Question

Найти производную n-го порядка функции y=cos(2x)

Answer 1

Посмотрите предложенный вариант:
1. y'= -2sin2x; y''= -4cos2x; y'''= 8sin2x; y''''=16cos2x; y'''''= -32sin2x
2. Получается, что (y)ⁿ=2ⁿcos($\frac{ \pi *n}{2} +2x$)

Answer 2

Finding the nth Derivative of y=cos(2x)

To find the nth derivative of the function y=cos(2x), we can use the general formula for the nth derivative of a function. The nth derivative of a function f(x) is denoted as f^(n)(x) and can be found using the formula:

f^(n)(x) = d^n/dx^n [f(x)]

For the function y=cos(2x), the first few derivatives are as follows: - First derivative: y' = -2sin(2x) - Second derivative: y'' = -4cos(2x) - Third derivative: y''' = 8sin(2x) - Fourth derivative: y'''' = 16cos(2x)

The pattern for the nth derivative of y=cos(2x) alternates between cosine and sine functions, with the coefficient being a power of 2.

The nth derivative of y=cos(2x) can be expressed as: y^(n)(x) = 2^n * cos(2x) if n is even y^(n)(x) = 2^n * sin(2x) if n is odd

This pattern continues for higher values of n.

[[10 #]] provides the derivatives of the function f(x)=3x*sin(x*e^2x) up to the 3rd derivative, which demonstrates the application of the general formula for finding the nth derivative of a function.