Answer
-$2$
Work Step by Step
$f(x)$ = $\frac{x+3}{x+1}$, $x$ $\gt$ -$1$, $a$ = $2$
$\frac{x+3}{x+1}$ = $2$
$x+3$ = $2(x+1)$
$x+3$ = $2x+2$
$x$ = $1$
$f(1)$ = $2$ $so$ $f^{-1}(2)$= $1$
$(f^{-1})$'$(2)$ = $\frac{1}{f'(f^{-1}(2))}$ = $\frac{1}{f'(1)}$
$f'(x)$ = $\frac{(x+1)\frac{d(x+3)}{dx}-(x+3)\frac{d(x+1)}{dx}}{(x+1)^{2}}$ = $\frac{(x+1)(1)-(x+3)(1)}{(x+1)^{2}}$ = $\frac{x+1-x-3}{(x+1)^{2}}$ = $\frac{-2}{(x+1)^{2}}$
$f'(1)$ = $\frac{-2}{(1+1)^{2}}$ = -$\frac{1}{2}$
$so$ $(f^{-1})$'$(2)$ = $\frac{1}{f'(1)}$ = $\frac{1}{-\frac{1}{2}}$ = -$2$
