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: 28


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

Work Step by Step

Recall that $(e^x)'=e^x$ Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$ Since $ h(y)=\frac{1+e^y}{1-e^y}$, then the derivative, using the quotient rule, is given by $$ h'(y)=\frac{(1-e^y)(e^y)-(1+e^y)(-e^y)}{(1-e^y)^2}=\frac{2e^y}{(1-e^y)^2}.$$
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