If $y(x) = \Large \int \limits_{\frac{\pi^2}{16}}^{\normalsize x^2} \large\frac{\cos x . \cos \sqrt \theta d\theta}{1+\sin^2 \sqrt \theta}$, find $\;y'(\pi)$

Question

(A) $\large\frac{\pi^2}{8}$ $\quad$ (B) $\large\frac{\pi^2}{16}$ $\quad$ (C) $2 \pi$ $\quad$ (D) $\large\frac{\pi^2}{16}$$ - 2 \pi$