Answer
$undefined$
Work Step by Step
$f(x)$ = $cos(2x)$, $0$ $\leq$ $x$ $\leq$ $\frac{\pi}{2}$, $a$ = $1$
$cos(2x)$ = $1$
$2x$ = $0$
$x$ = $0$
$f(0)$ = $1$ $so$ $f^{-1}$$(1)$ = $0$
$(f^{-1})$'$(1)$ = $\frac{1}{f'(f^{-1}(1))}$ = $\frac{1}{f'(0)}$
$f'(x)$ = -$2sin(2x)$
$f'(0)$ = -$2sin(2\times0)$ =-$2sin(0)$ = $0$
$so$ $(f^{-1})$'$(1)$ = $\frac{1}{f'(0)}$ = $\frac{1}{0}$ = $undefined$
