Answer is: option2

\( \frac{17}{8} \)

Solution:

We are given the function:

\[ g(x) = \frac{x^2 - 1}{f(x)} \] Applying the quotient rule: \[ g'(x) = \frac{(2x) f(x) - (x^2 - 1) f'(x)}{f(x)^2} \]

Substituting \( x = 4 \), \( f(4) = 2 \), and \( f'(4) = \frac{1}{2} \):

\[ g'(4) = \frac{(2(4))(2) - ((4^2 - 1) (\frac{1}{2}))}{2^2} \] \[ = \frac{(8)(2) - (15 \times \frac{1}{2})}{4} \] \[ = \frac{16 - \frac{15}{2}}{4} \] \[ = \frac{\frac{32}{2} - \frac{15}{2}}{4} \] \[ = \frac{\frac{17}{2}}{4} \] \[ = \frac{17}{8} \] The correct answer is: \[ \boxed{\frac{17}{8}} \quad \]