Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 4 - Calculating the Derivative - 4.5 Derivatives of Logarithmic Functions - 4.5 Exercises: 8

Answer

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

Work Step by Step

\[\begin{gathered} y = \ln \sqrt {2x + 1} \hfill \\ Write\,\,\sqrt {2x + 1} \,\,as\,\,\,{\left( {2x + 1} \right)^{1/2}} \hfill \\ y = \ln \,\,\,{\left( {2x + 1} \right)^{1/2}} \hfill \\ Use\,\,the\,\,\log \,\,property \hfill \\ \ln {a^n} = n\ln a \hfill \\ y = \frac{1}{2}\ln \,\left( {2x + 1} \right) \hfill \\ Differentiating \hfill \\ {y^,} = \,\,{\left[ {\frac{1}{2}\ln \,\left( {2x + 1} \right)} \right]^,} \hfill \\ {y^,} = \,\,\frac{1}{2}\,\,\,\left[ {\frac{{\,{{\left( {2x + 1} \right)}^,}}}{{2x + 1}}} \right] \hfill \\ {y^,} = \,\,\frac{1}{2}\,\left( {\frac{2}{{2x + 1}}} \right) \hfill \\ {y^,} = \,\frac{1}{{2x + 1}} \hfill \\ \end{gathered} \]
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.