Learn to take derivatives of functions. The derivative characterizes the rate of change of a function at a certain point lying on the graph of this function. IN in this case The graph can be either a straight or curved line. That is, the derivative characterizes the rate of change of a function at a specific point in time. Remember general rules, by which derivatives are taken, and only then proceed to the next step.
- Read the article.
- How to take the simplest derivatives, for example, derivative exponential equation, described. The calculations presented in next steps, will be based on the methods described therein.
Learn to distinguish between tasks in which slope needs to be calculated through the derivative of the function. Problems do not always ask you to find the slope or derivative of a function. For example, you may be asked to find the rate of change of a function at point A(x,y). You may also be asked to find the slope of the tangent at point A(x,y). In both cases it is necessary to take the derivative of the function.
Take the derivative of the function given to you. There is no need to build a graph here - you only need the equation of the function. In our example, take the derivative of the function. Take the derivative according to the methods outlined in the article mentioned above:
- Derivative:
Substitute the coordinates of the point given to you into the found derivative to calculate the slope. The derivative of a function is equal to the slope at a certain point. In other words, f"(x) is the slope of the function at any point (x,f(x)). In our example:
- Find the slope of the function f (x) = 2 x 2 + 6 x (\displaystyle f(x)=2x^(2)+6x) at point A(4,2).
- Derivative of the function:
- f ′ (x) = 4 x + 6 (\displaystyle f"(x)=4x+6)
- Substitute the value of the “x” coordinate of this point:
- f ′ (x) = 4 (4) + 6 (\displaystyle f"(x)=4(4)+6)
- Find the slope:
- Slope function f (x) = 2 x 2 + 6 x (\displaystyle f(x)=2x^(2)+6x) at point A(4,2) is equal to 22.
If possible, check your answer on a graph. Remember that the slope cannot be calculated at every point. Differential calculus is considering complex functions and complex graphs, where the slope cannot be calculated at every point, and in some cases the points do not lie on the graphs at all. If possible, use a graphing calculator to check that the slope of the function you are given is correct. IN otherwise draw a tangent to the graph at the point given to you and think about whether the slope value you found corresponds to what you see on the graph.
- The tangent will have the same slope as the graph of the function at a certain point. To draw a tangent at a given point, move left/right on the X axis (in our example, 22 values to the right), and then up one on the Y axis. Mark the point, and then connect it to the point given to you. In our example, connect the points with coordinates (4,2) and (26,3).
You will need
- - mathematical reference book;
- - notebook;
- - a simple pencil;
- - pen;
- - protractor;
- - compass.
Instructions
Please note that the graph of the differentiable function f(x) at the point x0 is no different from the tangent segment. Therefore, it is quite close to the segment l, to the one passing through the points (x0; f(x0)) and (x0+Δx; f(x0 + Δx)). To specify a straight line passing through point A with coefficients (x0; f(x0)), specify its slope. Moreover, it is equal to Δy/Δx secant tangent (Δх→0), and also tends to the number f‘(x0).
If there are no values for f‘(x0), then there is no tangent, or it runs vertically. Based on this, the derivative of the function at the point x0 is explained by the existence of a non-vertical tangent, which is in contact with the graph of the function at the point (x0, f(x0)). In this case, the angular coefficient of the tangent is equal to f "(x0). The geometric derivative, that is, the angular coefficient of the tangent, becomes clear.
That is, in order to find the slope of the tangent, you need to find the value of the derivative of the function at the point of tangency. Example: find the angular coefficient of the tangent to the function y = x³ at the point with abscissa X0 = 1. Solution: Find the derivative of this function y΄(x) = 3x²; find the value of the derivative at the point X0 = 1. у΄(1) = 3 × 1² = 3. The angle coefficient of the tangent at the point X0 = 3.
Draw additional tangents in the figure so that they touch the graph of the function at the points: x1, x2 and x3. Mark the angles formed by these tangents with the abscissa axis (the angle is counted in the positive direction - from the axis to the tangent line). For example, angle α1 will be acute, angle (α2) will be obtuse, and the third (α3) will be equal to zero, since the tangent line drawn is parallel axis OH. In this case, tangent obtuse angle There is negative meaning, and tangent acute angle– positive, at tg0 and the result equal to zero.
A tangent to a given circle is a straight line that has only one common point with this circle. A tangent to a circle is always perpendicular to its radius drawn to the point of tangency. If two tangents are drawn from one point that does not belong to the circle, then the distances from this point to the points of tangency will always be the same. Tangents to circles are being built different ways, depending on their location relative to each other.
Instructions
Constructing a tangent to one circle.
1. Construct a circle of radius R and take A, which the tangent will pass through.
2. A circle is constructed with a center in the middle of the segment OA and radii equal to this segment.
3. The intersection of two tangent points drawn through point A to a given circle.
External tangent to two circles.
2. Draw a circle of radius R – r with center at point O.
3. A tangent from O1 is drawn to the resulting circle, the point of tangency is designated M.
4. Radius R passing through point M to point T – the tangent point of the circle.
5. Through the center O1 of the small circle, a radius r is drawn parallel to R of the large circle. The radius r points to point T1 – the point of tangency of the small circle.
circles.
