Let \( f(x)=3 \sec x \). Find the requested derivatives. \( f^{\prime}(x)= \) \( f^{\prime \prime}(x)= \) Note: You can earn partial credit on this problem.

To find the derivatives of the function \( f(x) = 3 \sec x \), we will first compute the first derivative \( f'(x) \) and then the second derivative \( f''(x) \). ### Step 1: Find the first derivative \( f'(x) \) The derivative of \( \sec x \) is given by: \[ \frac{d}{dx}(\sec x) = \sec x \tan x \] Using this, we can find \( f'(x) \): \[ f'(x) = 3 \frac{d}{dx}(\sec x) = 3 \sec x \tan x \] Thus, the first derivative is: \[ f'(x) = 3 \sec x \tan x \] ### Step 2: Find the second derivative \( f''(x) \) To find the second derivative, we need to differentiate \( f'(x) = 3 \sec x \tan x \). We will use the product rule, which states that if \( u(x) = \sec x \) and \( v(x) = \tan x \), then: \[ \frac{d}{dx}(uv) = u'v + uv' \] Calculating \( u' \) and \( v' \): - \( u' = \sec x \tan x \) - \( v' = \sec^2 x \) Now applying the product rule: \[ f''(x) = 3 \left( \sec x \tan x \cdot \tan x + \sec x \cdot \sec^2 x \right) \] \[ = 3 \left( \sec x \tan^2 x + \sec^3 x \right) \] Thus, the second derivative is: \[ f''(x) = 3 \sec x \tan^2 x + 3 \sec^3 x \] ### Final Answers The derivatives are: \[ f'(x) = 3 \sec x \tan x \] \[ f''(x) = 3 \sec x \tan^2 x + 3 \sec^3 x \]