18. submit answer get help practice similar find the 31 th derivative of y = cos(3x) by finding the first…

Explanation:

Step1: Find first - few derivatives

Let (y = \cos(3x)). Using the chain - rule ((u(v(x)))^\prime=u^\prime(v(x))\cdot v^\prime(x)), where (u = \cos(u_1)) and (u_1 = 3x), (v^\prime(x)=3). (y^\prime=-3\sin(3x)) (y^{\prime\prime}=- 3^2\cos(3x)) (y^{\prime\prime\prime}=3^3\sin(3x)) (y^{(4)} = 3^4\cos(3x))

Step2: Identify the pattern

The pattern for the (n)th derivative of (y=\cos(3x)) is: If (n = 4k), (y^{(n)}=3^{n}\cos(3x)); if (n = 4k + 1), (y^{(n)}=-3^{n}\sin(3x)); if (n = 4k + 2), (y^{(n)}=-3^{n}\cos(3x)); if (n = 4k+3), (y^{(n)}=3^{n}\sin(3x)), where (k = 0,1,2,\cdots)

Step3: Determine (k) for (n = 31)

Divide (31) by (4): (31 = 4\times7+3)

Step4: Find the 31st derivative

Since (n = 31) and (n = 4k + 3) with (k = 7), (y^{(31)}=3^{31}\sin(3x))

Answer:

(3^{31}\sin(3x))