A new inverse trigonometric function question

Taking the first derivative equal to zero yields:

$\Large f'(x) = \frac{1}{2\sqrt{1-x^2}} \cdot [ \frac{1}{\sqrt{\arcsin(x)}} - \frac{1}{\sqrt{\arccos(x)}}] = 0$ ;

or $\arcsin(x) = \arccos(x) \Rightarrow \arcsin(x) = \arcsin(\sqrt{1-x^2}) \Rightarrow x = \sqrt{1-x^2} \Rightarrow 2x^2 = 1 \Rightarrow x = \frac{1}{\sqrt{2}}$

The second derivative at $x = \frac{1}{\sqrt{2}}$ yields:

$f''(1/\sqrt{2}) = -\frac{8}{\pi^{3/2}} < 0$ (a maximum). Hence, $M = \sqrt{\pi/4} + \sqrt{\pi/4} = 2 \cdot \sqrt{\pi}/2 = \sqrt{\pi}$ .

The minimum value, $m,$ occurs when $\arcsin(x) = 1, \arccos(x)= 0 \Rightarrow x = 1$ , or $\sqrt{\pi/2} + \sqrt{0} = \sqrt{\pi/2}$ . Finally, $(\frac{M}{m})^2 = (\sqrt{\pi/(\pi/2)})^2 = (\sqrt{2})^2 = \boxed{2}.$