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: 30

Answer

$$f'\left( x \right) = - \frac{{2x{{\left( {{{\tan }^{ - 1}}\left( {{x^2} + 4} \right)} \right)}^{ - 2}}}}{{1 + {{\left( {{x^2} + 4} \right)}^2}}}$$

Work Step by Step

$$\eqalign{ & f\left( x \right) = \frac{1}{{{{\tan }^{ - 1}}\left( {{x^2} + 4} \right)}} \cr & {\text{rewrite the function}} \cr & f\left( x \right) = {\left[ {{{\tan }^{ - 1}}\left( {{x^2} + 4} \right)} \right]^{ - 1}} \cr & {\text{differentiate both sides }} \cr & f'\left( x \right) = \frac{d}{{dx}}\left( {{{\left[ {{{\tan }^{ - 1}}\left( {{x^2} + 4} \right)} \right]}^{ - 1}}} \right) \cr & {\text{using the chain rule}} \cr & f'\left( x \right) = - {\left[ {{{\tan }^{ - 1}}\left( {{x^2} + 4} \right)} \right]^{ - 2}}\frac{d}{{dx}}\left( {{{\tan }^{ - 1}}\left( {{x^2} + 4} \right)} \right) \cr & {\text{use }}\frac{d}{{du}}\left[ {{{\tan }^{ - 1}}u} \right] = \frac{1}{{1 + {u^2}}}\frac{{du}}{{dx}}.{\text{ let }}u = {x^2} + 4 \cr & {\text{then}} \cr & f'\left( x \right) = - {\left[ {{{\tan }^{ - 1}}\left( {{x^2} + 4} \right)} \right]^{ - 2}}\left( {\frac{1}{{1 + {{\left( {{x^2} + 4} \right)}^2}}}} \right)\frac{d}{{dx}}\left[ {{x^2} + 4} \right] \cr & f'\left( x \right) = - {\left[ {{{\tan }^{ - 1}}\left( {{x^2} + 4} \right)} \right]^{ - 2}}\left( {\frac{1}{{1 + {{\left( {{x^2} + 4} \right)}^2}}}} \right)\left( {2x} \right) \cr & {\text{simplifying}} \cr & f'\left( x \right) = - \frac{{2x{{\left( {{{\tan }^{ - 1}}\left( {{x^2} + 4} \right)} \right)}^{ - 2}}}}{{1 + {{\left( {{x^2} + 4} \right)}^2}}} \cr} $$
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