\( f(x)=3^{x} \sec 3 x+17 \), find \( f^{\prime}(x) \) \( 3^{x} \sec 3 x(\ln 3+3 \tan x) \) \( 3^{x} \sec 3 x(1+3 \tan x) \) \( 3^{x} \sec 3 x(1+3 \tan 3 x) \) \( 3^{x} \sec 3 x(\ln 3+3 \tan 3 x) \)

To find the derivative \( f'(x) \) of the function \( f(x) = 3^x \sec(3x) + 17 \), we will use the product rule and the chain rule. 1. **Identify the components**: - Let \( u = 3^x \) and \( v = \sec(3x) \). - Then, \( f(x) = u v + 17 \). 2. **Differentiate \( u \)**: \[ u' = \frac{d}{dx}(3^x) = 3^x \ln(3) \] 3. **Differentiate \( v \)**: Using the chain rule: \[ v = \sec(3x) \implies v' = \sec(3x) \tan(3x) \cdot \frac{d}{dx}(3x) = 3 \sec(3x) \tan(3x) \] 4. **Apply the product rule**: The product rule states that \( (uv)' = u'v + uv' \): \[ f'(x) = u'v + uv' = (3^x \ln(3)) \sec(3x) + (3^x)(3 \sec(3x) \tan(3x)) \] 5. **Factor out common terms**: \[ f'(x) = 3^x \sec(3x) \left( \ln(3) + 3 \tan(3x) \right) \] Thus, the derivative \( f'(x) \) is: \[ f'(x) = 3^x \sec(3x) \left( \ln(3) + 3 \tan(3x) \right) \] The correct answer is: \[ \boxed{3^{x} \sec(3x)(\ln 3 + 3 \tan 3x)} \]