Answer
$g'(t)=(2^t)(2t+t^2\ln 2)$
Work Step by Step
$g(t)=t^22^t$
$u=t^2$
$u’=2t$
$v=2^t$
$v’=(\ln 2)(3^t)$
$g’(t)=u’v+uv’$
$=(2t)(2^t)+(t^2)(\ln 2)(2^t)$
$=(2^t)(2t+t^2\ln 2)$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.5 Exercises - Page 362: 43
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