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: 7

Answer

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

Work Step by Step

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