Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.3 Product and Quotient Rules - Exercises - Page 121: 10



Work Step by Step

Using the quotient rule, we have $\frac{dw}{dz}=\frac{(\sqrt {z}+z)\frac{d}{dz}(z^{2})-(z^{2})\frac{d}{dz}(\sqrt {z}+z)}{(\sqrt {z}+z)^{2}}$ $\frac{(\sqrt {z}+z)2z-(z^{2})(\frac{1}{2\sqrt {z}}+1)}{(\sqrt {z}+z)^{2}}$ $=\frac{2z\sqrt {z}+2z^{2}-\frac{z\sqrt {z}}{2}-z^{2}}{(\sqrt {z}+z)^{2}}$ $=\frac{z^{2}+\frac{3z\sqrt {z}}{2}}{(\sqrt {z}+z)^{2}}$ $\frac{dw}{dz}|_{z=9}=\frac{9^{2}+\frac{3\times9\times\sqrt {9}}{2}}{(\sqrt {9}+9)^{2}}=\frac{243}{288}=\frac{27}{32}$
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.