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: 28

Answer

\[{g^,}\,\left( x \right) = \frac{{120{x^3} - 186{x^2} - 56x - 141}}{{\,{{\left( {6x - 7} \right)}^2}}}\]

Work Step by Step

\[\begin{gathered} g\left( x \right) = \frac{{\,\left( {2{x^2} + 3} \right)\,\left( {5x + 2} \right)}}{{6x - 7}} \hfill \\ Use\,\,the\,\,quotient\,\,rule\,\,to\,\,find\,\,the\,\,derivative \hfill \\ {g^,}\,\left( x \right) = \frac{{\,\left( {6x - 7} \right)\,\,{{\left[ {\,\left( {2{x^2} + 3} \right)\,\left( {5x + 2} \right)} \right]}^,} - \,\left( {2{x^2} + 3} \right)\,\left( {5x + 2} \right)\,{{\left( {6x - 7} \right)}^,}}}{{\,{{\left( {6x - 7} \right)}^2}}} \hfill \\ Use\,\,the\,\,product\,\,rule\,\,to\,\,find\,\,{\left[ {\,\left( {2{x^2} + 3} \right)\,\left( {5x + 2} \right)} \right]^,} \hfill \\ {g^,}\,\left( x \right) = \frac{{\,{{\left( {6x - 7} \right)}^2}\,\,\left[ {\,\left( {2{x^2} + 3} \right)\,\left( 5 \right) + \,\left( {5x + 2} \right)\,\left( {4x} \right)} \right] - \,\left( {2{x^2} + 3} \right)\,\left( {5x + 2} \right)\,\left( 6 \right)}}{{\,{{\left( {6x - 7} \right)}^2}}} \hfill \\ Simplify\,\,by\,\,multiplying\,\,and\,\,combining\,\,terms \hfill \\ {g^,}\,\left( x \right) = \frac{{\,{{\left( {6x - 7} \right)}^2}\,\,\left[ {10{x^2} + 15 + 20{x^2} + 8x} \right] - \,\left( {2{x^2} + 3} \right)\,\left( {30x + 12} \right)}}{{\,{{\left( {6x - 7} \right)}^2}}} \hfill \\ {g^,}\,\left( x \right) = \frac{{\,{{\left( {6x - 7} \right)}^2}\,\left( {30{x^2} + 8x + 15} \right) - \,\left( {60{x^3} + 24{x^2} + 90x + 36} \right)}}{{\,{{\left( {6x - 7} \right)}^2}}} \hfill \\ {g^,}\,\left( x \right) = \frac{{180{x^3} + 48{x^2} + 90x - 210{x^2} - 56x - 105 - 60{x^3} - 24{x^2} - 90x - 36}}{{\,{{\left( {6x - 7} \right)}^2}}} \hfill \\ {g^,}\,\left( x \right) = \frac{{120{x^3} - 186{x^2} - 56x - 141}}{{\,{{\left( {6x - 7} \right)}^2}}} \hfill \\ \end{gathered} \]
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