#### Answer

$v'(t) = 1+t^{-2}$

#### Work Step by Step

$v(t) = t - \frac{1}{t}$
Therefore,
$v'(t) = (1)t^{1-1}-(-1)t^{-1-1}$
$v'(t) = 1+t^{-2}$

Published by
Pearson

ISBN 10:
0321999584

ISBN 13:
978-0-32199-958-0

$v'(t) = 1+t^{-2}$

$v(t) = t - \frac{1}{t}$
Therefore,
$v'(t) = (1)t^{1-1}-(-1)t^{-1-1}$
$v'(t) = 1+t^{-2}$

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