Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.3 - The Fundamental Theorem of Calculus - 5.3 Exercises: 14

Answer

$$h'(x) = \frac{\sqrt{x}}{2(x^2 + 1)}$$

Work Step by Step

$$h(x) = \int\limits^{\sqrt{x}}_1 {\frac{z^2}{z^4 + 1}}dz$$ Using FTC 1, plug the upper bound into z and multiply by the derivative of the upper bound (chain rule). $$h'(x) = \frac{(\sqrt{x})^2}{(\sqrt{x})^4 + 1} \times (\sqrt x)'$$ Simplify. $$h'(x) = \frac{x}{x^2 + 1} \times (\frac{1}{2\sqrt x})$$ $$h'(x) = \frac{\sqrt{x}}{2(x^2 + 1)}$$
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