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: 5

Answer

\[{k^,}\,\left( t \right) = 4{t^3} - 4t\]

Work Step by Step

\[\begin{gathered} k\,\left( t \right) = \,{\left( {{t^2} - 1} \right)^2} \hfill \\ Write\,\,as\,a\,\,product\,\, \hfill \\ k\,\left( t \right) = \,\left( {{t^2} - 1} \right)\,\left( {{t^2} - 1} \right) \hfill \\ Use\,\,the\,\,product\,\,rule\,\,to\,\,find\,\,{k^,}\,\left( t \right) \hfill \\ {k^,}\,\left( t \right) = \left( {{t^2} - 1} \right)\,\left( {{t^2} - 1} \right){\,^,} + \left( {{t^2} - 1} \right)\,\left( {{t^2} - 1} \right){\,^,} \hfill \\ {k^,}\,\left( t \right) = 2\left( {{t^2} - 1} \right)\,\left( {{t^2} - 1} \right){\,^,} \hfill \\ Then \hfill \\ {k^,}\,\left( t \right) = 2\,\left( {{t^2} - 1} \right)\,\left( {2t} \right) \hfill \\ Multiplying \hfill \\ {k^,}\,\left( t \right) = 4t\,\left( {{t^2} - 1} \right) \hfill \\ {k^,}\,\left( t \right) = 4{t^3} - 4t \hfill \\ \hfill \\ \end{gathered} \]
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