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: 40


$\dfrac{s^5}{2}$ and $\dfrac{5ts^4}{2}$

Work Step by Step

Consider $\dfrac{dw}{dt}=\dfrac{dw}{dx}\dfrac{dx}{dt}+\dfrac{d w}{d y}\dfrac{dy}{dt}$ $\implies xys^2+(x^2/2)(-s/t^2)=\dfrac{s^5}{2}$ Now, $\dfrac{dw}{ds}=xy(2ts)+(\dfrac{x^2}{2})(\dfrac{1}{t})$ $\implies (ts^2)(\dfrac{s}{t})(2ts) +(ts^2)^2(2)(1/t)=\dfrac{5ts^4}{2}$
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.