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