Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 14: Partial Derivatives - Section 14.4 - The Chain Rule - Exercises 14.4 - Page 817: 30


$-9$ and $-4$

Work Step by Step

Consider $\dfrac{\partial z}{ \partial x}=-\dfrac{F_x}{F_z}$ $\implies \dfrac{\partial z}{ \partial x}=-\dfrac{-(1/x^2)}{(-1/z^2)}=-(\dfrac{z^2}{x^2})$ For the point $(2,3,6)$, we have Now, $\dfrac{\partial z}{ \partial x}=-\dfrac{6^2}{2^2}=-9$ and, $\dfrac{\partial z}{ \partial y}=-\dfrac{F_y}{F_z}$ $\implies \dfrac{\partial z}{ \partial y}=-\dfrac{-(1/y^2)}{(-1/z^2)}=-(\dfrac{z^2}{y^2})$ Also, $\dfrac{\partial z}{ \partial y}(2,3,6)=-\dfrac{36}{9}=-4$
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.