## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 6 - Applications of Integration - 6.10 Hyperbolic Functions - 6.10 Exercises: 48

#### Answer

$$f'\left( t \right) = \frac{1}{{\left( {1 - t} \right)\sqrt t }}$$

#### Work Step by Step

\eqalign{ & f\left( t \right) = 2{\tanh ^{ - 1}}\sqrt t \cr & {\text{find the derivative}} \cr & f'\left( t \right) = \frac{d}{{dt}}\left( {2{{\tanh }^{ - 1}}\sqrt t } \right) \cr & f'\left( t \right) = 2\frac{d}{{dt}}\left( {{{\tanh }^{ - 1}}\sqrt t } \right) \cr & {\text{use derivatives of the inverse hyperbolic functions}} \cr & f'\left( t \right) = 2\left( {\frac{1}{{1 - {{\left( {\sqrt t } \right)}^2}}}} \right)\frac{d}{{dt}}\left( {\sqrt t } \right) \cr & f'\left( t \right) = 2\left( {\frac{1}{{1 - {{\left( {\sqrt t } \right)}^2}}}} \right)\left( {\frac{1}{{2\sqrt t }}} \right) \cr & {\text{simplify}} \cr & f'\left( t \right) = \left( {\frac{1}{{1 - t}}} \right)\left( {\frac{1}{{\sqrt t }}} \right) \cr & f'\left( t \right) = \frac{1}{{\left( {1 - t} \right)\sqrt t }} \cr}

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