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: 16

Answer

\[{y^,} = \frac{{{x^2} + 6x - 12}}{{\,{{\left( {x + 3} \right)}^2}}}\]

Work Step by Step

\[\begin{gathered} y = \frac{{{x^2} - 4x}}{{x + 3}} \hfill \\ Use\,\,the\,\,quotient\,\,rule\,\,to\,\,find\,\,{y^,}\, \hfill \\ {y^,} = \frac{{\,\left( {x + 3} \right)\,{{\left( {{x^2} - 4x} \right)}^,} - \,\left( {{x^2} - 4x} \right)\,{{\left( {x + 3} \right)}^,}}}{{\,{{\left( {x + 3} \right)}^2}}} \hfill \\ Then \hfill \\ {y^,} = \frac{{\left( {x + 3} \right)\,\left( {2x - 4} \right) - \,\left( {{x^2} - 4x} \right)\,\left( 1 \right)}}{{\,{{\left( {x + 3} \right)}^2}}} \hfill \\ Simplify\,\,by\,\,multiplying\,\,and\,\,combining\,\,terms \hfill \\ {y^,} = \frac{{2{x^2} - 4x + 6x - 12 - {x^2} + 4x}}{{\,{{\left( {x + 3} \right)}^2}}} \hfill \\ {y^,} = \frac{{{x^2} + 6x - 12}}{{\,{{\left( {x + 3} \right)}^2}}} \hfill \\ \end{gathered} \]
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