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: 32

Answer

$$\,\,\,\,\,y = \frac{1}{{2\sqrt 3 }}x - \frac{1}{{\sqrt 3 }} + \frac{\pi }{6}$$

Work Step by Step

$$\eqalign{ & f\left( x \right) = {\sin ^{ - 1}}\left( {\frac{x}{4}} \right);{\text{ }}\left( {2,\frac{\pi }{6}} \right) \cr & {\text{differentiate }}f\left( x \right) \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {{{\sin }^{ - 1}}\left( {\frac{x}{4}} \right)} \right] \cr & {\text{use }}\frac{d}{{du}}\left[ {{{\sin }^{ - 1}}u} \right] = \frac{1}{{\sqrt {1 - {u^2}} }}\frac{{du}}{{dx}} \cr & f'\left( x \right) = \frac{1}{{\sqrt {1 - {{\left( {x/4} \right)}^2}} }}\frac{d}{{dx}}\left[ {\frac{x}{4}} \right] \cr & f'\left( x \right) = \frac{1}{{\sqrt {\frac{{16 - {x^2}}}{{16}}} }}\left( {\frac{1}{4}} \right) \cr & f'\left( x \right) = \frac{4}{{\sqrt {16 - {x^2}} }}\left( {\frac{1}{4}} \right) \cr & f'\left( x \right) = \frac{1}{{\sqrt {16 - {x^2}} }} \cr & \cr & {\text{find the slope at the point }}\left( {2,\frac{\pi }{6}} \right) \cr & m = f'\left( 2 \right) = \frac{1}{{\sqrt {16 - {{\left( 2 \right)}^2}} }} \cr & \,\,m = \frac{1}{{\sqrt {12} }} \cr & \,{\text{find the equation of the tangent line at the point }}\left( {2,\frac{\pi }{6}} \right) \cr & \,\,\,\,\,y - {y_1} = m\left( {x - {x_1}} \right) \cr & \,\,\,\,\,y - \frac{\pi }{6} = \frac{1}{{\sqrt {12} }}\left( {x - 2} \right) \cr & \,\,\,\,\,y - \frac{\pi }{6} = \frac{1}{{\sqrt {12} }}x - \frac{2}{{\sqrt {12} }} \cr & \,\,\,\,\,y = \frac{1}{{2\sqrt 3 }}x - \frac{2}{{2\sqrt 3 }} + \frac{\pi }{6} \cr & \,\,\,\,\,y = \frac{1}{{2\sqrt 3 }}x - \frac{1}{{\sqrt 3 }} + \frac{\pi }{6} \cr} $$
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