Question

question
find $\frac{d^{74}}{dx^{74}}(-15cos(x))$.
provide your answer below:
$\frac{d^{74}}{dx^{74}}(-15cos(x))=square$

Explanation:

Step1: Recall derivative rules of cosine

The first - derivative of $y = \cos(x)$ is $y^\prime=-\sin(x)$, the second - derivative $y^{\prime\prime}=-\cos(x)$, the third - derivative $y^{\prime\prime\prime}=\sin(x)$, and the fourth - derivative $y^{(4)}=\cos(x)$. The derivatives of $\cos(x)$ have a cycle of 4.

Step2: Divide the order of derivative by 4

We want to find the 74 - th derivative of $- 15\cos(x)$. Divide 74 by 4: $74\div4 = 18$ with a remainder. Using the division formula $n = 4k + r$, where $n = 74$, $k = 18$, and $r=2$.

Step3: Determine the 74 - th derivative

Since the cycle of derivatives of $\cos(x)$ has a period of 4, the $n$ - th derivative of $\cos(x)$ is the same as the $r$ - th derivative in the cycle when $n = 4k + r$. For $r = 2$, the derivative of $\cos(x)$ is $-\cos(x)$. And since we have $y=-15\cos(x)$, by the constant - multiple rule of differentiation $(cf(x))^\prime = cf^\prime(x)$, the 74 - th derivative of $-15\cos(x)$ is $-15\times(-\cos(x))$.

Answer:

$15\cos(x)$