University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 12 - Section 12.1 - Curves in Space and Their Tangents - Exercises - Page 650: 29


a) $\dfrac{d(cu)}{dt}=c\dfrac{d(u)}{dt}$ b) $\dfrac{d(F(t)u)}{dt}=F(t)u'+F'(t)u$

Work Step by Step

Consider, $u=\lt f(t), g(t), h(t) \gt$ a) Apply product rule to get $\dfrac{d(cu)}{dt}=\dfrac{d(c)}{dt}\lt f(t), g(t), h(t) \gt=c\dfrac{d)}{dt}\lt f(t), g(t), h(t) \gt$ Thus, $\dfrac{d(cu)}{dt}=c\dfrac{d(u)}{dt}$ b) Apply product rule to get $\dfrac{d(F(t)u)}{dt}=\dfrac{d}{dt}(f(t)\lt f(t), g(t), h(t) \gt)=F(t)\lt f'(t), g'(t), h'(t) \gt+F'(t)\lt f(t), g(t), h(t) \gt$ Thus, $\dfrac{d(F(t)u)}{dt}=F(t)u'+F'(t)u$ Hence, a) $\dfrac{d(cu)}{dt}=c\dfrac{d(u)}{dt}$ b) $\dfrac{d(F(t)u)}{dt}=F(t)u'+F'(t)u$
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