Calculus (3rd Edition)

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


$$ g'(t) =2(2-t)e^{4t-t^2}.$$

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 $$ g'(t)=e^{4t-t^2}(4t-t^2)'=(4-2t)e^{4t-t^2}=2(2-t)e^{4t-t^2}.$$
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