derivative value. Two point method
1. Using the derivative graph, examine the function. The function y=f(x) decreases on the intervals (x1;x2) and (x3;x4). Using the graph of the derivative y=f ‘(x) you can also compare the values of the function y=f(x).
Let's denote these points as A (x1; y1) and B (x2; y2). Write down the coordinates correctly - this is the key point of the solution, and any mistake here leads to an incorrect answer.
IN physical sense derivative is the rate of change of any process. Material point moves rectilinearly according to the law x(t) = t²-13t+23, where x is the distance from the reference point in meters, t is the time in seconds measured from the beginning of the movement.
Tangent to a circle, ellipse, hyperbola, parabola.
Let me remind you that it sounds like this: a function is called increasing/decreasing on an interval if a larger argument of the function corresponds to a larger/smaller value of the function. But please look at your solution to problem 7089. There, when specifying increasing intervals, boundaries are not included. Please note that the derivative graph is given. As usual: the punctured point does not lie on the graph, the values in it do not exist and are not considered. Well-prepared children distinguish between the concepts “derivative” and “second derivative.” You are confusing: if the derivative were 0, then at the point the function could have a minimum or maximum. Negative values the derivative corresponds to the intervals over which the function f(x) decreases.
Up to this point, we have been working on finding equations for tangents to graphs of single-valued functions of the form y = f(x) at various points.
The figure below shows three actually different secants (points A and B are different), but they coincide and are given by one equation. But still, if we start from the definition, then the straight line and its secant line coincide. Let's start finding the coordinates of the tangent points. Please pay attention to it, since later we will use it when calculating the ordinates of the tangent points. A hyperbola with a center at a point and vertices and is given by equality (the figure below on the left), and with vertices and by equality (the figure below on the right). A logical question arises: how to determine which function a point belongs to. To answer it, we substitute the coordinates into each equation and see which of the equalities turns into an identity.
Sometimes students ask what a tangent to the graph of a function is. This is a straight line that has a single common point with the graph in this section, and as shown in our figure. It looks like a tangent to a circle. We'll find it. We remember that tangent acute angle V right triangle equal to the ratio of the opposite side to the adjacent side. On the graph, this corresponds to a sharp break, when it is impossible to draw a tangent at a given point. How to find the derivative if the function is given not by a graph, but by a formula?
Definition. Let the function \(y = f(x)\) be defined in a certain interval containing the point \(x_0\). Let's give the argument an increment \(\Delta x \) such that it does not leave this interval. Let's find the corresponding increment of the function \(\Delta y \) (when moving from the point \(x_0 \) to the point \(x_0 + \Delta x \)) and compose the relation \(\frac(\Delta y)(\Delta x) \). If there is a limit to this ratio at \(\Delta x \rightarrow 0\), then the specified limit is called derivative of a function\(y=f(x) \) at the point \(x_0 \) and denote \(f"(x_0) \).
$$ \lim_(\Delta x \to 0) \frac(\Delta y)(\Delta x) = f"(x_0) $$
The symbol y is often used to denote the derivative. Note that y" = f(x) is a new function, but naturally related to the function y = f(x), defined at all points x at which the above limit exists . This function is called like this: derivative of the function y = f(x).
Geometric meaning of derivative is as follows. If it is possible to draw a tangent to the graph of the function y = f(x) at the point with abscissa x=a, which is not parallel to the y-axis, then f(a) expresses the slope of the tangent:
\(k = f"(a)\)
Since \(k = tg(a) \), then the equality \(f"(a) = tan(a) \) is true.
Now let’s interpret the definition of derivative from the point of view of approximate equalities. Let the function \(y = f(x)\) have a derivative at a specific point \(x\):
$$ \lim_(\Delta x \to 0) \frac(\Delta y)(\Delta x) = f"(x) $$
This means that near the point x the approximate equality \(\frac(\Delta y)(\Delta x) \approx f"(x)\), i.e. \(\Delta y \approx f"(x) \cdot\Delta x\). The meaningful meaning of the resulting approximate equality is as follows: the increment of the function is “almost proportional” to the increment of the argument, and the coefficient of proportionality is the value of the derivative in given point X. For example, for the function \(y = x^2\) the approximate equality \(\Delta y \approx 2x \cdot \Delta x \) is valid. If we carefully analyze the definition of a derivative, we will find that it contains an algorithm for finding it.
Let's formulate it.
How to find the derivative of the function y = f(x)?
1. Fix the value of \(x\), find \(f(x)\)
2. Give the argument \(x\) an increment \(\Delta x\), go to a new point \(x+ \Delta x \), find \(f(x+ \Delta x) \)
3. Find the increment of the function: \(\Delta y = f(x + \Delta x) - f(x) \)
4. Create the relation \(\frac(\Delta y)(\Delta x) \)
5. Calculate $$ \lim_(\Delta x \to 0) \frac(\Delta y)(\Delta x) $$
This limit is the derivative of the function at point x.
