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: 61

Answer

\[\left( {{f}^{-1}} \right)'\left( x \right)=\frac{1}{2\sqrt{x+4}}\]

Work Step by Step

\[\begin{align} & f\left( x \right)={{x}^{2}}-4,\text{ for }x>0 \\ & \text{Write }f\left( x \right)\text{ as }y \\ & y={{x}^{2}}-4 \\ & \text{Interchange }x\text{ and }y \\ & x={{y}^{2}}-4 \\ & \text{Solve for }y \\ & {{y}^{2}}=\sqrt{x+4} \\ & y=\sqrt{x+4} \\ & \text{Write }y\text{ as }{{f}^{-1}}\left( x \right) \\ & {{f}^{-1}}\left( x \right)=\sqrt{x+4} \\ & \text{Compute the derivative} \\ & \left( {{f}^{-1}} \right)'\left( x \right)=\frac{d}{dx}\left[ \sqrt{x+4} \right] \\ & \left( {{f}^{-1}} \right)'\left( x \right)=\frac{1}{2\sqrt{x+4}} \\ \end{align}\]
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