Answer
$$ g'(t) =2(2-t)e^{4t-t^2}.$$
Work Step by Step
Recall that $(e^x)'=e^x$
Since $ g(t)=e^{4t-t^2}$, then the derivative, using the chain rule, is given by
$$ g'(t)=e^{4t-t^2}(4t-t^2)'=(4-2t)e^{4t-t^2}=2(2-t)e^{4t-t^2}.$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 7 - Exponential Functions - Chapter Review Exercises - Page 386: 19
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