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 222: 44

Answer

\[\left( {{f}^{-1}} \right)'\left( 8 \right)=\frac{1}{12}\]

Work Step by Step

\[\begin{align} & f\left( x \right)={{x}^{3}};\text{ }\left( 8,2 \right) \\ & \text{We have the point }\left( {{x}_{0}},{{y}_{0}} \right)\text{ and }f\left( x \right)=x^3 \\ & \text{Use the formula of the theorem 3}\text{.23} \\ & \left( {{f}^{-1}} \right)'\left( {{y}_{0}} \right)=\frac{1}{f'\left( {{x}_{0}} \right)} \\ & \text{Where }{{x}_{0}}={{f}^{-1}}\left( {{y}_{0}} \right)\text{ for the point }\left( 8,2 \right)\Rightarrow {{x}_{0}}=2\text{ and }{{y}_{0}}=8 \\ & \left( {{f}^{-1}} \right)'\left( 8 \right)=\frac{1}{f'\left( 2 \right)} \\ & \text{Calculate }f'\left( 2 \right) \\ & f\left( x \right)={{x}^{3}} \\ & f'\left( x \right)=3{{x}^{2}} \\ & f'\left( 2 \right)=3{{\left( 2 \right)}^{2}} \\ & f'\left( 2 \right)=12 \\ & \text{Therefore,} \\ & \left( {{f}^{-1}} \right)'\left( 8 \right)=\frac{1}{f'\left( 2 \right)} \\ & \left( {{f}^{-1}} \right)'\left( 8 \right)=\frac{1}{12} \\ \end{align}\]
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