If $$y = \int_x^{{x^2}} {\sqrt {5 - {t^2}} dt} $$     then the value of $$\frac{{dy}}{{dx}}$$  at $$x = \sqrt 2 $$   is :

Solution :
$$\eqalign{ & y = \int_x^{{x^2}} {\sqrt {5 - {t^2}} dt} - \int_0^x {\sqrt {5 - {t^2}} dt} \cr & {\text{Differentiating w}}{\text{.r}}{\text{.t}}{\text{. }}x, \cr & \frac{{dy}}{{dx}} = \sqrt {5 - {{\left( {{x^2}} \right)}^2}} .2x - \sqrt {5 - {x^2}} \cr & {\text{Put }}x = \sqrt 2 \cr} $$