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: 18

Answer

\[{y^,} = \frac{{ - 32{x^2} + 10x - 40}}{{\,{{\left( {4{x^2} - 5} \right)}^2}}}\]

Work Step by Step

\[\begin{gathered} y = \frac{{ - {x^2} + 8x}}{{4{x^2} - 5}} \hfill \\ Use\,\,the\,\,quotient\,\,rule\,\,to\,\,find\,\,{y^,} \hfill \\ {y^,} = \frac{{\,\left( {4{x^2} - 5} \right)\,{{\left( { - {x^2} + 8x} \right)}^,} - \,\left( { - {x^2} + 8x} \right)\,{{\left( {4{x^2} - 5} \right)}^,}}}{{\,{{\left( {4{x^2} - 5} \right)}^2}}}\, \hfill \\ Then, \hfill \\ {y^,} = \frac{{\,\,\,\left( {4{x^2} - 5} \right)\,\left( { - 2x + 8} \right) - \,\left( { - {x^2} + 8x} \right)\,\left( {8x} \right)}}{{\,{{\left( {4{x^2} - 5} \right)}^2}}} \hfill \\ Simplify\,\,by\,\,multiplying\,\,and\,\,combining\,\,terms \hfill \\ {y^,} = \frac{{ - 8{x^3} + 32{x^2} + 10x - 40 + 8{x^3} - 64{x^2}}}{{\,{{\left( {4{x^2} - 5} \right)}^2}}} \hfill \\ {y^,} = \frac{{ - 32{x^2} + 10x - 40}}{{\,{{\left( {4{x^2} - 5} \right)}^2}}} \hfill \\ \end{gathered} \]
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