Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 6 - Applications of Integration - 6.10 Hyperbolic Functions - 6.10 Exercises - Page 503: 15

Answer

$${e^x} = {e^x}$$

Work Step by Step

$$\eqalign{ & \cosh x + \sinh x = {e^x} \cr & {\text{definition of }}sinhx{\text{ and }}\cosh x \cr & \frac{{{e^x} + {e^{ - x}}}}{2} + \frac{{{e^x} - {e^{ - x}}}}{2} = {e^x} \cr & {\text{add}} \cr & \frac{{{e^x} + {e^{ - x}} + {e^x} - {e^{ - x}}}}{2} = {e^x} \cr & {\text{simplify}} \cr & \frac{{{e^x} + {e^x}}}{2} = {e^x} \cr & \frac{{2{e^x}}}{2} = {e^x} \cr & {e^x} = {e^x} \cr} $$
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