Answer
\[\left( {{f}^{-1}} \right)'\left( x \right)=\frac{1}{2\sqrt{x+4}}\]
Work Step by Step
\[\begin{align}
& f\left( x \right)={{x}^{2}}-4,\text{ for }x>0 \\
& \text{Write }f\left( x \right)\text{ as }y \\
& y={{x}^{2}}-4 \\
& \text{Interchange }x\text{ and }y \\
& x={{y}^{2}}-4 \\
& \text{Solve for }y \\
& {{y}^{2}}=\sqrt{x+4} \\
& y=\sqrt{x+4} \\
& \text{Write }y\text{ as }{{f}^{-1}}\left( x \right) \\
& {{f}^{-1}}\left( x \right)=\sqrt{x+4} \\
& \text{Compute the derivative} \\
& \left( {{f}^{-1}} \right)'\left( x \right)=\frac{d}{dx}\left[ \sqrt{x+4} \right] \\
& \left( {{f}^{-1}} \right)'\left( x \right)=\frac{1}{2\sqrt{x+4}} \\
\end{align}\]
