By computing the first few derivatives and looking for a pattern, find \( \frac{d^{969}}{d x^{969}}(\cos x) \). \( \frac{d^{969}}{d x^{969}}(\cos x)=\square \)

To find the \( 969 \)-th derivative of \( \cos x \), we start by computing the first few derivatives of \( \cos x \): 1. The first derivative: \[ \frac{d}{dx}(\cos x) = -\sin x \] 2. The second derivative: \[ \frac{d^2}{dx^2}(\cos x) = \frac{d}{dx}(-\sin x) = -\cos x \] 3. The third derivative: \[ \frac{d^3}{dx^3}(\cos x) = \frac{d}{dx}(-\cos x) = \sin x \] 4. The fourth derivative: \[ \frac{d^4}{dx^4}(\cos x) = \frac{d}{dx}(\sin x) = \cos x \] From these calculations, we can observe a pattern in the derivatives of \( \cos x \): - \( \frac{d^0}{dx^0}(\cos x) = \cos x \) - \( \frac{d^1}{dx^1}(\cos x) = -\sin x \) - \( \frac{d^2}{dx^2}(\cos x) = -\cos x \) - \( \frac{d^3}{dx^3}(\cos x) = \sin x \) - \( \frac{d^4}{dx^4}(\cos x) = \cos x \) The derivatives repeat every four steps. Thus, we can summarize the results as follows: - If \( n \equiv 0 \mod 4 \), then \( \frac{d^n}{dx^n}(\cos x) = \cos x \) - If \( n \equiv 1 \mod 4 \), then \( \frac{d^n}{dx^n}(\cos x) = -\sin x \) - If \( n \equiv 2 \mod 4 \), then \( \frac{d^n}{dx^n}(\cos x) = -\cos x \) - If \( n \equiv 3 \mod 4 \), then \( \frac{d^n}{dx^n}(\cos x) = \sin x \) Next, we need to determine \( 969 \mod 4 \): \[ 969 \div 4 = 242 \quad \text{remainder } 1 \] Thus, \( 969 \equiv 1 \mod 4 \). According to our pattern, since \( 969 \equiv 1 \mod 4 \), we have: \[ \frac{d^{969}}{dx^{969}}(\cos x) = -\sin x \] Therefore, the final answer is: \[ \frac{d^{969}}{dx^{969}}(\cos x) = -\sin x \]