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 816: 6


$e^{t-1}-(1+\ln t) \cos (t \ln t)$ and $0$

Work Step by Step

a. Using chain rule. $\dfrac{dw}{dt}=\dfrac{\partial w}{\partial x}\dfrac{dx}{dt}+\dfrac{\partial w}{\partial y}\dfrac{dy}{dt}+\dfrac{\partial w}{\partial z}\dfrac{dz}{dt}$ and, $(-y) \cos (xy)-\dfrac{x \cos xy}{t}+e^{(t-1)}=e^{(t-1)}-(1+\ln t) \cos (t \ln t)$ Now, by using direct differentiation. we have, $w=2-\sin xy$ and, $w=e^{(t-1)}- \sin (t \ln t)$ and $\dfrac{dw}{dt}=e^{t-1}-(1+\ln t) \cos (t \ln t)$ b. $\dfrac{dw}{dt}(1)=e^{(t-1)}-(1+\ln t) \cos (t \ln t)=e^{(1-1)}-(1+\ln 1) \cos ( \ln 1)=1-1=0$
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