Answer
$$y' = \frac{-tan\sqrt x}{2}$$
Work Step by Step
$$y = \int\limits^{\frac{\pi}{4}}_{\sqrt x} {\theta tan\theta}d\theta$$
Switch the bounds, making the integral negative.
$$y = -\int\limits_{\frac{\pi}{4}}^{\sqrt x} {\theta tan\theta}d\theta$$
Substitute in the upper bound for $\theta$ and multiply by the derivative of the upper bound.
$$y' = - (\sqrt{x}tan(\sqrt x))(\sqrt x)'$$
Simplify.
$$y' = - (\sqrt{x}tan(\sqrt x))(\frac{1}{2\sqrt x})'$$
$$y' = \frac{-tan\sqrt x}{2}$$
