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: 20

Answer

\[{k^,}\,\left( t \right) = \frac{{ - 7{x^2} - 14}}{{\,{{\left( {{x^2} - 2} \right)}^2}}}\]

Work Step by Step

\[\begin{gathered} k\,\left( x \right) = \frac{{{x^2} + 7x - 2}}{{{x^2} - 2}} \hfill \\ Use\,\,the\,\,quotient\,\,rule\,\,to\,\,find\,\,{k^,}\,\left( x \right) \hfill \\ {k^,}\,\left( x \right) = \frac{{\,\left( {{x^2} - 2} \right)\,{{\left( {{x^2} + 7x - 2} \right)}^,} - \,\left( {{x^2} + 7x - 2} \right)\,{{\left( {{x^2} - 2} \right)}^,}}}{{\,{{\left( {{x^2} - 2} \right)}^2}}} \hfill \\ Then \hfill \\ {k^,}\,\left( x \right) = \frac{{\left( {{x^2} - 2} \right)\,\left( {2x + 7} \right) - \left( {{x^2} + 7x - 2} \right)\,\,\left( {2x} \right)}}{{{{\left( {{x^2} - 2} \right)}^2}}} \hfill \\ Simplify\,\,by\,\,multiplying\,\,and\,\,combining\,\,terms \hfill \\ {k^,}\,\left( x \right) = \frac{{2{x^3} + 7{x^2} - 4x - 14 - 2{x^3} - 14{x^2} + 4x}}{{\,{{\left( {{x^2} - 2} \right)}^2}}} \hfill \\ {k^,}\,\left( t \right) = \frac{{ - 7{x^2} - 14}}{{\,{{\left( {{x^2} - 2} \right)}^2}}} \hfill \\ \end{gathered} \]
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