University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.11 - Linearization and Differentials - Exercises - Page 199: 25



Work Step by Step

We evaluate the function: $y=\sin{5\sqrt{x}}$ on differentiating the above: $\frac{dy}{dx}=\frac{d\sin{5\sqrt{x}}}{dx}$ or $\frac{dy}{dx}=\cos({5\sqrt{x}})\frac{d(5\sqrt{x})}{dx}$ or $\frac{dy}{dx}=\frac{5}{2\sqrt{x}}\cos({5\sqrt{x}})$ or ${dy}=(\frac{5}{2\sqrt{x}}\cos({5\sqrt{x}})){dx}$ The final answer is: ${dy}=(\frac{5}{2\sqrt{x}}\cos({5\sqrt{x}})){dx}$
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