Answer
\[\left( {{f}^{-1}} \right)'\left( x \right)=-1\]
Work Step by Step
\[\begin{align}
& f\left( x \right)=\left| x+2 \right|,\text{ for }x\le -2 \\
& \text{By the definition of the absolute value} \\
& f\left( x \right)=-\left( x+2 \right),\text{ for }x\le -2 \\
& f\left( x \right)=-\left( x+2 \right) \\
& y=-\left( x+2 \right) \\
& \text{Interchange }x\text{ and }y \\
& x=-\left( y+2 \right) \\
& \text{Solve for }y \\
& x=-y-2 \\
& y=-x-2 \\
& \text{Write }y\text{ as }{{f}^{-1}}\left( x \right) \\
& {{f}^{-1}}\left( x \right)=-x-2 \\
& \text{Compute the derivative} \\
& \left( {{f}^{-1}} \right)'\left( x \right)=\frac{d}{dx}\left[ -x-2 \right] \\
& \left( {{f}^{-1}} \right)'\left( x \right)=-1 \\
\end{align}\]
