## Calculus with Applications (10th Edition)

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