Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.8 Hyperbolic Functions And Hanging Cables - Exercises Set 6.8 - Page 481: 26

Answer

$$y' = \frac{{{{\operatorname{sech} }^2}x}}{{\sqrt {1 + {{\tanh }^2}x} }}$$

Work Step by Step

$$\eqalign{ & y = {\sinh ^{ - 1}}\left( {\tanh x} \right) \cr & {\text{find the derivatvive}} \cr & y' = \left( {{{\sinh }^{ - 1}}\left( {\tanh x} \right)} \right)' \cr & {\text{use the theorem 6}}{\text{.8}}{\text{.5 }}\left( {{\text{ see page 479}}} \right){\text{ }} \cr & \frac{d}{{dx}}\left[ {{{\sinh }^{ - 1}}u} \right] = \frac{1}{{\sqrt {1 + {u^2}} }}\frac{{du}}{{dx}} \cr & y' = \frac{1}{{\sqrt {1 + {{\left( {\tanh x} \right)}^2}} }}\left( {\tanh x} \right)' \cr & y' = \frac{1}{{\sqrt {1 + {{\tanh }^2}x} }}\left( {{{\operatorname{sech} }^2}x} \right) \cr & {\text{simplifying}} \cr & y' = \frac{{{{\operatorname{sech} }^2}x}}{{\sqrt {1 + {{\tanh }^2}x} }} \cr} $$
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