Answer
\[\left( {{f}^{-1}} \right)'\left( x \right)=\frac{5}{{{\left( 1-x \right)}^{2}}}\]
Work Step by Step
\[\begin{align}
& f\left( x \right)=\frac{x}{x+5} \\
& \text{Write }f\left( x \right)\text{ as }y \\
& y=\frac{x}{x+5} \\
& \text{Interchange }x\text{ and }y \\
& x=\frac{y}{y+5} \\
& \text{Solve for }y \\
& xy+5x=y \\
& xy-y=-5x \\
& y\left( x-1 \right)=-5x \\
& y=\frac{5x}{1-x} \\
& \text{Write }y\text{ as }{{f}^{-1}}\left( x \right) \\
& {{f}^{-1}}\left( x \right)=\frac{5x}{1-x} \\
& \text{Compute the derivative} \\
& \left( {{f}^{-1}} \right)'\left( x \right)=\frac{d}{dx}\left[ \frac{5x}{1-x} \right] \\
& \text{By the quotient rule} \\
& \left( {{f}^{-1}} \right)'\left( x \right)=\frac{\left( 1-x \right)\left( 5 \right)-5x\left( -1 \right)}{{{\left( 1-x \right)}^{2}}} \\
& \left( {{f}^{-1}} \right)'\left( x \right)=\frac{5-5x+5x}{{{\left( 1-x \right)}^{2}}} \\
& \left( {{f}^{-1}} \right)'\left( x \right)=\frac{5}{{{\left( 1-x \right)}^{2}}} \\
\end{align}\]