Internal tangent to two circles.
1. Two circles of radius R and r are constructed.
2. Draw a circle of radius R + r with center at point O.
3. A tangent is drawn to the resulting circle from point O1, the point of tangency is designated by the letter M.
4. Ray OM intersects the first circle at point T - at the point of tangency of the great circle.
5. Through the center O1 of the small circle, a radius r is drawn parallel to the ray OM. The radius r points to point T1 – the point of tangency of the small circle.
6. Straight line TT1 – tangent to the given circles.
Sources:
- internal tangent
Angular closet – perfect option for empty corners in the apartment. In addition, the corner configuration closet ov gives the interior a classic atmosphere. As finishing corners closet Any material that is suitable for this purpose can be used.
You will need
- Fiberboard, MDF, screws, nails, saw blade, frieze.
Instructions
Cut a template 125 mm wide and 1065 mm long from plywood or fiberboard. The edges must be filed at an angle of 45 degrees. By ready-made template determine the dimensions of the side walls, as well as the place where it will be located closet.
Connect the lid to the side walls and triangular shelves. The cover must be secured to the upper edges of the side walls using screws. For structural strength, additional glue is used. Attach the shelves to the slats.
Angle the saw blade at a 45-degree angle and bevel the leading edge of the side walls along the guide bar. Attach fixed shelves to MDF strips. Connect the side walls with screws. Make sure there are no gaps.
Make marks in the wall, between which place the frame of the corner closet A. Attach using screws closet to Wall. The length of the dowel should be 75 mm.
Cut out the front frame from a solid MDF board. Using a circular saw, cut the openings in it using a ruler. Finish the corners.
Find the abscissa value of the tangent point, which is denoted by the letter “a”. If it coincides with a given tangent point, then "a" will be its x-coordinate. Determine the value functions f(a) by substituting into the equation functions abscissa value.
Determine the first derivative of the equation functions f’(x) and substitute the value of point “a” into it.
Take general equation tangent, which is defined as y = f(a) = f (a)(x – a), and substitute the found values of a, f(a), f "(a) into it. As a result, the solution to the graph and the tangent will be found.
Solve the problem in a different way if the given tangent point does not coincide with the tangent point. In this case, it is necessary to substitute “a” instead of numbers in the tangent equation. After that, instead of the letters “x” and “y”, substitute the coordinate value given point. Solve the resulting equation in which “a” is the unknown. Plug the resulting value into the tangent equation.
Write an equation for a tangent with the letter “a” if the problem statement specifies the equation functions and equation parallel line relative to the desired tangent. After this we need the derivative functions, to the coordinate at point “a”. Substitute the appropriate value into the tangent equation and solve the function.
When composing the equation of a tangent to the graph of a function, the concept of “abscissa of the point of tangency” is used. This value may be specified initially in the conditions of the task or it must be determined independently.
Instructions
Draw the x and y coordinate axes on a piece of paper. Explore given equation for the graph of a function. If it is , then two values for the parameter y are enough for any x, then plot the found points on the coordinate axis and connect them with a line. If the graph is nonlinear, then make a table of the dependence of y on x and select at least five points to construct the graph.
Determine the value of the abscissa of the tangent point for the case when the given tangent point does not coincide with the graph of the function. We set the third parameter with the letter “a”.
Write down the equation of the function f(a). To do this, substitute a instead of x in the original equation. Find the derivative of the function f(x) and f(a). Substitute the required data into the general tangent equation, which has the form: y = f(a) + f "(a)(x – a). As a result, obtain an equation that consists of three unknown parameters.
Substitute into it, instead of x and y, the coordinates of the given point through which the tangent passes. After this, find the solution to the resulting equation for all a. If it is square, then there will be two values for the abscissa of the tangent point. This is that the tangent passes twice near the graph of the function.
Draw a graph given function and , which are specified according to the conditions of the problem. In this case, it is also necessary to specify the unknown parameter a and substitute it into the equation f(a). Equate the derivative f(a) to the derivative of the equation of a parallel line. This comes from the condition of parallelism of the two. Find the roots of the resulting equation, which will be the abscissa of the point of tangency.
The straight line y=f(x) will be tangent to the graph shown in the figure at point x0 if it passes through the point with coordinates (x0; f(x0)) and has an angular coefficient f"(x0). Find such a coefficient, Knowing the features of a tangent, it’s not difficult.
You will need
- - mathematical reference book;
- - a simple pencil;
- - notebook;
- - protractor;
- - compass;
- - pen.
Instructions
If the value f‘(x0) does not exist, then either there is no tangent, or it runs vertically. In view of this, the presence of a derivative of the function at the point x0 is due to the existence of a non-vertical tangent tangent to the graph of the function at the point (x0, f(x0)). In this case, the angular coefficient of the tangent will be equal to f "(x0). Thus, it becomes clear geometric meaning derivative – calculation of the slope of the tangent.
