Determine the derivative of the following composite logarithmic trigonometric function and simplify as much as possible. \( f(x)=-2 \ln (\cos (3 x)) \) \( f^{\prime}(x)=\square \)

To find the derivative of the function \( f(x) = -2 \ln(\cos(3x)) \), we can use the chain rule. First, we identify the outer function and the inner function. The outer function is \( -2 \ln(u) \) where \( u = \cos(3x) \). The derivative of \( -2 \ln(u) \) with respect to \( u \) is \( -\frac{2}{u} \). Now, we need to differentiate the inner function \( \cos(3x) \). The derivative of \( \cos(3x) \) is \( -3\sin(3x) \) due to the chain rule. Now, applying the chain rule, we have: \[ f'(x) = -\frac{2}{\cos(3x)} \cdot (-3\sin(3x)) \] This simplifies to: \[ f'(x) = \frac{6\sin(3x)}{\cos(3x)} \] Recognizing that \( \frac{\sin(3x)}{\cos(3x)} = \tan(3x) \), we find: \[ f'(x) = 6\tan(3x) \] So, the simplified derivative is \( f'(x) = 6\tan(3x) \).