Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 4 - Calculating the Derivative - 4.2 Derivatives of Products and Quotients - 4.2 Exercises - Page 216: 17

Answer

\[{f^,}\,\left( t \right) = \frac{{2t}}{{\,{{\left( {{t^2} + 3} \right)}^2}}}\]

Work Step by Step

\[\begin{gathered} f\,\left( t \right) = \frac{{4{t^2} + 11}}{{{t^2} + 3}} \hfill \\ Use\,\,the\,\,quotient\,\,rule\,\,to\,\,find\,\,{f^,}\,\left( t \right) \hfill \\ {f^,}\,\left( t \right) = \frac{{\,\left( {{t^2} + 3} \right)\,{{\left( {4{t^2} + 11} \right)}^,} - \,\left( {4{t^2} + 11} \right)\,{{\left( {{t^2} + 3} \right)}^,}}}{{\,{{\left( {{t^2} + 3} \right)}^2}}} \hfill \\ Then, \hfill \\ {f^,}\,\left( t \right) = \frac{{\,\left( {{t^2} + 3} \right)\,\left( {8t} \right) - \left( {4{t^2} + 11} \right)\,\left( {2t} \right)}}{{\,{{\left( {{t^2} + 3} \right)}^2}}} \hfill \\ Simplify\,\,by\,\,multiplying\,\,and\,\,combining\,\,terms \hfill \\ {f^,}\,\left( t \right) = \frac{{8{t^3} + 24t - 8{t^3} - 22t}}{{\,{{\left( {{t^2} + 3} \right)}^2}}} \hfill \\ {f^,}\,\left( t \right) = \frac{{2t}}{{\,{{\left( {{t^2} + 3} \right)}^2}}} \hfill \\ \end{gathered} \]
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