Find the 35th derivative of the function \( f(x)=\cos (x) \). The answer is function

To find the 35th derivative of the function \( f(x) = \cos(x) \), we first need to observe the pattern in the derivatives of \( \cos(x) \). 1. The first derivative is: \[ f'(x) = -\sin(x) \] 2. The second derivative is: \[ f''(x) = -\cos(x) \] 3. The third derivative is: \[ f'''(x) = \sin(x) \] 4. The fourth derivative is: \[ f^{(4)}(x) = \cos(x) \] From this, we can see that the derivatives of \( \cos(x) \) repeat every four derivatives: - \( f^{(0)}(x) = \cos(x) \) - \( f^{(1)}(x) = -\sin(x) \) - \( f^{(2)}(x) = -\cos(x) \) - \( f^{(3)}(x) = \sin(x) \) - \( f^{(4)}(x) = \cos(x) \) (and the cycle repeats) To find the 35th derivative, we can determine the position of 35 in the cycle of 4. We do this by calculating \( 35 \mod 4 \): \[ 35 \div 4 = 8 \quad \text{(remainder 3)} \] Thus, \( 35 \mod 4 = 3 \). This means that the 35th derivative corresponds to the 3rd derivative in our cycle. From our earlier calculations, we know: \[ f^{(3)}(x) = \sin(x) \] Therefore, the 35th derivative of \( f(x) = \cos(x) \) is: \[ \boxed{\sin(x)} \]