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: 20


$$ g'(t)= (2t-1)e^{1/t}.$$

Work Step by Step

Recall that $(e^x)'=e^x$ Recall the product rule: $(uv)'=u'v+uv'$ Since $ g(t)=t^2e^{1/t}$, then the derivative, using the product and chain rules, is given by $$ g'(t)=(t^2)'e^{1/t}+t^2(e^{1/t})'=2t e^{1/t}+t^2e^{1/t} (\frac{-1}{t^2})=(2t-1)e^{1/t}.$$
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