JEE Advance - Mathematics (2004 - No. 10)

If $$y\left( x \right) = \int\limits_{{x^2}/16}^{{x^2}} {{{\cos x\cos \sqrt \theta } \over {1 + {{\sin }^2}\sqrt \theta }}d\theta ,} $$ then find $${{dy} \over {dx}}$$ at $$x = \pi $$