Answer
\[\left( {{f}^{-1}} \right)'\left( 2 \right)=4\]
Work Step by Step
\[\begin{align}
& f\left( x \right)=\sqrt{x};\text{ }\left( 2,4 \right) \\
& \text{We have the point }\left( {{x}_{0}},{{y}_{0}} \right)=\left( 2,4 \right)\text{ and }f\left( x \right)=\sqrt{x} \\
& \text{Use the formula of the theorem 3}\text{.23} \\
& \left( {{f}^{-1}} \right)'\left( {{y}_{0}} \right)=\frac{1}{f'\left( {{x}_{0}} \right)} \\
& \text{Where }{{x}_{0}}={{f}^{-1}}\left( {{y}_{0}} \right)\text{ for the point }\left( 2,4 \right)\Rightarrow {{x}_{0}}=4\text{ and }{{y}_{0}}=2 \\
& \left( {{f}^{-1}} \right)'\left( 2 \right)=\frac{1}{f'\left( 4 \right)} \\
& \text{Calculate }f'\left( 4 \right) \\
& f\left( x \right)=\sqrt{x} \\
& f'\left( x \right)=\frac{1}{2\sqrt{x}} \\
& f'\left( 4 \right)=\frac{1}{2\sqrt{4}} \\
& \text{Therefore,} \\
& \left( {{f}^{-1}} \right)'\left( 2 \right)=\frac{1}{\frac{1}{2\sqrt{4}}} \\
& \left( {{f}^{-1}} \right)'\left( 2 \right)=2\sqrt{4} \\
& \left( {{f}^{-1}} \right)'\left( 2 \right)=4 \\
\end{align}\]
