Calculus with Applications (10th Edition)

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

Chapter 4 - Calculating the Derivative - Chapter Review - Review Exercises: 19

Answer

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

Work Step by Step

\[\begin{gathered} y = \frac{{{x^2} - x + 1}}{{x - 1}} \hfill \\ Find\,\,the\,\,derivative\,\,of\,\,the\,\,function \hfill \\ {y^,} = \frac{d}{{dx}}\,\,\left[ {\frac{{{x^2} - x + 1}}{{x - 1}}} \right] \hfill \\ Use\,\,the\,\,quotient\,\,rule \hfill \\ {y^,} = \frac{{\,\left( {x - 1} \right)\,{{\left( {{x^2} - x + 1} \right)}^,} - \,\left( {{x^2} - x + 1} \right)\,{{\left( {x - 1} \right)}^,}}}{{\,{{\left( {x - 1} \right)}^2}}} \hfill \\ Then \hfill \\ {y^,} = \frac{{\,\left( {x - 1} \right)\,\left( {2x - 1} \right) - \,\left( {{x^2} - x + 1} \right)\,\left( 1 \right)}}{{\,{{\left( {x - 1} \right)}^2}}} \hfill \\ Simplify\,\,by\,\,multiplying\,\,and\,\,combining\,\,terms\, \hfill \\ {y^,} = \frac{{2{x^2} - x - 2x + 1 - {x^2} + x - 1}}{{\,{{\left( {x - 1} \right)}^2}}} \hfill \\ {y^,} = \frac{{{x^2} - 2x}}{{\,{{\left( {x - 1} \right)}^2}}} \hfill \\ \end{gathered} \]
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