Answer
$$\left( {{f^{ - 1}}} \right)'\left( 8 \right) = \frac{1}{4}$$
Work Step by Step
$$\eqalign{
& {\text{We know that }}f\left( 2 \right) = 8{\text{ and }}f'\left( 2 \right) = 4.{\text{ Calculate }}\left( {{f^{ - 1}}} \right)'\left( 8 \right) \cr
& {\text{From the theorem 3}}{\text{.23 }}\left( {{\text{Derivative of the Inverse Function}}} \right) \cr
& \left( {{f^{ - 1}}} \right)'\left( {{y_0}} \right) = \frac{1}{{f'\left( {{x_0}} \right)}},{\text{ where }}{y_0} = f\left( {{x_0}} \right) \cr
& {\text{Comparing:}}\left( {{f^{ - 1}}} \right)'\left( {{y_0}} \right) = \left( {{f^{ - 1}}} \right)'\left( 8 \right) \Rightarrow {y_0} = 8{\text{ and }}{x_0} = \left( {{f^{ - 1}}} \right)\left( {{y_0}} \right) \cr
& {\text{We have that: }}f\left( 2 \right) = 8{\text{ and }}f'\left( 2 \right) = 4 \Rightarrow {x_0} = 2,{\text{ then}} \cr
& {\text{Applying }}\left( {{f^{ - 1}}} \right)'\left( {{y_0}} \right) = \frac{1}{{f'\left( {{x_0}} \right)}} \cr
& \left( {{f^{ - 1}}} \right)'\left( 8 \right) = \frac{1}{{f'\left( 2 \right)}} \cr
& \left( {{f^{ - 1}}} \right)'\left( 8 \right) = \frac{1}{4} \cr} $$
