Answer
$\frac{1}{5}$
Work Step by Step
$f(x)$ = $x^{3}+2x-1$, $a$ = $2$
$x^{3}+2x-1$ = $2$
$x^{3}+2x-3$ = $0$
$(x-1)$$(x^{2}+x+3)$ = $0$
$x$ = $1$
$f(1)$ = $2$ $so$ $f^{-1}(2)$ = $1$
$(f^{-1})$$'(2)$ = $\frac{1}{f'(f^{-1}(2))}$ = $\frac{1}{f'(1)}$
$f'(x)$ = $3x^{2}+2$
$f'(1)$ = $3(1)^{2}+2$ = $5$
$so$ $(f^{-1})$$'(2)$ = $\frac{1}{f'(1)}$ = $\frac{1}{5}$