Answer
$\frac{2\sqrt 3}{3}$
Work Step by Step
$f(x)$ = $sinx$, -$\frac{\pi}{2}$ $\leq$ $x$ $\leq$ $\frac{\pi}{2}$, $a$ = $\frac{1}{2}$
$sinx$ = $\frac{1}{2}$
$x$ = $\frac{\pi}{6}$
$f(\frac{\pi}{6})$ = $\frac{1}{2}$ $so$ $f^{-1}(\frac{1}{2})$ = $\frac{\pi}{6}$
$(f^{-1})$'$(\frac{1}{2})$ = $\frac{1}{f'(f^{-1}(\frac{1}{2}))}$ = $\frac{1}{f'(\frac{\pi}{6})}$
$f'(x)$ = $cosx$
$f'(\frac{\pi}{6})$ = $cos(\frac{\pi}{6})$ = $\frac{\sqrt 3}{2}$
$so$ $(f^{-1})$'$(\frac{1}{2})$ = $\frac{1}{f'(\frac{\pi}{6})}$ = $\frac{1}{\frac{\sqrt 3}{2}}$ = $\frac{2}{\sqrt 3}$ = $\frac{2\sqrt 3}{3}$