If a function y = f(x) has a derivative at a point x, then it is called differentiable at a point x. The procedure for finding the derivative of the function y = f(x) is called differentiation functions y = f(x).
Let us discuss the following question: how are continuity and differentiability of a function at a point related to each other?
Let the function y = f(x) be differentiable at the point x. Then a tangent can be drawn to the graph of the function at point M(x; f(x)), and, recall, the angular coefficient of the tangent is equal to f "(x). Such a graph cannot “break” at point M, i.e. the function must be continuous at point x.
These were “hands-on” arguments. Let us give a more rigorous reasoning. If the function y = f(x) is differentiable at the point x, then the approximate equality \(\Delta y \approx f"(x) \cdot \Delta x\) holds. If in this equality \(\Delta x \) tends to zero, then \(\Delta y \) will tend to zero, and this is the condition for the continuity of the function at a point.
So, if a function is differentiable at a point x, then it is continuous at that point.
The reverse statement is not true. For example: function y = |x| is continuous everywhere, in particular at the point x = 0, but the tangent to the graph of the function at the “junction point” (0; 0) does not exist. If at some point a tangent cannot be drawn to the graph of a function, then the derivative does not exist at that point.
One more example. The function \(y=\sqrt(x)\) is continuous on the entire number line, including at the point x = 0. And the tangent to the graph of the function exists at any point, including at the point x = 0. But at this point the tangent coincides with the y-axis, i.e., it is perpendicular to the abscissa axis, its equation has the form x = 0. Slope coefficient such a line does not have, which means that \(f"(0) \) does not exist either
So, we got acquainted with a new property of a function - differentiability. How can one conclude from the graph of a function that it is differentiable?
The answer is actually given above. If at some point it is possible to draw a tangent to the graph of a function that is not perpendicular to the abscissa axis, then at this point the function is differentiable. If at some point the tangent to the graph of a function does not exist or it is perpendicular to the abscissa axis, then at this point the function is not differentiable.
Rules of differentiation
The operation of finding the derivative is called differentiation. When performing this operation, you often have to work with quotients, sums, products of functions, as well as “functions of functions,” that is, complex functions. Based on the definition of derivative, we can derive differentiation rules that make this work easier. If C is a constant number and f=f(x), g=g(x) are some differentiable functions, then the following are true differentiation rules:
$$ f"_x(g(x)) = f"_g \cdot g"_x $$
Table of derivatives of some functions
$$ \left(\frac(1)(x) \right) " = -\frac(1)(x^2) $$ $$ (\sqrt(x)) " = \frac(1)(2\ sqrt(x)) $$ $$ \left(x^a \right) " = a x^(a-1) $$ $$ \left(a^x \right) " = a^x \cdot \ln a $$ $$ \left(e^x \right) " = e^x $$ $$ (\ln x)" = \frac(1)(x) $$ $$ (\log_a x)" = \frac (1)(x\ln a) $$ $$ (\sin x)" = \cos x $$ $$ (\cos x)" = -\sin x $$ $$ (\text(tg) x) " = \frac(1)(\cos^2 x) $$ $$ (\text(ctg) x)" = -\frac(1)(\sin^2 x) $$ $$ (\arcsin x) " = \frac(1)(\sqrt(1-x^2)) $$ $$ (\arccos x)" = \frac(-1)(\sqrt(1-x^2)) $$ $$ (\text(arctg) x)" = \frac(1)(1+x^2) $$ $$ (\text(arcctg) x)" = \frac(-1)(1+x^2) $ $The content of the article
DERIVATIVE– derivative of the function y = f(x), given on a certain interval ( a, b) at point x of this interval is called the limit to which the ratio of the increment of the function tends f at this point to the corresponding increment of the argument when the increment of the argument tends to zero.
The derivative is usually denoted as follows:
Other designations are also widely used:
Instant speed.
Let the point M moves in a straight line. Distance s moving point, counted from some initial position M 0 , depends on time t, i.e. s there is a function of time t: s= f(t). Let at some point in time t moving point M was at a distance s from the starting position M 0, and at some next moment t+D t found herself in a position M 1 - on distance s+D s from the initial position ( see pic.).
Thus, over a period of time D t distance s changed by the amount D s. In this case they say that during the time interval D t magnitude s received increment D s.
