Answer
$$ c'(t)= -t^{-2}i-2e^{2t}k.$$
Work Step by Step
Since $ c(t)=t^{-1}i-e^{2t}k $, then by using the chain rule, the derivative $ c'(t)$ is given by
$$ c'(t)=-t^{-2}i- (2t)'e^{2t}k=-t^{-2}i-2e^{2t}k.$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 14 - Calculus of Vector-Valued Functions - 14.2 Calculus of Vector-Valued Functions - Exercises - Page 720: 11
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