## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 5 - Section 5.3 - The Fundamental Theorem of Calculus - 5.3 Exercises - Page 400: 17

#### 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. $$y' = - (\sqrt{x}tan(\sqrt x))(\frac{1}{2\sqrt x})'$$ $$y' = \frac{-tan\sqrt x}{2}$$

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