The average speed cannot in all cases accurately characterize the speed of movement of a point M at a point in time t. If, for example, the body at the beginning of the interval D t moved very quickly, and at the end very slowly, then average speed will not be able to reflect the specified features of the point’s movement and give an idea of the true speed of its movement at the moment t. To more accurately express the true speed using the average speed, you need to take a shorter period of time D t. Most fully characterizes the speed of movement of a point at the moment t the limit to which the average speed tends at D t® 0. This limit is called the speed of movement in this moment:
Thus, the speed of movement at a given moment is called the limit of the path increment ratio D s to time increment D t, when the time increment tends to zero. Because
Geometric meaning of the derivative. Tangent to the graph of a function.
The construction of tangent lines is one of those problems that led to the birth of differential calculus. The first published work related to differential calculus, written by Leibniz, was entitled New method maxima and minima, as well as tangents, for which neither fractional nor irrational quantities, and a special type of calculus for this, serve as an obstacle.
Let the curve be the graph of the function y =f(x) in a rectangular coordinate system ( cm. rice.).
At some value x function matters y =f(x). These values x And y the point on the curve corresponds M 0(x, y). If the argument x give increment D x, then the new value of the argument x+D x corresponds to the new function value y+ D y = f(x + D x). The corresponding point of the curve will be the point M 1(x+D x,y+D y). If you draw a secant M 0M 1 and denoted by j the angle formed by a transversal with the positive direction of the axis Ox, it is immediately clear from the figure that .
If now D x tends to zero, then the point M 1 moves along the curve, approaching the point M 0, and angle j changes with D x. At Dx® 0 the angle j tends to a certain limit a and the straight line passing through the point M 0 and the component with the positive direction of the x-axis, angle a, will be the desired tangent. Its slope is:
Hence, f´( x) = tga
those. derivative value f´( x) at given value argument x equals the tangent of the angle formed by the tangent to the graph of the function f(x) at the corresponding point M 0(x,y) with positive axis direction Ox.
Differentiability of functions.
Definition. If the function y = f(x) has a derivative at the point x = x 0, then the function is differentiable at this point.
Continuity of a function having a derivative. Theorem.
If the function y = f(x) is differentiable at some point x = x 0, then it is continuous at this point.
Thus, the function cannot have a derivative at discontinuity points. The opposite conclusion is incorrect, i.e. from the fact that at some point x = x 0 function y = f(x) is continuous does not mean that it is differentiable at this point. For example, the function y = |x| continuous for everyone x(–Ґ x x = 0 has no derivative. At this point there is no tangent to the graph. There is a right tangent and a left one, but they do not coincide.
Some theorems on differentiable functions. Theorem on the roots of the derivative (Rolle's theorem). If the function f(x) is continuous on the segment [a,b], is differentiable at all interior points of this segment and at the ends x = a And x = b goes to zero ( f(a) = f(b) = 0), then inside the segment [ a,b] exists, according to at least one, point x= With, a c b, in which the derivative fў( x) goes to zero, i.e. fў( c) = 0.
Finite increment theorem (Lagrange's theorem). If the function f(x) is continuous on the interval [ a, b] and is differentiable at all interior points of this segment, then inside the segment [ a, b] there is at least one point With, a c b that
f(b) – f(a) = fў( c)(b– a).
Theorem on the ratio of the increments of two functions (Cauchy's theorem). If f(x) And g(x) – two functions continuous on the segment [a, b] and differentiable at all interior points of this segment, and gў( x) does not vanish anywhere inside this segment, then inside the segment [ a, b] there is such a point x = With, a c b that
Derivatives of various orders.
Let the function y =f(x) is differentiable on some interval [ a, b]. Derivative values f ў( x), generally speaking, depend on x, i.e. derivative f ў( x) is also a function of x. When differentiating this function, we obtain the so-called second derivative of the function f(x), which is denoted f ўў ( x).
Derivative n- th order of function f(x) is called the (first order) derivative of the derivative n- 1- th and is denoted by the symbol y(n) = (y(n– 1))ў.
Differentials of various orders.
Function differential y = f(x), Where x– independent variable, yes dy = f ў( x)dx, some function from x, but from x only the first factor can depend f ў( x), the second factor ( dx) is the increment of the independent variable x and does not depend on the value of this variable. Because dy there is a function from x, then we can determine the differential of this function. The differential of the differential of a function is called the second differential or second-order differential of this function and is denoted d 2y:
d(dx) = d 2y = f ўў( x)(dx) 2 .
Differential n- of the first order is called the first differential of the differential n- 1- th order:
d n y = d(dn–1y) = f(n)(x)dx(n).
Partial derivative.
