Answer
\[\left( {{f}^{-1}} \right)'\left( 8 \right)=\frac{1}{12}\]
Work Step by Step
\[\begin{align}
& f\left( x \right)={{x}^{3}};\text{ }\left( 8,2 \right) \\
& \text{We have the point }\left( {{x}_{0}},{{y}_{0}} \right)\text{ and }f\left( x \right)=x^3 \\
& \text{Use the formula of the theorem 3}\text{.23} \\
& \left( {{f}^{-1}} \right)'\left( {{y}_{0}} \right)=\frac{1}{f'\left( {{x}_{0}} \right)} \\
& \text{Where }{{x}_{0}}={{f}^{-1}}\left( {{y}_{0}} \right)\text{ for the point }\left( 8,2 \right)\Rightarrow {{x}_{0}}=2\text{ and }{{y}_{0}}=8 \\
& \left( {{f}^{-1}} \right)'\left( 8 \right)=\frac{1}{f'\left( 2 \right)} \\
& \text{Calculate }f'\left( 2 \right) \\
& f\left( x \right)={{x}^{3}} \\
& f'\left( x \right)=3{{x}^{2}} \\
& f'\left( 2 \right)=3{{\left( 2 \right)}^{2}} \\
& f'\left( 2 \right)=12 \\
& \text{Therefore,} \\
& \left( {{f}^{-1}} \right)'\left( 8 \right)=\frac{1}{f'\left( 2 \right)} \\
& \left( {{f}^{-1}} \right)'\left( 8 \right)=\frac{1}{12} \\
\end{align}\]
