Question

question
find $\frac{d^{44}y}{dx^{44}}$ of $y = - 10sin(x)-8cos(x)$.
provide your answer below:
$\frac{d^{44}y}{dx^{44}}=square$

Explanation:

Step1: Recall derivative rules for sine and cosine

The derivative of $\sin(x)$ is $\cos(x)$ and the derivative of $\cos(x)$ is $-\sin(x)$. The second - derivative of $\sin(x)$ is $-\sin(x)$ and of $\cos(x)$ is $-\cos(x)$, the third - derivative of $\sin(x)$ is $-\cos(x)$ and of $\cos(x)$ is $\sin(x)$, and the fourth - derivative of $\sin(x)$ is $\sin(x)$ and of $\cos(x)$ is $\cos(x)$. The derivatives of $\sin(x)$ and $\cos(x)$ have a cycle of 4.

Step2: Divide the order of derivative by 4

Since we want to find the 44th - derivative, and $44\div4 = 11$ with a remainder of 0. This means that for $y_1=- 10\sin(x)$ and $y_2=-8\cos(x)$, the 44th - derivative of $\sin(x)$ is $\sin(x)$ and the 44th - derivative of $\cos(x)$ is $\cos(x)$.

Step3: Apply the linearity of differentiation

If $y = y_1 + y_2$ where $y_1=-10\sin(x)$ and $y_2=-8\cos(x)$, then $\frac{d^{44}y}{dx^{44}}=\frac{d^{44}y_1}{dx^{44}}+\frac{d^{44}y_2}{dx^{44}}$.
We know that $\frac{d^{44}}{dx^{44}}(-10\sin(x))=-10\sin(x)$ and $\frac{d^{44}}{dx^{44}}(-8\cos(x))=-8\cos(x)$.
So $\frac{d^{44}y}{dx^{44}}=-10\sin(x)-8\cos(x)$.

Answer:

$-10\sin(x)-8\cos(x)$