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: 14

Answer

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

Work Step by Step

\[\begin{gathered} y = \frac{{9 - 7t}}{{1 - t}} \hfill \\ Use\,\,the\,\,quotient\,\,rule\,\,to\,\,find\,\,{y^,}\, \hfill \\ {y^,}\, = \frac{{\,\left( {1 - t} \right)\,{{\left( {9 - 7t} \right)}^,} - \,\left( {9 - 7t} \right)\,{{\left( {1 - t} \right)}^,}}}{{\,{{\left( {1 - t} \right)}^2}}} \hfill \\ Then \hfill \\ {y^,} = \frac{{\,\,\left( {1 - t} \right)\,\left( { - 7} \right) - \,\left( {9 - 7t} \right)\,\left( { - 1} \right)}}{{\,\,{{\left( {1 - t} \right)}^2}}} \hfill \\ Simplify\,\,by\,\,multiplying\,\,and\,\,combining\,\,terms \hfill \\ {y^,} = \frac{{ - 7 + 7t - 7t + 9}}{{\,{{\left( {1 - t} \right)}^2}}} \hfill \\ {y^,} = \frac{2}{{\,{{\left( {1 - t} \right)}^2}}} \hfill \\ \end{gathered} \]
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