Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 3 - Section 3.4 - The Chain Rule - 3.4 Exercises - Page 206: 11

Answer

$g'\left( t \right) = - \frac{4}{{{{\left( {2t + 1} \right)}^3}}}$

Work Step by Step

$$\eqalign{ & g\left( t \right) = \frac{1}{{{{\left( {2t + 1} \right)}^2}}} \cr & {\text{Rewrite the function}}{\text{, recall that }}\frac{1}{{{u^n}}} = {u^{ - n}},{\text{ let }}u = 2t + 1 \cr & g\left( t \right) = {\left( {2t + 1} \right)^{ - 2}} \cr & {\text{Differentiate}} \cr & g'\left( t \right) = \frac{d}{{dt}}\left[ {{{\left( {2t + 1} \right)}^{ - 2}}} \right] \cr & {\text{Use The Power Rule Combined with the Chain Rule}} \cr & \frac{d}{{dx}}{\left[ {g\left( x \right)} \right]^n} = n{\left[ {g\left( x \right)} \right]^{n - 1}}g'\left( x \right) \cr & {\text{Therefore}}{\text{,}} \cr & g'\left( t \right) = - 2{\left( {2t + 1} \right)^{ - 2 - 1}}\frac{d}{{dt}}\left[ {2t + 1} \right] \cr & {\text{Computing derivatives}} \cr & g'\left( t \right) = - 2{\left( {2t + 1} \right)^{ - 3}}\left( 2 \right) \cr & {\text{Multiply and simplify}} \cr & g'\left( t \right) = - 4{\left( {2t + 1} \right)^{ - 3}} \cr & g'\left( t \right) = - \frac{4}{{{{\left( {2t + 1} \right)}^3}}} \cr} $$
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