Explanation

(b)
We have, $f(x) = \sqrt{1 + \cos^{2}{(x^{2})}}\ldots(i)$
On differentiating Eq. (i) w.r.t. $x$, we get
$f^{'}(x) = \frac{- 2\sin{x^{2}\cos x^{2}}}{\sqrt{1 + \cos^{2}x^{2}}}(x)$
$\Rightarrow f^{'}(x) = \frac{- \sin{2x^{2}}}{\sqrt{1 + \cos^{2}x^{2}}}(x)\ldots(ii)$
Put, $x = \frac{\sqrt{\pi}}{2}$ in Eq. (ii), we get
$f^{'}\left( \frac{\sqrt{\pi}}{2} \right) = - \frac{\sqrt{\pi}}{2}.\frac{\sin{2\left( \frac{\pi}{4} \right)}}{\sqrt{1 + \frac{1}{2}}}$
$= - \frac{\sqrt{\pi}}{2}.\frac{\sin\frac{\pi}{2}}{\sqrt{\frac{3}{2}}} = - \sqrt{\frac{\pi}{6}}$