Answer
\[{f^,}\,\left( x \right) = \,\frac{{{e^{\sqrt x }}}}{{2\sqrt x \,\left( {\sqrt {x + 5} } \right)}} + \frac{{{e^{\sqrt x }}\ln \,\left( {\sqrt x + 5} \right)}}{{2\sqrt x }}\]
Work Step by Step
\[\begin{gathered}
f\,\left( x \right) = {e^{\sqrt x }}\ln \,\left( {\sqrt x + 5} \right) \hfill \\
Find\,\,the\,\,derivative \hfill \\
{f^,}\,\left( x \right) = \,\,{\left[ {{e^{\sqrt x }}\ln \,\left( {\sqrt x + 5} \right)} \right]^,} \hfill \\
Use\,\,the\,\,product\,\,rule \hfill \\
{f^,}\,\left( x \right) = {e^{\sqrt x }}\,\,{\left[ {\ln \,\left( {\sqrt {x + 5} } \right)} \right]^,} + \ln \,\left( {\sqrt x + 5} \right)\,\,{\left[ {{e^{\sqrt x }}} \right]^,} \hfill \\
Use\,\,the\,\,formulas \hfill \\
\frac{d}{{dx}}\,\,\left[ {{e^{g\,\left( x \right)}}} \right] = {e^{g\,\left( x \right)}}{g^,}\,\left( x \right)\,,\,\frac{d}{{dx}}\,\,\left[ {\ln g\,\left( x \right)} \right] = \frac{{{g^,}\,\left( x \right)}}{{g\,\left( x \right)}} \hfill \\
Then \hfill \\
{f^,}\,\left( x \right) = {e^{\sqrt x }}\,\,\left( {\frac{{\frac{1}{{2\sqrt x }}}}{{\sqrt x + 5}}} \right) + \ln \,\left( {\sqrt x + 5} \right)\,\left( {\frac{1}{{2\sqrt x }}{e^{\sqrt x }}} \right) \hfill \\
Simplifying \hfill \\
{f^,}\,\left( x \right) = \,\frac{{{e^{\sqrt x }}}}{{2\sqrt x \,\left( {\sqrt {x + 5} } \right)}} + \frac{{{e^{\sqrt x }}\ln \,\left( {\sqrt x + 5} \right)}}{{2\sqrt x }} \hfill \\
\end{gathered} \]
