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: 26

Answer

\[\,{g^,}\,\left( y \right) = \frac{{2.8{y^{0.4}} - 2.5{y^{1.5}} - 1.1{y^{2.9}}}}{{\,{{\left( {{y^{2.5}} + 2} \right)}^2}}}\]

Work Step by Step

\[\begin{gathered} g\,\left( y \right) = \frac{{{y^{1.4}} + 1}}{{{y^{2.5}} + 2}} \hfill \\ Use\,\,the\,\,quotient\,\,rule\,\,to\,\,find\,\,{g^,}\,\left( y \right) \hfill \\ \,{g^,}\,\left( y \right) = \frac{{\,\left( {{y^{2.5}} + 2} \right)\,{{\left( {{y^{1.4}} + 1} \right)}^,} - \left( {{y^{1.4}} + 1} \right){{\left( {{y^{2.5}} + 2} \right)}^,}\,}}{{{{\left( {{y^{2.5}} + 2} \right)}^2}\,}} \hfill \\ Then \hfill \\ \,{g^,}\,\left( y \right) = \frac{{\left( {{y^{2.5}} + 2} \right)\,\left( {1.4{y^{0.4}}} \right) - \,\left( {{y^{1.4}} + 1} \right)\,\left( {2.5{y^{1.5}}} \right)}}{{\,{{\left( {{y^{2.5}} + 2} \right)}^2}}} \hfill \\ Simplify\,\,by\,\,multiplying\,\,and\,\,combining\,\,terms \hfill \\ \,{g^,}\,\left( y \right) = \frac{{1.4{y^{2.9}} + 2.8{y^{0.4}} - 2.5{y^{2.9}} - 2.5{y^{1.5}}}}{{\,{{\left( {{y^{2.5}} + 2} \right)}^2}}} \hfill \\ \,{g^,}\,\left( y \right) = \frac{{2.8{y^{0.4}} - 2.5{y^{1.5}} - 1.1{y^{2.9}}}}{{\,{{\left( {{y^{2.5}} + 2} \right)}^2}}} \hfill \\ \end{gathered} \]
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