If a function depends not on one, but on several arguments x i(i varies from 1 to n,i= 1, 2,… n),f(x 1,x 2,… x n), then in differential calculus the concept of partial derivative is introduced, which characterizes the rate of change of a function of several variables when only one argument changes, for example, x i. 1st order partial derivative with respect to x i is defined as an ordinary derivative, and it is assumed that all arguments except x i, keep constant values. For partial derivatives, the notation is introduced
The 1st order partial derivatives defined in this way (as functions of the same arguments) can, in turn, also have partial derivatives, these are second order partial derivatives, etc. Taken by different arguments such derivatives are called mixed. Continuous mixed derivatives of the same order do not depend on the order of differentiation and are equal to each other.
Anna Chugainova
In this lesson we will learn to apply formulas and rules of differentiation.
Examples. Find derivatives of functions.
1. y=x 7 +x 5 -x 4 +x 3 -x 2 +x-9. Applying the rule I, formulas 4, 2 and 1. We get:
y’=7x 6 +5x 4 -4x 3 +3x 2 -2x+1.
2. y=3x 6 -2x+5. We solve similarly, using the same formulas and formula 3.
y’=3∙6x 5 -2=18x 5 -2.
Applying the rule I, formulas 3, 5
And 6
And 1.
Applying the rule IV, formulas 5 And 1 .
In the fifth example, according to the rule I the derivative of the sum is equal to the sum of the derivatives, and we just found the derivative of the 1st term (example 4 ), therefore, we will find derivatives 2nd And 3rd terms, and for 1st summand we can immediately write the result.
Let's differentiate 2nd And 3rd terms according to the formula 4
. To do this, we transform the roots of the third and fourth powers in the denominators to powers with negative exponents, and then, according to 4
formula, we find derivatives of powers.
Look at this example and the result. Did you catch the pattern? Fine. This means we have a new formula and can add it to our derivatives table.
Let's solve the sixth example and derive another formula.
Let's use the rule IV and formula 4
. Let's reduce the resulting fractions.
Let's look at this function and its derivative. You, of course, understand the pattern and are ready to name the formula:
Learning new formulas!
Examples.
1. Find the increment of the argument and the increment of the function y= x 2, if the initial value of the argument was equal to 4 , and new - 4,01 .
Solution.
New argument value x=x 0 +Δx. Let's substitute the data: 4.01=4+Δx, hence the increment of the argument Δх=4.01-4=0.01. The increment of a function, by definition, is equal to the difference between the new and previous values of the function, i.e. Δy=f (x 0 +Δx) - f (x 0). Since we have a function y=x2, That Δу=(x 0 +Δx) 2 - (x 0) 2 =(x 0) 2 +2x 0 · Δx+(Δx) 2 - (x 0) 2 =2x 0 · Δx+(Δx) 2 =
2 · 4 · 0,01+(0,01) 2 =0,08+0,0001=0,0801.
Answer: argument increment Δх=0.01; function increment Δу=0,0801.
The function increment could be found differently: Δy=y (x 0 +Δx) -y (x 0)=y(4.01) -y(4)=4.01 2 -4 2 =16.0801-16=0.0801.
2. Find the angle of inclination of the tangent to the graph of the function y=f(x) at the point x 0, If f "(x 0) = 1.
Solution.
The value of the derivative at the point of tangency x 0 and is the value of the tangent of the tangent angle ( geometric meaning derivative). We have: f "(x 0) = tanα = 1 → α = 45°, because tg45°=1.
Answer: the tangent to the graph of this function forms an angle with the positive direction of the Ox axis equal to 45°.
3. Derive the formula for the derivative of the function y=x n.
Differentiation is the action of finding the derivative of a function.
When finding derivatives, use formulas that were derived based on the definition of a derivative, in the same way as we derived the formula for the derivative degree: (x n)" = nx n-1.
These are the formulas.
Table of derivatives It will be easier to memorize by pronouncing verbal formulations:
1. Derivative constant value equal to zero.
2. X prime is equal to one.
3. The constant factor can be taken out of the sign of the derivative.
4. The derivative of a degree is equal to the product of the exponent of this degree by a degree with the same base, but the exponent is one less.
5. The derivative of a root is equal to one divided by two equal roots.
6. The derivative of one divided by x is equal to minus one divided by x squared.
7. The derivative of the sine is equal to the cosine.
8. The derivative of the cosine is equal to minus sine.
9. The derivative of the tangent is equal to one divided by the square of the cosine.
10. The derivative of the cotangent is equal to minus one divided by the square of the sine.
We teach differentiation rules.
1.
The derivative of an algebraic sum is equal to the algebraic sum of the derivatives of the terms.
2. The derivative of a product is equal to the product of the derivative of the first factor and the second plus the product of the first factor and the derivative of the second.
3. The derivative of “y” divided by “ve” is equal to a fraction in which the numerator is “y prime multiplied by “ve” minus “y multiplied by ve prime”, and the denominator is “ve squared”.
4. A special case of the formula 3.
Let's learn together!
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