## Calculus: Early Transcendentals 8th Edition

$$y' = \frac{-tan\sqrt x}{2}$$
$$y = \int\limits^{\frac{\pi}{4}}_{\sqrt x} {\theta tan\theta}d\theta$$ 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. $$y' = - (\sqrt{x}tan(\sqrt x))(\frac{1}{2\sqrt x})'$$ $$y' = \frac{-tan\sqrt x}{2}$$