Determine the general one. This kind of information can be obtained by referring to census data. To determine the general fertility, mortality, marriage and divorce rates, you will need to find the product general population and billing period. Write the resulting number into the denominator.
Put on the numerator the indicator corresponding to the desired relative. For example, if you are faced with determining the total fertility rate, then in place of the numerator there should be a number that reflects the total number of births for the period of interest to you. If your goal is the mortality rate or marriage rate, then in place of the numerator put the number of deaths in the calculation period or the number of marriages, respectively.
Multiply the resulting number by 1000. This will be the overall coefficient you are looking for. If you are faced with the task of finding the overall growth rate, then subtract the mortality rate from the birth rate.
Video on the topic
Sources:
- General vital rates
The main indicator of extraction efficiency is coefficient distribution. It is calculated by the formula: Co/Sw, where Co is the concentration of the extracted substance in the organic solvent (extractor), and St is the concentration of the same substance in water, after equilibrium has reached. How can you experimentally find the distribution coefficient?
Instructions
We determine the angular coefficient of the tangent to the curve at point M.
The curve representing the graph of the function y = f(x) is continuous in a certain neighborhood of the point M (including the point M itself).
If the value f‘(x0) does not exist, then either there is no tangent, or it runs vertically. In view of this, the presence of a derivative of the function at the point x0 is due to the existence of a non-vertical tangent tangent to the graph of the function at the point (x0, f(x0)). In this case, the angular coefficient of the tangent will be equal to f "(x0). Thus, the geometric meaning of the derivative becomes clear - the calculation of the angular coefficient of the tangent.
Find the abscissa value of the tangent point, which is denoted by the letter “a”. If it coincides with a given tangent point, then "a" will be its x-coordinate. Determine the value functions f(a) by substituting into the equation functions abscissa value.
Determine the first derivative of the equation functions f’(x) and substitute the value of point “a” into it.
Take the general tangent equation, which is defined as y = f(a) = f (a)(x – a), and substitute the found values of a, f(a), f "(a) into it. As a result, the solution to the graph will be found and tangent.
Solve the problem in a different way if the given tangent point does not coincide with the tangent point. In this case, it is necessary to substitute “a” instead of numbers in the tangent equation. After that, instead of the letters “x” and “y”, substitute the value of the coordinates of the given point. Solve the resulting equation in which “a” is the unknown. Plug the resulting value into the tangent equation.
Write an equation for a tangent with the letter “a” if the problem statement specifies the equation functions and the equation of a parallel line relative to the desired tangent. After this we need the derivative functions, to the coordinate at point “a”. Substitute the appropriate value into the tangent equation and solve the function.
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Consider the following figure:
It depicts a certain function y = f(x), which is differentiable at point a. Point M with coordinates (a; f(a)) is marked. Through arbitrary point P(a + ∆x; f(a + ∆x)) graph is drawn by a secant MR.
If now point P is shifted along the graph to point M, then straight line MR will rotate around point M. In this case, ∆x will tend to zero. From here we can formulate the definition of a tangent to the graph of a function.
Tangent to the graph of a function
The tangent to the graph of a function is the limiting position of the secant as the increment of the argument tends to zero. It should be understood that the existence of the derivative of the function f at the point x0 means that at this point of the graph there is tangent to him.
In this case, the angular coefficient of the tangent will be equal to the derivative of this function at this point f’(x0). This is the geometric meaning of the derivative. The tangent to the graph of a function f differentiable at point x0 is a certain straight line passing through the point (x0;f(x0)) and having an angular coefficient f’(x0).
Tangent equation
Let's try to obtain the equation of the tangent to the graph of some function f at point A(x0; f(x0)). The equation of a straight line with slope k has next view:
Since our slope coefficient is equal to the derivative f’(x0), then the equation will take the following form: y = f’(x0)*x + b.
Now let's calculate the value of b. To do this, we use the fact that the function passes through point A.
f(x0) = f’(x0)*x0 + b, from here we express b and get b = f(x0) - f’(x0)*x0.
We substitute the resulting value into the tangent equation:
y = f’(x0)*x + b = f’(x0)*x + f(x0) - f’(x0)*x0 = f(x0) + f’(x0)*(x - x0).
y = f(x0) + f’(x0)*(x - x0).
Consider the following example: find the equation of the tangent to the graph of the function f(x) = x 3 - 2*x 2 + 1 at point x = 2.
2. f(x0) = f(2) = 2 2 - 2*2 2 + 1 = 1.
3. f’(x) = 3*x 2 - 4*x.
4. f’(x0) = f’(2) = 3*2 2 - 4*2 = 4.
5. Substitute the obtained values into the tangent formula, we get: y = 1 + 4*(x - 2). Opening the brackets and bringing similar terms we get: y = 4*x - 7.
Answer: y = 4*x - 7.
General scheme for composing the tangent equation to the graph of the function y = f(x):
1. Determine x0.
2. Calculate f(x0).
3. Calculate f’(x)