University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 13 - Section 13.4 - The Chain Rule - Exercises - Page 711: 5

Answer

$4t \tan^{-1} t+1$ and $\pi+1$

Work Step by Step

a) Using chain rule. $\dfrac{dw}{dt}=\dfrac{\partial w}{\partial x}\dfrac{dx}{dt}+\dfrac{\partial w}{\partial y}\dfrac{dy}{dt}+\dfrac{\partial w}{\partial z}\dfrac{dz}{dt}$ or, $=\dfrac{4yte^x}{t^2+1}+\dfrac{2e^x}{t^2+1}-\dfrac{e^t}{z}$ or, $\dfrac{dw}{dt}=4t \tan^{-1} t+1$ Using direct differentiation. since, $w=2ye^x-\ln z$ or, $w=(2 \tan^{-1} t) (t^1+1)-t$ and $\dfrac{dw}{dt}=4t \tan^{-1} t+1$ Now, b) $\dfrac{dw}{dt}(1)=4(1) \tan^{-1} (1)+1=4(1) (\dfrac{\pi}{4})+1=\pi+1$
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