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: 31


${dy}=\frac{1}{2\sqrt {x}}e^{\sqrt{x}}{dx}$

Work Step by Step

We evaluate the function: $y=e^{\sqrt{x}}$ $y=e^{x^{\frac{1}{2}}}$ On differentiating the above: $\frac{dy}{dx}=e^{x^{\frac{1}{2}}}{\frac{dx^{\frac{1}{2}}}{dx}}$ or $\frac{dy}{dx}=\frac{1}{2\sqrt {x}}e^{x^{\frac{1}{2}}}$ or ${dy}=\frac{1}{2\sqrt {x}}e^{\sqrt{x}}{dx}$ The final answer is: ${dy}=\frac{1}{2\sqrt {x}}e^{\sqrt{x}}{dx}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.