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: 27


$$ h'(y)=- 2^{1-y}\ln 2.$$

Work Step by Step

Recall that $(a^x)'=a^x\ln{a}$ Since $ h(y)=2^{1-y}$, then the derivative, using the chain rule, is given by $$ h'(y)=2^{1-y}(1-y)' \ln 2=- 2^{1-y}\ln 2.$$
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