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: 12

Answer

$$\tanh \left( { - x} \right) = - \tanh x$$

Work Step by Step

$$\eqalign{ & \tanh \left( { - x} \right) = - \tanh x \cr & {\text{use the hyperbolic function tan}}\theta = \frac{{{e^\theta } - {e^{ - \theta }}}}{{{e^\theta } + {e^{ - \theta }}}} \cr & \frac{{{e^{ - x}} - {e^{ - \left( { - x} \right)}}}}{{{e^{ - x}} + {e^{ - \left( { - x} \right)}}}} = \frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}} \cr & {\text{simplify}} \cr & \frac{{{e^{ - x}} - {e^x}}}{{{e^{ - x}} + {e^x}}} = \frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}} \cr & {\text{then}} \cr & \frac{{{e^{ - x}} - {e^x}}}{{{e^{ - x}} + {e^x}}} = - \frac{{ - \left( {{e^x} - {e^{ - x}}} \right)}}{{{e^x} + {e^{ - x}}}} \cr & \frac{{{e^{ - x}} - {e^x}}}{{{e^{ - x}} + {e^x}}} = - \frac{{{e^{ - x}} - {e^x}}}{{{e^x} + {e^{ - x}}}} \cr & {\text{by the definition of hyperbolic functions}} \cr & \tanh \left( { - x} \right) = - \tanh x \cr} $$
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