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: 3

Answer

\[{y^,} = 8x - 20\]

Work Step by Step

\[\begin{gathered} y = \,{\left( {2x - 5} \right)^2} \hfill \\ Write\,as\,\,a\,\,\,product \hfill \\ y = \,\left( {2x - 5} \right)\,\left( {2x - 5} \right) \hfill \\ Use\,\,the\,\,product\,\,rule\,\,to\,\,find\,\,{y^,} \hfill \\ {y^,} = \,\left( {2x - 5} \right)\,{\left( {2x - 5} \right)^,} + \,\left( {2x - 5} \right)\,{\left( {2x - 5} \right)^,} \hfill \\ {y^,} = 2\,\left( {2x - 5} \right)\,{\left( {2x - 5} \right)^,} \hfill \\ Then \hfill \\ {y^,} = 2\,\left( {2x - 5} \right)\,\left( 2 \right) \hfill \\ Multiplying \hfill \\ {y^,} = 4\,\left( {2x - 5} \right) \hfill \\ {y^,} = 8x - 20 \hfill \\ \end{gathered} \]
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.