If $$f\left( x \right) = \int_{ - 1}^1 {\frac{{\sin \,x}}{{1 + {t^2}}}dt} $$     then $$f'\left( {\frac{\pi }{3}} \right)$$  is :

Solution :
$$\eqalign{ & f\left( x \right) = \sin \,x.\int_{ - 1}^1 {\frac{{dt}}{{1 + {t^2}}}} \cr & = \sin \,x.\left[ {{{\tan }^{ - 1}}t} \right]_{ - 1}^1 \cr & = \sin \,x\left\{ {\frac{\pi }{4} - \left( { - \frac{\pi }{4}} \right)} \right\} \cr & = \frac{\pi }{2}\sin \,x \cr & \therefore f'\left( x \right) = \frac{\pi }{2}\cos \,x \cr & \therefore f'\left( {\frac{\pi }{3}} \right) = \frac{\pi }{2}.\cos \,\frac{\pi }{3} = \frac{\pi }{4} \cr} $$