Answer
\[{f^,}\,\left( x \right) = \,\frac{{\,\left( {\sqrt x + 1} \right){e^{\sqrt x }}\ln \,\left( {\sqrt x + 1} \right) - {e^{\sqrt x }}}}{{2\sqrt x \,\left( {\sqrt x + 1} \right)\,\,{{\left[ {\ln \,\left( {\sqrt x + 1} \right)} \right]}^2}}}\]
Work Step by Step
\[\begin{gathered}
f\,\left( x \right) = \frac{{{e^{\sqrt x }}}}{{\ln \,\left( {\sqrt x + 1} \right)}} \hfill \\
Find\,\,the\,\,derivative\,\,of\,\,the\,\,\,function \hfill \\
{f^,}\,\left( x \right) = \frac{{\,\,\left[ {\ln \,\left( {\sqrt x + 1} \right)} \right]\,{{\left( {{e^{\sqrt x }}} \right)}^,} - {e^{\sqrt x }}\,{{\left( {\ln \,\left( {\sqrt x + 1} \right)} \right)}^,}}}{{\,\,{{\left[ {\ln \,\left( {\sqrt x + 1} \right)} \right]}^,}}} \hfill \\
Use\,\,the\,\,formulas\, \hfill \\
\frac{d}{{dx}}\,\,\left[ {\ln g\,\left( x \right)} \right] = \frac{{{g^,}\,\left( x \right)}}{{g\,\left( x \right)}}\,,\,\frac{d}{{dx}}\,\,\left[ {{e^{g\,\left( x \right)}}} \right] = {e^{g\,\left( x \right)}}{g^,}\,\left( x \right) \hfill \\
Then \hfill \\
{f^,}\,\left( x \right) = \frac{{\,\,\left[ {\ln \,\left( {\sqrt x + 1} \right)} \right]\,\left( {\,\frac{{{e^{\sqrt x }}}}{{2\sqrt x }}} \right) - {e^{\sqrt x }}\,\left( {\frac{{\frac{1}{{2\sqrt x }}}}{{\sqrt x + 1}}} \right)}}{{\,\,{{\left[ {\ln \,\left( {\sqrt x + 1} \right)} \right]}^2}}} \hfill \\
Simplifying \hfill \\
{f^,}\,\left( x \right) = \frac{{\frac{{{e^{\sqrt x }}\ln \,\left( {\sqrt x + 1} \right)}}{{2\sqrt x }} - \frac{{{e^{\sqrt x }}}}{{2\sqrt x \,\left( {\sqrt x + 1} \right)}}}}{{\,\,{{\left[ {\ln \,\left( {\sqrt x + 1} \right)} \right]}^2}}} \hfill \\
{f^,}\,\left( x \right) = \,\frac{{\,\left( {\sqrt x + 1} \right){e^{\sqrt x }}\ln \,\left( {\sqrt x + 1} \right) - {e^{\sqrt x }}}}{{2\sqrt x \,\left( {\sqrt x + 1} \right)\,\,{{\left[ {\ln \,\left( {\sqrt x + 1} \right)} \right]}^2}}} \hfill \\
\end{gathered} \]
