Calculus: Early Transcendentals (2nd Edition)

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

Chapter 3 - Derivatives - 3.5 Derivatives of Trigonometric Functions - 3.5 Exercises - Page 171: 81

Answer

$$\eqalign{ & \left. a \right){\text{ }}f\left( x \right) = \sin x,{\text{ }}a = \frac{\pi }{6} \cr & \left. b \right){\text{ }}\frac{{\sqrt 3 }}{2} \cr} $$

Work Step by Step

$$\eqalign{ & \mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {\frac{\pi }{6} + h} \right) - \frac{1}{2}}}{h} \cr & {\text{The derivative of a continuous }}f\left( x \right){\text{ at the point }}a{\text{ is given by}} \cr & f'\left( a \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {a + h} \right) - f\left( a \right)}}{h} \cr & {\text{Then comparing both expressions we have,}} \cr & \underbrace {\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {\frac{\pi }{6} + h} \right) - \frac{1}{2}}}{h}}_{\mathop {\lim }\limits_{h \to 0} \frac{{f\left( {a + h} \right) - f\left( a \right)}}{h}} \Rightarrow f\left( x \right) = \sin x{\text{ and }}a = \frac{\pi }{6} \cr & {\text{Therefore,}} \cr & \left. a \right){\text{ One possible }}f\left( x \right){\text{ and }}a{\text{ is:}} \cr & f\left( x \right) = \sin x,{\text{ }}a = \frac{\pi }{6} \cr & \cr & \left. b \right){\text{ Evaluating the limit}} \cr & \mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {\frac{\pi }{6} + h} \right) - \frac{1}{2}}}{h} \Rightarrow f'\left( x \right) = \frac{d}{{dx}}\left[ {\sin x} \right] \cr & {\text{ }} \Rightarrow f'\left( x \right) = \cos x \cr & {\text{Evaluating at }}a = \frac{\pi }{6} \cr & f'\left( {\frac{\pi }{6}} \right) = \cos \left( {\frac{\pi }{6}} \right) = \frac{{\sqrt 3 }}{2} \cr & {\text{,Then}} \cr & \mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {\frac{\pi }{6} + h} \right) - \frac{1}{2}}}{h} = \frac{{\sqrt 3 }}{2} \cr} $$
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