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: 5


a. $4t \tan^{-1} t+1$ ; b. $\pi+1$

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, $\dfrac{4yte^x}{(t^2+1)}+\dfrac{2e^x}{(t^2+1)}-(\dfrac{e^t}{z})=(4t) \tan^{-1} t+1$ Now, by using direct differentiation. we have, $w=2ye^x-\ln z$ Also, $w=(2 \tan^{-1} t) (t^1+1)-t$ and $\dfrac{dw}{dt}=(4t) \tan^{-1} t+1$ b. $\dfrac{dw}{dt}(1)=4t \tan^{-1} t+1=4(1) \tan^{-1} (1)+1=4(1) (\dfrac{\pi}{4})+1=\pi+1$
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