Ryabushko A.P. IDZ 6.2 version 15

IDZ - 6.2. Solving problems in differential calculus.

No. 1.15. The equation 4sin²(x+y) = x is given. It is necessary to find the first and second derivatives of the function y with respect to the variable x.

Solution: Differentiate this equation with respect to x using the rule for differentiating a complex function: (4sin²(x+y))' = x' 8sin(x+y)cos(x+y)(y' + 1) = 1 y' = ( 1 - 8sin(x+y)cos(x+y)) / (8sin(x+y)cos(x+y))

We differentiate the resulting expression with respect to x using the quotient differentiation rule: y'' = [(8sin(x+y)cos(x+y))(8sin(x+y)cos(x+y)) - (1 - 8sin( x+y)cos(x+y))(8cos(x+y)cos(x+y) - 8sin(x+y)sin(x+y))]/(8sin(x+y)cos(x +y))^3 y'' = [16sin²(x+y) - 8cos²(x+y)] / (8sin(x+y)cos(x+y))^2 y'' = [2sin²(2x + 2y) - 1] / (sin(2x + 2y))^2

Ответ: y' = (1 - 8sin(x+y)cos(x+y)) / (8sin(x+y)cos(x+y)), y'' = [2sin²(2x + 2y) - 1 ] / (sin(2x + 2y))^2.

No. 2.15. The equations x = arctan t, y = Ln(1 + t²) are given. It is necessary to find the first and second derivatives of the function y with respect to the variable x.

Solution: Let's represent y as a function of t: y(t) = Ln(1 + t²). Then we express t in terms of x: t(x) = tan(x).

Let's find the first derivative of y with respect to x using the rule for differentiating a complex function: y' = y'(t) * t'(x) = 2t / (1 + t²)

Let's find the second derivative of y with respect to x: y'' = (2(1 + t²) - 4t²) / (1 + t²)^2 = -2 / (1 + t²)^2

We replace t with tg(x) and get: y' = 2tg(x) / (1 + tg²(x)), y'' = -2cos²(x) / (1 + tg²(x))^2

Ответ: y' = 2tg(x) / (1 + tg²(x)), y'' = -2cos²(x) / (1 + tg²(x))^2.

No. 3.15. Given the function y = x sin2x and the argument x₀ = -π/4. It is necessary to find the third derivative of the function y at the point x₀.

Solution: First derivative of the function y: y' = sin2x + 2xcos2x. The second derivative of the function y: y'' = 2cos2x - 4xsin2x. Third derivative of the function y: y''' = -12cos2x - 8xsin2x.

Substitute x₀ = -π/4 and get: y'''(-π/4) = -12cos(π/2) - 8(-π/4)sin(π/2) = 12

Answer: y'''(-π/4) = 12.

No. 4.15. Write a formula for the nth order derivative of the function y = 5ˣ.

Solution: First derivative of the function y: y' = 5ln(5)5^x. Second derivative of the function y: y'' = (5ln(5))^25^x. Third derivative of the function y: y''' = (5ln(5))^35^x. ...nth derivative of the function y: yⁿ = (5ln(5))^n5^x.

Answer: yⁿ = (5ln(5))^n5^x.

No. 5.15. Find the equation of the normal to the curve y = 6tg5x at the point with the abscissa x = π/20.

Solution: Find the first derivative of the function y: y' = 65sec²(5x). At point x = π/20 the value of y': y'(π/20) = 65sec²(π/4) = 30.

The equation of the tangent to the curve y = 6tg5x at point x₀ has the form: y - y(x₀) = y'(x₀)*(x - x₀)

Let's substitute x₀ = π/20 and y'(π/20) = 30: y - 6tg(5π/20) = 30(x - π/20)

Let's simplify the equation using tg(π/4) = 1: y = 30x - 6

The equation of the normal to the curve y = 6tg5x at the point x = π/20 will be perpendicular to the tangent and pass through the point (π/20, 6tg(π/20)): y - 6tg(π/20) = (-1/30)( x - π/20)

Simple equation using tg(π/4) = 1: y = (-1/30)x + (7/3)

Answer: the equation of the normal to the curve y = 6tg5x at the point with the abscissa x = π/20 has the form y = (-1/30)x + (7/3).

No. 6.15. Two material points move along the Ox axis with the laws of motion x₁ = 3t² - 8 and x₂ = 2t² + 5t + 6. It is necessary to find the speed with which these points move away from each other at the moment of meeting.

Solution: The distance between points x₁ and x₂ at the moment of meeting will be equal to |x₁ - x₂|. |x₁ - x₂| = |(3t² - 8) - (2t² + 5t + 6)| = |t² - 5t - 14| = |(t - 7)(t + 2)|

The moment of meeting is determined from the condition x₁ = x₂: 3t² - 8 = 2t² + 5t + 6 t² + 5t - 14 = 0 (t - 2)(t + 7) = 0 t₁ = 2, t₂ = -7

From the conditions of the problem it follows that the points collide, i.e. moving in opposite directions. Therefore, at the moment of meeting, the speed with which they move away from each other will be equal to the sum of their speeds. Let's find the speeds of the points at the moment of meeting: v₁ = x₁'(t₁) = 12t₁ = 24 v₂ = x₂'(t₁) = 4t₁ + 5 = 13

The speed of points moving away from each other at the moment of meeting will be equal to |v₁ - v₂|: |v₁ - v₂| = |24 - 13| = 11

Answer: the speed with which two material points move away from each other at the moment of meeting is equal to 11.

This digital product is a solution to problems in differential calculus within the framework of Individual Homework (IH) number 6.2, option 15, from the author Ryabushko A.P. The product includes solutions to problems of finding derivatives of functions, the equation of the normal to a curve, as well as a problem of determining the speed of material points moving away from each other.

This product is designed in a beautiful html format, which will allow you to conveniently and quickly familiarize yourself with solutions to problems. You can also easily save this file on your device and use it for educational purposes.

By purchasing this digital product, you receive ready-made solutions to problems in differential calculus, which will allow you to save time and avoid unnecessary headaches when completing such tasks.

This product is a solution to problems in differential calculus, option 15 from task IDZ 6.2. The task requires finding the first and second derivatives of functions with respect to the variable x for three functions: 4sin²(x+y) = x, y = Ln(1 + t²), where t = tan(x), and y = x sin2x. It is also necessary to find the third derivative of the function y = x sin2x at the point x₀ = -π/4 and the speed with which two material points move away from each other at the moment of meeting. Solutions to problems are presented in text format.

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