Answer
\[{y^,} = \frac{{2{e^{2x - 1}}}}{{2x - 1}} + 2{e^{2x - 1}}\ln \,\left( {2x - 1} \right)\]
Work Step by Step
\[\begin{gathered}
y = {e^{2x - 1}}\ln \,\left( {2x - 1} \right) \hfill \\
Differentiate \hfill \\
{y^,} = \,\,{\left[ {{e^{2x - 1}}\ln \,\left( {2x - 1} \right)} \right]^,} \hfill \\
Use\,\,the\,\,product\,\,rule\, \hfill \\
{y^,} = {e^{2x - 1}}\,{\left( {\ln \,\left( {2x - 1} \right)} \right)^,} + \ln \,\left( {2x - 1} \right)\,{\left( {{e^{2x - 1}}} \right)^,} \hfill \\
Use\,\,the\,\,\,formula \hfill \\
\frac{d}{{dx}}\,\,\left[ {\ln g\,\left( x \right)} \right] = \frac{{{g^,}\,\left( x \right)}}{{g\,\left( x \right)}},\frac{d}{{dx}}\,\,\left[ {{e^{g\,\left( x \right)}}} \right] = {e^{g\,\left( x \right)}}{g^,}\,\left( x \right) \hfill \\
Then \hfill \\
{y^,} = {e^{2x - 1}}\,\left( {\frac{2}{{2x - 1}}} \right) + \ln \,\left( {2x - 1} \right)\,\left( {2{e^{2x - 1}}} \right) \hfill \\
Multipliying \hfill \\
{y^,} = \frac{{2{e^{2x - 1}}}}{{2x - 1}} + 2{e^{2x - 1}}\ln \,\left( {2x - 1} \right) \hfill \\
\end{gathered} \]
