Answer
\[\left( {{f}^{-1}} \right)'\left( x \right)=-\frac{2}{{{x}^{3}}}\]
Work Step by Step
\[\begin{align}
& f\left( x \right)={{x}^{-1/2}},x>0 \\
& \text{Write }f\left( x \right)\text{ as }y \\
& y={{x}^{-1/2}} \\
& \text{Interchange }x\text{ and }y \\
& x={{y}^{-1/2}} \\
& \text{Solve for }y \\
& {{\left( x \right)}^{-2}}={{\left( {{y}^{-1/2}} \right)}^{-2}} \\
& {{\left( x \right)}^{-2}}=y \\
& y={{x}^{-2}} \\
& \text{Write }y\text{ as }{{f}^{-1}}\left( x \right) \\
& {{f}^{-1}}\left( x \right)={{x}^{-2}} \\
& \text{Compute the derivative} \\
& \left( {{f}^{-1}} \right)'\left( x \right)=\frac{d}{dx}\left[ {{x}^{-2}} \right] \\
& \left( {{f}^{-1}} \right)'\left( x \right)=-2{{x}^{-3}} \\
& \left( {{f}^{-1}} \right)'\left( x \right)=-\frac{2}{{{x}^{3}}} \\
\end{align}\]
