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: 24

Answer

\[{y^,} = 2{x^{ - 1/2}} + \frac{3}{2}{x^{ - 3/2}}\]

Work Step by Step

\[\begin{gathered} y = \frac{{4x - 3}}{{\sqrt x }} \hfill \\ \sqrt x \,\,can\,\,be\,\,written\,\,as\,\,{x^{1/2}} \hfill \\ y = \frac{{4x - 3}}{{{x^{1/2}}}} \hfill \\ Use\,\,the\,\,product\,\,rule\,\,to\,\,find\,\,{y^,} \hfill \\ {y^,} = \frac{{{x^{1/2}}\,{{\left( {4x - 3} \right)}^,} - \left( {4x - 3} \right)\,{{\left( {{x^{1/2}}\,} \right)}^,}}}{{\,{{\left( {{x^{1/2}}\,} \right)}^2}}} \hfill \\ Then \hfill \\ {y^,} = \frac{{{x^{1/2}}\,\left( 4 \right) - \,\left( {4x - 3} \right)\,\left( {\frac{1}{2}{x^{ - 1/2}}} \right)}}{{\,{{\left( {{x^{1/2}}} \right)}^2}}} \hfill \\ Simplify\,\,by\,\,multiplying\,\,and\,\,combining\,\,terms \hfill \\ {y^,} = \frac{{4{x^{1/2}} - 2{x^{1/2}} + \frac{3}{2}{x^{1/2}}}}{x} \hfill \\ {y^,} = \frac{{2{x^{1/2}} + \frac{3}{2}{x^{ - 1/2}}}}{x} \hfill \\ {y^,} = 2{x^{ - 1/2}} + \frac{3}{2}{x^{ - 3/2}} \hfill \\ \end{gathered} \]
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