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: 15

Answer

\[{y^,} = \frac{{{x^2} - 2x - 1}}{{\,{{\left( {x - 1} \right)}^2}}}\]

Work Step by Step

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

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.