Calculus: Early Transcendentals (2nd Edition)

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

Chapter 3 - Derivatives - 3.10 Derivatives of Inverse Trigonometric Functions - 3.10 Execises - Page 221: 37

Answer

$$2$$

Work Step by Step

$$\eqalign{ & {\text{Let }}f\left( x \right) = \frac{1}{2}x + 8;\,\,\,\,\,\underbrace {\left( {10,4} \right)}_{\left( {{x_0},{y_0}} \right)} \cr & {\text{Calculate }}f'\left( {10} \right) \cr & f'\left( x \right) = \frac{1}{2} \cr & f'\left( {10} \right) = \frac{1}{2} \cr & {\text{Find the derivative of the inverse function}}{\text{, using the THEOREM 3}}{\text{.23}} \cr & \left( {{f^{ - 1}}} \right)'\left( {{y_0}} \right) = \frac{1}{{f'\left( {{x_0}} \right)}},\,\,\,\,{\text{Where }}{y_0} = f\left( {{x_0}} \right) \cr & {\text{Then}}{\text{,}} \cr & \left( {{f^{ - 1}}} \right)'\left( 4 \right) = \frac{1}{{f'\left( {10} \right)}} \cr & \left( {{f^{ - 1}}} \right)'\left( 4 \right) = \frac{1}{{1/2}} \cr & \left( {{f^{ - 1}}} \right)'\left( 4 \right) = 2 \cr} $$
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