Answer
$4$
Work Step by Step
$f(x)$ = $\sqrt {x-4}$, $a$ = $2$
$\sqrt {x-4}$ = $2$
$x-4$ = $4$
$x$ = $8$
$f(8)$ = $2$ $so$ $f^{-1}(2)$ = $8$
$(f^{-1})$'$(2)$ = $\frac{1}{f'(f^{-1}(2))}$ = $\frac{1}{f'(8)}$
$f'(x)$ = $\frac{1}{2\sqrt {x-4}}$
$f'(8)$ = $\frac{1}{2\sqrt {8-4}}$ = $\frac{1}{4}$
$(f^{-1})$'$(2)$ = $\frac{1}{f'(8)}$ = $\frac{1}{\frac{1}{4}}$ = $4$
